3.1. Preparation

In the proofs, we analyze the g.f. of the underlying processes. To prove distributional convergence for nonnegative integer-valued processes, it is enough to prove the pointwise convergence of the g.f., and show that the limit is a g.f. as well [Reference Feller7, p. 280].

Recall that $f_n(s) = \mathbb{E} s^{\xi_n}$ represents the offspring g.f. in generation n. For the composite g.f., introduce the notation $f_{n,n}(s) = s$ and, for $j < n$ , $f_{j,n}(s) \;:\!=\; f_{j+1} \circ \cdots \circ f_n(s)$ ; and for the corresponding means $\overline{f}_{n,n} = 1$ and $\overline{f}_{j,n} \;:\!=\; \overline{f}_{j+1} \cdots \overline{f}_n$ , $j <n$ . Then it is well known that $\mathbb{E} s^{X_n} = f_{0,n}(s)$ and $\mathbb{E} X_n = \overline{f}_{0,n}$ .

For a g.f. f, with mean $\overline{f}$ and $f^{\prime\prime}(1) < \infty$ , define the shape function as

(4) \begin{equation} \varphi(s) = \frac{1}{1 - f(s)} - \frac{1}{\overline{f} (1-s)}, \quad 0 \leq s < 1, \qquad \varphi(1) = \frac{f^{\prime\prime}(1)}{2 \, (\overline{f})^2}.\end{equation}

Let $\varphi_j$ be the shape function of $f_j$ . By the definition of $f_{j,n}$ ,

\[ \frac{1}{1 - f_{j,n} (s)} = \frac{1}{\overline{f}_{j+1} (1 - f_{j+1,n}(s))} + \varphi_{j+1}(f_{j+1,n}(s)).\]

Therefore, iteration gives ([Reference Kersting15, Lemma 5], [Reference Kersting and Vatutin16, Proposition 1.3])

(5) \begin{equation} \frac{1}{1 - f_{j,n} (s)} = \frac{1}{\overline{f}_{j,n}(1 - s)} + \varphi_{j,n}(s),\end{equation}

where

(6) \begin{equation} \varphi_{j,n}(s)\;:\!=\;\sum_{k=j+1}^{n} \frac{\varphi_k(f_{k,n}(s))}{\overline{f}_{j,k-1}}.\end{equation}

The latter formulas show the usefulness of the shape function. The next statement gives precise upper and lower bounds on the shape function.

For further properties of shape functions we refer to [Reference Kersting and Vatutin16, Chapter 1] and [Reference Kersting15].

We frequently use the following extension of [Reference Györfi, Ispány, Pap and Varga12, Lemma 5], which is a version of the Silverman–Toeplitz theorem.

Lemma 2. Let $(\overline{f}_n)_{n\in\mathbb{N}}$ be a sequence of positive real numbers satisfying (C1), and define

\begin{align*} a_{n,j}^{(k)} & = (1-\overline{f}_j)\prod_{i=j+1}^{n}\overline{f}_i^k = (1 - \overline{f}_j) \overline{f}_{j,n}^k, \qquad n,j,k\in\mathbb{N}, \ j\leq n-1, \\ a_{n,n}^{(k)} & = 1 - \overline{f}_n. \end{align*}

If $(x_n)_{n\in\mathbb{N}}$ is bounded and $\lim_{n\to\infty,\overline{f}_n < 1}x_n = x \in \mathbb{R}$ , then, for all $k\in\mathbb{N}$ ,

(7) \begin{equation} \lim_{n \to \infty} \sum_{j=1}^n a_{n,j}^{(k)} x_j = \frac{x}{k}, \qquad \lim_{n \to \infty} \sum_{j=1}^n | a_{n,j}^{(k)}| x_j = \frac{x}{k}. \end{equation}

Proof. Let us define $A = \{ j \colon \overline{f}_j > 1 \}$ , $A_n = A \cap \{ 1, \ldots, n\}$ . First we show that the following conditions (similar to those of the Silverman–Toeplitz theorem) hold:

1. (i) $\lim_{n\to\infty} a_{n,j}^{(k)} = 0$ for all $j\in\mathbb{N}$ ;

2. (ii) $\lim_{n\to\infty} \sum_{j=1}^n a_{n,j}^{(k)} = {1}/{k}$ ;

3. (iii) $\sup_{n\in \mathbb{N}} \sum_{j=1}^n | a_{n,j}^{(k)} | < \infty$ ;

4. (iv) $\lim_{n \to \infty} \sum_{j \in A_n} |a_{n,j}^{(k)} | = 0$ .

It is easy to see that (i) holds, as, by (C1),

\[ |a_{n,j}^{(k)}| = |1 - \overline{f}_j|\prod_{\ell=j+1}^n\overline{f}_\ell^k \leq |1 - \overline{f}_j|\exp\Bigg\{{-}k\sum_{\ell=j+1}^n(1 - \overline{f}_\ell)\Bigg\} \to 0. \]

Here and later on, any nonspecified limit relation is meant as $n \to \infty$ . Since

\[ 0 \leq \overline{f}_{0,n} \leq \exp\Bigg\{{-}\sum_{j=1}^n(1-\overline{f}_j)\Bigg\} \to 0, \]

we have

\[ \sum_{j=1}^n a_{n,j}^{(1)} = \sum_{j=1}^n (1-\overline{f}_j) \overline{f}_{j,n} = \sum_{j=1}^n (\overline{f}_{j,n} - \overline{f}_{j-1,n}) = 1 - \overline{f}_{0,n} \to 1, \]

which is (ii) for $k = 1$ . Furthermore,

\[ \sum_{j=1}^n | a_{n,j}^{(1)} | = \sum_{j=1}^n |1-\overline{f}_j| \cdot \overline{f}_{j,n} = \sum_{j=1}^n \left[ (1-\overline{f}_j) + 2 (1 - \overline{f}_j)_- \right] \overline{f}_{j,n} < \infty, \]

i.e. (iii) holds for $k = 1$ . Note that $\sup_{j,n} \overline{f}_{j,n} < \infty$ by (C1). To see (iv), we have

\[ \sum_{j \in A_n} |a_{n,j}^{(k)} | = \sum_{j=1}^n (\overline{f}_j - 1 )_+ \overline{f}_{j,n}^k \to 0, \]

since $\lim_{n \to \infty} \overline{f}_{j,n} = 0$ , and $\big((\overline{f}_j - 1 )_+ \sup_{\ell,n} \overline{f}_{\ell,n}^k\big)_j$ is an integrable majorant, so Lebesgue’s dominated convergence theorem applies.

Before proving (ii) and (iii) for $k \geq 2$ we show (7) for $k = 1$ . Let $(x_n)_{n\in \mathbb{N}}$ be a sequence with the required properties. Then

\[ \sum_{j=1}^n a_{n,j}^{(1)} x_j - x = \sum_{j=1}^n a_{n,j}^{(1)}(x_j - x) + x\Bigg(\sum_{j=1}^n a_{n,j}^{(1)} - 1 \Bigg), \]

where the second term tends to 0 by (ii). For the first term we have

\[ \begin{split} \Bigg|\sum_{j=1}^n a_{n,j}^{(1)}(x_j -x)\Bigg| & = \Bigg|\sum_{j \in A_n}a_{n,j}^{(1)}(x_j -x) + \sum_{j \not \in A_n}a_{n,j}^{(1)}(x_j -x)\Bigg| \\ & \leq 2\sup_k|x_k| \cdot \sum_{j \in A_n}|a_{n,j}^{(1)}| + \sum_{j \not \in A_n}|a_{n,j}^{(1)}(x_j - x)|, \end{split} \]

where the first term tends to 0 by (iv), and the second tends to 0 by (i) and (iii). Thus, the first equation in (7) holds for $k =1$ . Furthermore, since $a_{j,n}^{(k)} \leq 0$ if and only if $j \in A_n$ , conditions (i)–(iv) remain true for $|a_{n,j}^{(k)}|$ , and thus

(8) \begin{equation} \lim_{n\to \infty} \sum_{j=1}^n |a_{n,j}^{(1)} | x_j = x. \end{equation}

Next, we prove (ii) and (iii) for arbitrary $k\geq 2$ . By the binomial theorem,

(9) \begin{equation} 1 - \overline{f}_{0,n}^k = k\sum_{j=1}^n a_{n,j}^{(k)} + \sum_{i=2}^k({-}1)^{i+1}\binom{k}{i}\sum_{j=1}^n(1-\overline{f}_j)^{i-1}a_{n,j}^{(k)}. \end{equation}

Moreover, for $i \geq 2$ ,

\[ \sum_{j=1}^n|(1-\overline{f}_j)^{i-1}a_{n,j}^{(k)}| = \sum_{j=1}^n|1-\overline{f}_j|^i\overline{f}_{j,n}^k \leq (\sup_{\ell,n}\overline{f}_{\ell,n})^{k-1}\sum_{j=1}^n|a_{n,j}^{(1)}| |1 - \overline{f}_j|^{i-1} \to 0 \]

by (8). Thus, all the terms in the second sum on the right-hand side of (9) tend to 0. Since $\overline{f}_{0,n} \to 0$ , (ii) follows. Then (iii) follows as for $k= 1$ . This completes the proof of (ii) and (iii) for $k \geq 2$ . Then (7) for $k \geq 2$ follows exactly as for $k = 1$ .

Lemma 3. Let $\varphi_n$ be the shape function of $f_n$ . Then, under the conditions of Theorem 1 with $\nu > 0$ ,

\[ \lim_{n\to\infty,\overline{f}_n < 1}\,\sup_{s \in [0,1]}\frac{|\varphi_n(1)-\varphi_n(s)|}{1-\overline{f}_n} = 0, \]

and the sequence $\sup_{s\in[0,1]}{|\varphi_n(1)-\varphi_n(s)|}/{|1-\overline{f}_n|}$ is bounded.

Proof. To ease the notation we suppress the lower index n. By the Taylor expansion,

\[ f(s) = 1 + \overline{f}(s-1) + \frac{1}{2}f^{\prime\prime}(1)(s-1)^2 + \frac{1}{6}f^{\prime\prime\prime}(t)(s-1)^3 \]

for some $ t\in(s,1)$ . Thus, recalling (4), the shape function can be written in the form

\[ \varphi(s) = \frac{\overline{f}(1-s)-1+f(s)}{\overline{f}(1-s)(1-f(s))} = \frac{f^{\prime\prime}(1)}{2\overline{f}^2}\frac{1 - ({f^{\prime\prime\prime}(t)}/{(3 f^{\prime\prime}(1))})(1-s)} {1 - ({f^{\prime\prime}(1)}/{(2\overline{f})})(1-s) + ({f^{\prime\prime\prime}(t)}/{(6\overline{f})})(1-s)^2}. \]

Therefore,

(10) \begin{equation} \frac{\varphi(1)-\varphi(s)}{1-\overline{f}} = \frac{f^{\prime\prime}(1)}{2\overline{f}^2(1-\overline{f})} \bigg(1 - \frac{1-({f^{\prime\prime\prime}(t)}/{(3f^{\prime\prime}(1))})(1-s)} {1-({f^{\prime\prime}(1)}/{(2\overline{f})})(1-s)+({f^{\prime\prime\prime}(t)}/{(6\overline{f})})(1-s)^2}\bigg). \end{equation}

Using the assumptions and the monotonicity of fʹʹʹ, uniformly in $s \in (0,1]$ ,

\[ \lim_{n\to\infty,\overline{f}_n < 1}\bigg[\frac{f_n^{\prime\prime\prime}(t)}{f_n^{\prime\prime}(1)} + \frac{f_n^{\prime\prime}(1)}{\overline{f}_n} + \frac{f_n^{\prime\prime\prime}(t)}{\overline{f}_n} \bigg] = 0; \]

thus, convergence on $\{n\colon\overline{f}_n < 1 \}$ follows.

The boundedness also follows from (10), since if $\overline{f}_n > 1$ then

\[ f^{\prime\prime}_n(1) = \mathbb{E}(\xi_n(\xi_n -1)) \geq \mathbb{E}(\xi_n - 1) = \overline{f}_n - 1, \]

and $f^{\prime\prime\prime}_n(1)/|1 - \overline{f}_n|$ is bounded by assumption (C3).

3.2. Proof of Theorem 1

Lemma 4. Under the conditions of Theorem 1, for $s \in [0,1)$ ,

\[ \lim_{n \to \infty}\frac{\overline{f}_{0,n}}{1-f_{0,n}(s)} = \frac{1}{1-s}+\frac{\nu}{2}. \]

Proof. Recalling (5) and (6), we have to show that

\[ \overline{f}_{0,n}\varphi_{0,n}(s) = \sum_{j=1}^n\overline{f}_{j-1,n}\varphi_j(f_{j,n}(s)) \to \frac{\nu}{2}. \]

First, let $\nu=0$ . Using Lemmas 1 and 2,

\[ \overline{f}_{0,n} \varphi_{0,n}(s) \leq \sum_{j=1}^n \overline{f}_{j,n} \frac{f_j^{\prime\prime}(1)}{\overline{f}_j} = \sum_{j=1}^n |a_{n,j}^{(1)} | \frac{f_j^{\prime\prime}(1)}{\overline{f}_j |1-\overline{f}_j|} \to 0. \]

If $\overline{f}_j = 1$ then by (C2) necessarily $f_j^{\prime\prime}(1) = 0$ , and $0/0$ in these types of sums are meant as 0.

For $\nu\in(0,\infty)$ write

(11) \begin{equation} \sum_{j=1}^n \overline{f}_{j-1,n} \varphi_j(f_{j,n}(s)) = \sum_{j=1}^n \overline{f}_{j-1,n} \varphi_j(1) - \sum_{j=1}^n \overline{f}_{j-1,n} (\varphi_j(1) - \varphi_j(f_{j,n}(s))). \end{equation}

By Lemma 2, for the first term we have

(12) \begin{equation} \sum_{j=1}^n \overline{f}_{j-1,n} \varphi_j(1) = \frac{1}{2} \sum_{j=1}^n a_{n,j}^{(1)} \frac{1}{\overline{f}_j} \frac{f_j^{\prime\prime}(1)}{1- \overline{f}_j} \to \frac{\nu}{2}. \end{equation}

For the second term in (11),

\[ \Bigg| \sum_{j=1}^n \overline{f}_{j-1,n} (\varphi_j(1) - \varphi_j(f_{j,n}(s))) \Bigg| \leq \sum_{j=1}^{n}a_{n,j}^{(1)}\overline{f}_j\sup_{s\in[0,1]}\frac{|\varphi_j(1)-\varphi_j(s)|}{|1-\overline{f}_j|} \to 0 \]

according to Lemmas 2 and 3. Combining with (11) and (12), the statement follows.

Proof of Theorem 1. We prove the convergence of the conditional g.f. For $s \in (0,1)$ we have, by the previous lemma,

\[ \mathbb{E}[s^{X_n}\mid X_n>0] = \frac{f_{0,n}(s)-f_{0,n}(0)}{1-f_{0,n}(0)} = 1 - \frac{1-f_{0,n}(s)}{1-f_{0,n}(0)} \to \frac{2}{2+\nu} \frac{s}{1-({\nu}/({\nu+2}))s}, \]

where the limit is the g.f. of the geometric distribution with parameter ${2}/({2+\nu})$ .

3.3. Proofs of Theorems 2 and 3

We need a strange version of the Riemann approximation, where the points in the partition are not necessarily increasing. It follows immediately from the uniform continuity.

Lemma 5. Assume that $x_{n,j} \in [0,1]$ , $j=0, \ldots, n$ , $n \in \mathbb{N}$ , such that $x_{n,0} = 0$ , $x_{n,n} = 1$ , $\lim_{n\to\infty}\sup_{j}|x_{n,j} - x_{n,j-1}| = 0$ , and $\sup_n\sum_{j=1}^n|x_{n,j} - x_{n,j-1}| < \infty$ . If f is continuous on [0, 1] then

\[ \lim_{n\to\infty}\sum_{j=1}^n f(u_{n,j})(x_{n,j} - x_{n,j-1}) = \int_0^1 f(x) \, \textrm{d}x, \]

with $u_{n,j} \in [\min (x_{n,j-1}, x_{n,j}), \max(x_{n,j-1}, x_{n,j})]$ .

We also need a simple lemma on an alternating sum involving binomial coefficients.

Lemma 6. For $x\in\mathbb{R}$ ,

\[ \sum_{i=1}^{k-1}\binom{k-1}{i}({-}1)^{i}\frac{(1+x)^i-1}{i} = \sum_{i=1}^{k-1}({-}1)^{i}\frac{x^i}{i}. \]

Proof. Both sides are polynomials of x, and equality holds for $x = 0$ , therefore it is enough to show that the derivatives are equal. Differentiating the left-hand side we obtain

\[ \sum_{i=1}^{k-1}\binom{k-1}{i}({-}1)^i(1+x)^{i-1} = \frac{1}{1+x}[(1 - (1+x))^{k-1} - 1]. \]

Differentiating the right-hand side and multiplying by $(1+x)$ , the statement follows.

The next statement is an easy consequence of the Taylor expansion of the g.f.

Lemma 7. ([Reference Györfi, Ispány, Pap and Varga12, Lemma 6].) Let $\varepsilon$ be a nonnegative integer-valued random variable with factorial moments $m_{k} \;:\!=\; \mathbb{E}[\varepsilon(\varepsilon-1)\cdots(\varepsilon-k+1)]$ , $k\in \mathbb{N}$ , $m_{0} \;:\!=\; 1$ , and with g.f. $h(s) = \mathbb{E}s^{\varepsilon}$ , $s\in[-1,1]$ . If $m_{\ell}<\infty$ for some $\ell \in \mathbb{N}$ with $|s|\leq 1$ , then

\[ h(s) = \sum_{k=0}^{\ell-1}\frac{m_{k}}{k!}(s-1)^k + R_{\ell}(s), \qquad |R_{\ell}(s)| \leq \frac{m_{\ell}}{\ell!}|s-1|^\ell. \]

Proof of Theorem 2. Recall that $f_n(s)=\mathbb{E}s^{\xi_n}$ , and let $g_n(s)=\mathbb{E}s^{Y_n}$ and $h_n(s)=\mathbb{E} s^{\varepsilon_n}$ . Then the branching property gives (see, e.g., [Reference González, Kersting, Minuesa and del Puerto9, Proposition 1]) $g_n(s)=\prod_{j=1}^{n}h_j(f_{j,n}(s))$ . We prove the convergence of the g.f., i.e. $\lim_{n \to \infty } g_n(s) = f_Y(s)$ , $s \in [0,1]$ . Fix $s \in [0,1)$ , and introduce $\widehat{g}_n(s)=\prod_{j=1}^{n}\textrm{e}^{h_j(f_{j,n}(s))-1}$ . Note that $\widehat g_n$ is a kind of accompanying law, its distribution is compound-Poisson, and therefore its limit is compound-Poisson too [Reference Steutel and van Harn19, Proposition 2.2]. By the convexity of the g.f.,

(13) \begin{equation} f_{j,n}(s) \geq 1+\overline{f}_{j,n}(s-1). \end{equation}

If $|a_k|, |b_k|< 1$ , $k=1,\ldots, n$ , then

\begin{equation*} \Bigg|\prod_{k=1}^{n}a_k - \prod_{k=1}^{n}b_k\Bigg| \leq \sum_{k=1}^{n}| a_k-b_k|. \end{equation*}

Consequently, using the latter inequality, the inequalities $|\textrm{e}^u-1-u|\leq u^2$ for $|u| \leq 1$ and $0\leq 1-h_j(s)\leq m_{j,1}(1-s)$ , and (13), we have

(14) \begin{align} |g_n(s)-\widehat{g}_n(s)| & \leq \sum_{j=1}^{n}|\textrm{e}^{h_j(f_{j,n}(s))-1}-h_j(f_{j,n}(s))| \nonumber \\ & \leq \sum_{j=1}^{n}(h_j(f_{j,n}(s))-1)^2\leq \sum_{j=1}^{n}\frac{m_{j,1}^2}{|1-\overline{f}_j|}|a_{n,j}^{(2)}|\to 0, \end{align}

where we used Lemma 2, since

\[ \lim_{n \to \infty,\overline{f}_n < 1}\frac{m_{n,1}}{1-\overline{f}_n} = \lambda_1 \quad \text{implies} \quad \lim_{n \to \infty, \overline{f}_n < 1} \frac{m_{n,1}^2}{1-\overline{f}_n}= 0. \]

Therefore, we need to show that

(15) \begin{equation} \lim_{n\to \infty} \sum_{j=1}^{n}(h_j(f_{j,n}(s))-1) = \log f_Y(s). \end{equation}

By Lemma 7,

(16) \begin{equation} h_j(s)= \sum_{k=0}^{K-1} \frac{m_{j,k}}{k!}(s-1)^k + R_{j,K}(s), \end{equation}

where $|R_{j,K}(s)| \leq ({m_{j,K}}/{K!})(1-s)^K$ , and thus

(17) \begin{align} \sum_{j=1}^{n}(h_j(f_{j,n}(s))-1) & = \sum_{j=1}^{n}\Bigg[\sum_{k=1}^{K-1}\frac{m_{j,k}}{k!}(f_{j,n}(s)-1)^k + R_{j,K}(f_{j,n}(s))\Bigg] \nonumber \\ & = \sum_{k=1}^{K-1}{({-}1)^k}\sum_{j=1}^{n}\frac{m_{j,k}}{k!(1 - \overline{f}_j)}a_{n,j}^{(k)} \bigg(\frac{1 - f_{j,n}(s)}{\overline{f}_{j,n}}\bigg)^k + \sum_{j=1}^{n}R_{j,K}(f_{j,n}(s)). \end{align}

By (13) and Lemma 2,

(18) \begin{align} \Bigg|\sum_{j=1}^{n} R_{j,K}(f_{j,n}(s))\Bigg| & \leq \sum_{j=1}^{n}\frac{m_{j,K}}{K!}|f_{j,n}(s)-1|^K \nonumber \\ & \leq \sum_{j=1}^{n}\frac{m_{j,K}}{K!}\overline{f}_{j,n}^K(1-s)^K = (1-s)^K\sum_{j=1}^{n}\frac{m_{j,K}}{K!|1-\overline{f}_j|}|a_{n,j}^{(K)}| \to 0. \end{align}

Up to this point, everything works for $\nu \geq 0$ . Now assume that $\nu > 0$ . Then, by Lemmas 2 and 3,

(19) \begin{equation} \Bigg|\sum_{i=j+1}^{n}a_{n,i}^{(1)}\overline{f}_i\frac{\varphi_i(1)-\varphi_i(f_{i,n}(s))}{1-\overline{f}_i}\Bigg| \leq \sup_{k\geq1}\overline{f}_k\sum_{i=1}^n|a_{n,i}^{(1)}| \sup_{t\in[0,1]}\frac{|\varphi_i(1)-\varphi_i(t)|}{|1-\overline{f}_i|} \to 0, \end{equation}

and similarly

(20) \begin{equation} \Bigg|\sum_{i=j+1}^{n}a_{n,i}^{(1)}\overline{f}_i\frac{\varphi_i(1)}{1-\overline{f}_i} - \frac{\nu}{2}\sum_{i=j+1}^{n}a_{n,i}^{(1)}\Bigg| \leq \sum_{i=1}^{n}|a_{n,i}^{(1)}|\bigg|\frac{1}{\overline{f}_i} \frac{f_i^{\prime\prime}(1)}{2(1-\overline{f}_i)}-\frac{\nu}{2}\bigg| \to 0. \end{equation}

Putting

(21) \begin{equation} \varepsilon_{j,n} = \sum_{i=j+1}^{n}a_{n,i}^{(1)}\overline{f}_i\frac{\varphi_i(f_{i,n}(s))}{1-\overline{f}_i} - \frac{\nu}{2}(1 - \overline{f}_{j,n}), \end{equation}

by (5) and (6) we have

(22) \begin{equation} \frac{\overline{f}_{j,n}}{1-f_{j,n}(s)} = \frac{1}{1 - s} + \sum_{i=j+1}^{n}a_{n,i}^{(1)}\overline{f}_i\frac{\varphi_i(f_{i,n}(s))}{1-\overline{f}_i} = \frac{1}{1-s} + \frac{\nu}{2}(1 - \overline{f}_{j,n}) + \varepsilon_{j,n}. \end{equation}

Noting that $\sum_{i=j+1}^{n}a_{n,i}^{(1)} = 1-\overline{f}_{j,n}$ , (19), (20), and the triangle inequality imply that, for $\varepsilon_{j,n}$ in (21),

(23) \begin{equation} \max_{j\leq n}|\varepsilon_{j,n}| = \max_{j\leq n} \bigg|\frac{\overline{f}_{j,n}}{1-f_{j,n}(s)} - \frac{1}{1-s} - \frac{\nu}{2}(1 - \overline{f}_{j,n})\bigg| \to 0. \end{equation}

The latter further implies that

(24) \begin{equation} \limsup_{n \to \infty} \max_{j \leq n} \frac{1- f_{j,n}(s)}{\overline{f}_{j,n}} \leq 1-s, \end{equation}

and that for n large enough, by the mean value theorem and (22),

\[ \bigg|\bigg(\frac{\overline{f}_{j,n}}{1-f_{j,n}(s)}\bigg)^{-k} - \bigg(\frac{1}{1-s} + \frac{\nu}{2}(1 - \overline{f}_{j,n})\bigg)^{-k}\bigg| \leq k\varepsilon_{j,n}. \]

Thus, by (24),

(25) \begin{align} \Bigg|\sum_{j=1}^{n}a_{n,j}^{(k)} & \bigg[\frac{m_{j,k}}{k!(1 - \overline{f}_j)}\bigg(\frac{1 - f_{j,n}(s)}{\overline{f}_{j,n}}\bigg)^k - \lambda_k\bigg(\frac{1}{1-s} + \frac{\nu}{2}{(1 - \overline{f}_{j,n})}\bigg)^{-k}\bigg]\Bigg| \nonumber \\ & \leq \sum_{j=1}^{n}|a_{n,j}^{(k)}|\Bigg[ \bigg|\frac{m_{j,k}}{k!(1 - \overline{f}_j)} - \lambda_k\bigg| \bigg(\frac{1 - f_{j,n}(s)}{\overline{f}_{j,n}}\bigg)^k \nonumber \\ & \qquad\qquad\qquad + \lambda_k\bigg|\bigg(\frac{1 - f_{j,n}(s)}{\overline{f}_{j,n}}\bigg)^k - \bigg(\frac{1}{1-s} + \frac{\nu}{2}{(1 - \overline{f}_{j,n})}\bigg)^{-k}\bigg|\Bigg] \nonumber \\ & \leq \sum_{j=1}^{n}|a_{n,j}^{(k)}|\bigg|\frac{m_{j,k}}{k!(1 - \overline{f}_j)} - \lambda_k\bigg| + \lambda_k\sum_{j=1}^{n}|a_{n,j}^{(k)}|k\max_{j\leq n}\varepsilon_{j,n} \to 0, \end{align}

where the second inequality holds for n large enough.

Furthermore, by Lemma 5,

(26) \begin{equation} \lim_{n\to\infty}\sum_{j=1}^{n}a_{n,j}^{(k)}\bigg(\frac{1}{1-s}+\frac{\nu}{2}(1-\overline{f}_{j,n})\bigg)^{-k} = \int_{0}^{1}\frac{y^{k-1}}{({1}/({1-s})+({\nu}/{2})(1-y))^k}\,\textrm{d}y, \end{equation}

since the left-hand side is a Riemann approximation of the right-hand side corresponding to the partition $\{\overline{f}_{j,n}\}_{j=-1}^n$ , with $\overline{f}_{-1,n} \;:\!=\; 0$ and $\overline{f}_{j,n} - \overline{f}_{j-1,n} = a_{n,j}^{(1)}\to 0$ uniformly in j according to Lemma 2, while $\overline{f}_{0,n} \to 0$ by (C1). Changing variables $u=\nu (1-s)(1-y)+2$ and using the binomial theorem,

(27) \begin{align} \int_{0}^{1} & \frac{y^{k-1}}{({1}/({1-s})+({\nu}/{2})(1-y))^k}\,\textrm{d}y \nonumber \\ & = 2^k(1-s)^k\int_{0}^{1}\frac{y^{k-1}}{(2+\nu(1-s)(1-y))^k}\,\textrm{d}y \nonumber \\ & = \frac{2^k}{\nu^k}\int_{2}^{\nu(1-s)+2}\frac{(2+\nu(1-s)-u)^{k-1}}{u^k}\,\textrm{d}u \nonumber \\ & = \frac{2^k}{\nu^k}\int_2^{\nu(1-s)+2}\frac{1}{u^k} \sum_{i=0}^{k-1}\bigg[\binom{k-1}{i}(2+\nu(1-s))^i({-}u)^{k-1-i}\bigg]\,\textrm{d}u \nonumber \\ & = \frac{2^k}{\nu^k}\sum_{i=0}^{k-1}\bigg[\binom{k-1}{i}(2+\nu(1-s))^i ({-}1)^{k-1-i}\int_2^{\nu(1-s)+2}u^{-(i+1)}\,\textrm{d}u\bigg] \nonumber \\ & = ({-}1)^{k+1}\frac{2^k}{\nu^k}\bigg(\log\bigg(1+\frac{\nu}{2}(1-s)\bigg) + \sum_{i=1}^{k-1}\binom{k-1}{i}({-}1)^{i}\frac{(1+({\nu}/{2})(1-s))^{i}-1}{i}\bigg) \nonumber \\ & = ({-}1)^{k+1}\frac{2^k}{\nu^k}\Bigg( \log\bigg(1+\frac{\nu}{2}(1-s)\bigg) + \sum_{i=1}^{k-1}({-}1)^{i}\frac{\nu^i}{i2^i}(1-s)^i\Bigg), \end{align}

where the last equality follows from Lemma 6. Substituting back into (17), by (18), (25), (26), and (27) we obtain (15).

To finish the proof we need to handle the $\nu = 0$ case. Then the calculations are easier. We can still define $\varepsilon_{j,n}$ as in (21), and (20) remains true. Applying Lemma 1, we see that (23) and (24) hold true as well, therefore (25) follows with $\nu = 0$ everywhere. There is no need for the Riemann approximation, (15) follows from Lemma 2.

Proof of Theorem 3. Here we consider only the $\nu > 0$ case. For $\nu = 0$ the calculations are easier, and follow similarly to the previous proof.

Similarly to the proof of Theorem 2, (14) holds, so it is enough to prove that

\begin{equation*} \sum_{j=1}^{n}(h_j(f_{j,n}(s))-1) \to \log f_Y(s). \end{equation*}

Fix $s\in(0,1)$ . By Taylor’s theorem, for some $\xi \in (0, \nu(1-s)/2)$ ,

\[ \log\bigg(1+\frac{\nu}{2}(1-s)\bigg) + \sum_{i=1}^{k-1}({-}1)^{i}\frac{\nu^i}{i2^i}(1-s)^i = \frac{\nu^k(1-s)^k}{2^k}\frac{1}{(1 + \xi)^k}\frac{1}{k}, \]

so

\[ \frac{2^k}{\nu^{k}}\Bigg|\log\bigg(1+\frac{\nu}{2} (1-s)\bigg) + \sum_{i=1}^{k-1}({-}1)^{i}\frac{\nu^i}{i2^i}(1-s)^i\Bigg| \leq \frac{(1-s)^k}{k}. \]

Therefore, for any $\varepsilon > 0$ there exists $\ell$ large enough such that

(28) \begin{equation} \Bigg|\sum_{k=\ell}^{\infty}\frac{2^k\lambda_k}{\nu^k} \Bigg(\log\bigg(1 + \frac{\nu}{2}(1-s)\bigg) + \sum_{i=1}^{k-1}({-}1)^{i}\frac{\nu^i}{i2^i}(1-s)^i\Bigg)\Bigg| \leq \sum_{k=\ell}^\infty\frac{\lambda_k(1-s)^k}{k} < \varepsilon. \end{equation}

By Lemma 7, (16) holds with $K = \ell$ and therefore

\[ \begin{split} \sum_{j=1}^{n}(h_j(f_{j,n}(s))-1) & = \sum_{j=1}^{n}\Bigg[\sum_{k=1}^{\ell-1}\frac{m_{j,k}}{k!}(f_{j,n}(s)-1)^k + R_{j,\ell}(f_{j,n}(s))\Bigg] \\ & = \sum_{k=1}^{\ell-1}\sum_{j=1}^{n}\frac{m_{j,k}}{k!}(f_{j,n}(s)-1)^k + \sum_{j=1}^{n}R_{j,\ell}(f_{j,n}(s)). \end{split} \]

Moreover, by (13) and Lemma 2,

(29) \begin{equation} \Bigg|\sum_{j=1}^{n}R_{j,\ell}(f_{j,n}(s))\Bigg| \leq \sum_{j=1}^{n}\frac{m_{j,\ell}}{\ell!}|f_{j,n}(s)-1|^\ell \leq \sum_{j=1}^{n}\frac{m_{j,\ell}}{\ell!}\overline{f}_{j,n}^\ell(1-s)^\ell \to \frac{(1-s)^\ell}{\ell}\lambda_\ell \leq \varepsilon. \end{equation}

Summarizing,

\begin{align*} \Bigg|\sum_{j=1}^n(h_j(f_{j,n}(s)) & - 1) - \log f_Y(s)\Bigg| \\ & \leq \Bigg|\sum_{k=1}^{\ell-1}\sum_{j=1}^{n}\frac{m_{j,k}}{k!}(f_{j,n}(s)-1)^k \\ & \qquad + \sum_{k=1}^{\ell-1}\frac{2^k\lambda_k}{\nu^k} \Bigg(\log\bigg(1+\frac{\nu}{2}(1-s)\bigg) - \sum_{i=1}^{k-1}({-}1)^{i+1}\frac{\nu^i}{i2^i}(1-s)^i\Bigg)\Bigg| \\ & \quad + \Bigg|\sum_{k=\ell}^{\infty}\frac{2^k\lambda_k}{\nu^k} \Bigg(\log\bigg(1+\frac{\nu}{2}(1-s)\bigg)-\sum_{i=1}^{k-1}({-}1)^{i+1}\frac{\nu^i}{i2^i}(1-s)^i\Bigg)\Bigg| \\ & \quad + \Bigg|\sum_{j=1}^{n}R_{j,\ell}(f_{j,n}(s))\Bigg|, \end{align*}

where the first term on the right-hand side converges to 0 by the previous result, while the second and third terms are small for n large by (28) and (29). As $\varepsilon > 0$ is arbitrary, the proof is complete.

3.4. Proof of Theorem 4

Before the proof, we need two auxiliary lemmas.

Lemma 8. If condition (C4) of Theorem 2 or (C4ʹ) of Theorem 3 hold, then

$$q_i = \lim_{n\to\infty,\overline{f}_n < 1}\frac{\mathbb{P}(\varepsilon_n = i)}{1-\overline{f}_n}$$

exists for each $i = 1,2,\ldots$

Let $h_n[i] = \mathbb{P}(\varepsilon_n = i)$ . If (C4) holds, there are only finitely many $\lambda$ s, and the statement follows easily by backward induction. However, if (C4ʹ) holds, a more involved argument is needed, which works in both cases.

The kth moment of a random variable can be expressed in terms of its factorial moments as

\[ \mu_{n,k} \;:\!=\; \mathbb{E} \varepsilon_n^k = \sum_{i=1}^k \genfrac\{\}{0pt}{0}{k}{i} m_{n,i}, \]

where

$$\genfrac\{\}{0pt}{0}{k}{i} = \frac{1}{i!}\sum_{j=0}^i({-}1)^j\binom{i}{j}(i-j)^k$$

denotes the Stirling number of the second kind; on Stirling numbers, see [Reference Graham, Knuth and Patashnik10, Section 6.1]. Therefore,

(30) \begin{equation} \sum_{i=1}^\infty\frac{i^k h_n[i]}{1-\overline{f}_n} = \frac{\mu_{n,k}}{1-\overline{f}_n} = \sum_{i=1}^k\genfrac\{\}{0pt}{0}{k}{i}\frac{m_{n,i}}{1-\overline{f}_n} \to \sum_{i=1}^k\genfrac\{\}{0pt}{0}{k}{i}i!\lambda_i \;=\!:\; \mu_k. \end{equation}

Hence, the sequence $(h_n[i]/(1- \overline f_n))_{n\in\mathbb{N}}$ is bounded in n for all i. Therefore, any subsequence contains a further subsequence $(n_\ell)$ such that

(31) \begin{equation} \lim_{\ell\to\infty}\frac{h_{n_\ell}[i]}{1 - \overline f_{n_\ell}} = q_i \quad \text{for all}\ i\in\mathbb{N}. \end{equation}

To prove the statement we have to show that the sequence $(q_i)$ is unique, it does not depend on the subsequence. Note that the sequence $(q_i)_{i\in\mathbb{N}}$ is not necessarily a probability distribution.

By Fatou’s lemma and (30),

(32) \begin{equation} \mu_k = \lim_{\ell\to\infty}\frac{\mu_{n_\ell,k}}{1 - \overline f_{n_\ell}} \geq \sum_{i=1}^\infty\lim_{\ell\to\infty}\frac{h_{n_\ell}[i]}{1-\overline f_{n_\ell}}i^k = \sum_{i=1}^\infty q_i i^k. \end{equation}

Let $\varepsilon > 0$ and $k \in \mathbb{N}$ be arbitrary. Write $n_\ell = n$ , and put $K \;:\!=\; \lfloor{\mu_{k+1}}/{\varepsilon}\rfloor + 1$ , with $\lfloor\cdot\rfloor$ representing the lower integer part. Then

\[ \sum_{i=1}^\infty\frac{h_n[i]}{1 - \overline f_n}i^{k+1} \geq \sum_{i=K+1}^\infty\frac{h_n[i]}{1-\overline f_n}i^{k}K, \]

and hence, by (30) and the definition of K,

\[ \limsup_{n\to\infty}\sum_{i=K+1}^\infty\frac{h_n[i]}{1 - \overline f_n}i^{k} \leq \varepsilon. \]

Therefore, by (31),

\[ \mu_{k} = \lim_{n\to\infty}\sum_{i=1}^\infty\frac{h_n[i]}{1 - \overline f_n}i^{k} \leq \limsup_{n\to\infty}\sum_{i=1}^K\frac{h_n[i]}{1 - \overline f_n}i^k + \varepsilon \leq \sum_{i=1}^\infty q_i i^k + \varepsilon. \]

Since $\varepsilon > 0$ is arbitrary, by (32),

(33) \begin{equation} \mu_{k} = \sum_{i=1}^\infty q_i i^k. \end{equation}

Using the Stieltjes moment problem we show that the sequence $(\mu_k)$ uniquely determines $(q_i)$ . In order to do that, it is enough to show that Carleman’s condition [Reference Simon18, Theorem 5.6.6] is fulfilled, i.e. $\mu_k$ is not too large. Since $\limsup_{n\to\infty}\lambda_n^{1/n} \leq 1$ , for n large, $\lambda_n \leq 2^n$ . Furthermore, by trivial upper bounds,

\[ \genfrac\{\}{0pt}{0}{k}{i}i! = \sum_{\ell=0}^i({-}1)^\ell\binom{i}{\ell}(i - \ell)^k \leq i^k 2^i, \]

and thus, by (30), for some $C >0$ ,

\[ \mu_k = \sum_{i=0}^k\genfrac\{\}{0pt}{0}{k}{i}i!\lambda_i \leq C + \sum_{i=0}^ki^k4^i \leq C(4k)^k \leq k^{2k} \]

for k large enough, showing that Carleman’s condition holds. Hence the sequence $(q_i)_{i\in\mathbb{N}}$ is indeed unique, and the proof is complete.

Lemma 9. For any $L \geq 1$ , $n \geq 1$ , and $x \in \mathbb{R}$ ,

\[ \sum_{j=1}^L\sum_{\ell = 1}^j({-}1)^{\ell+j}\binom{L+n}{j+n}\binom{j-\ell + n - 1}{n-1}x^\ell = (1+x)^L - 1. \]

Proof. Changing the order of summation, it is enough to prove that, for $L\geq 1$ , $n \geq 1$ , and $1 \leq \ell \leq L$ ,

\begin{equation*} \sum_{j=\ell}^L({-}1)^{j+\ell}\binom{L+n}{j+n}\binom{j-\ell+n-1}{n-1} = \binom{L}{\ell}. \end{equation*}

We prove this by induction on L. It holds for $L = 1$ , and assuming for $L \geq 1$ , for $L+1$ we have, by the induction hypothesis for $(L,\ell, n)$ and $(L, \ell-1, n)$ ,

\begin{align*} \sum_{j=\ell}^{L+1}({-}1)^{j+\ell} & \binom{L+1+n}{j+n}\binom{j-\ell+n-1}{n-1} \\ & = \binom{L}{\ell} + \sum_{j=\ell}^{L+1}({-}1)^{j+\ell}\binom{L+n}{j+n-1}\binom{j-\ell+n-1}{n-1} \\ & = \binom{L}{\ell} + \sum_{j=\ell-1}^{L}({-}1)^{j+\ell-1}\binom{L+n}{j+n}\binom{j-(\ell-1)+n-1}{n-1} \\ & = \binom{L}{\ell} + \binom{L}{\ell-1} = \binom{L+1}{\ell}, \end{align*}

as claimed.

Proof of Theorem 4. In order to handle assumptions (i) and (ii) together, under (i) we use the notation $\lambda_k = 0$ for $k \geq K$ . Then the limiting g.f. $f_Y$ is given in (3).

First, assume that $\nu > 0$ . Substituting the Taylor series of the logarithm,

\begin{equation*} \log\bigg(1+\frac{\nu}{2}(1-s)\bigg) = \log\bigg(1+\frac{\nu}{2}\bigg) - \sum_{n=1}^\infty n^{-1}\bigg(\frac{\nu}{2+\nu}\bigg)^{n}s^n, \qquad s \in (0,1), \end{equation*}

into (3), expanding $(1-s)^i$ , and gathering together the powers of s, we have

(34) \begin{align} f_Y(s) & = \exp\!\Bigg\{{-}\!\sum_{k=1}^{\infty}\frac{2^k\lambda_k}{\nu^k} \Bigg(\!\log\bigg(1+\frac{\nu}{2}\bigg) - \sum_{n=1}^\infty\frac{1}{n}\bigg(\frac{\nu}{2 + \nu}\bigg)^n s^n + \sum_{i=1}^{k-1}({-}1)^{i}\frac{\nu^i}{i2^i}\sum_{j=0}^i\binom{i}{j}({-}s)^j\Bigg)\Bigg\} \nonumber \\ & = \exp\Bigg\{{-}\sum_{k=1}^\infty\frac{2^k\lambda_k}{\nu^k} \Bigg(\log\bigg(1+\frac{\nu}{2}\bigg) + \sum_{i=1}^{k-1}({-}1)^{i}\frac{\nu^{i}}{i2^{i}}\Bigg) \nonumber \\ & \qquad\qquad + \sum_{k=1}^\infty\frac{2^k\lambda_k}{\nu^k}\sum_{n=1}^\infty s^n \Bigg(n^{-1}\bigg(\frac{\nu}{2+\nu}\bigg)^{n} - \sum_{i=n}^{k-1}({-}1)^{i+n}\frac{\nu^i}{i 2^i}\binom{i}{n}\Bigg)\Bigg\}, \end{align}

where the empty sum is defined to be 0. We claim that the order of summation in (34) with respect to k and n can be interchanged. Indeed, this is clear under (i), since the sum in k is in fact finite.

Assume (ii). Then Taylor’s theorem applied to the function $(1+x)^{-n}$ with $n \leq k -1$ gives

\[ (1 + x)^{-n} = n\sum_{\ell=n}^{k-1}({-}1)^{\ell+n}\binom{\ell}{n}\frac{x^{\ell-n}}{\ell} + ({-}1)^{k-n}\binom{k-1}{k-n}(1 + \xi)^{-k} x^{k-n}, \]

with $\xi \in (0,x)$ . Therefore,

(35) \begin{equation} \Bigg|\sum_{i=n}^{k-1}({-}1)^{i+n}\binom{i}{n}\frac{x^{i-n}}{i} - n^{-1}(1 + x)^{-n}\Bigg| \leq n^{-1}\binom{k-1}{k-n}x^{k-n}, \end{equation}

thus, with $x = \nu / 2$ ,

\begin{align*} \sum_{k=1}^\infty\frac{2^k\lambda_k}{\nu^k}\sum_{n=1}^{k-1}s^n \Bigg|\sum_{i=n}^{k-1}({-}1)^{i+n}\binom{i}{n}\frac{\nu^{i}}{2^i i} - n^{-1}\bigg(\frac{\nu}{2+\nu}\bigg)^{n}\Bigg| & \leq \sum_{k=1}^\infty\frac{2^k\lambda_k}{\nu^k}\sum_{n=1}^{k-1}s^n \frac{\nu^n}{2^n}n^{-1}\binom{k-1}{k-n}\frac{\nu^{k-n}}{2^{k-n}} \\ & = \sum_{k=1}^\infty\lambda_k\sum_{n=1}^{k-1}s^n n^{-1}\binom{k-1}{k-n} \\ & \leq \sum_{k=1}^\infty\lambda_k(1+s)^k, \end{align*}

which is finite for $s \in (0,1)$ if $\limsup \lambda_n^{1/n} \leq \frac12$ . The other part is easy to handle, as

\[ \sum_{k=1}^\infty\frac{2^k\lambda_k}{\nu^k}\sum_{n=k}^\infty s^n n^{-1}\bigg(\frac{\nu}{2 + \nu}\bigg)^n \leq \sum_{k=1}^\infty\frac{2^k\lambda_k}{\nu^k}s^k\bigg(\frac{\nu}{2 + \nu}\bigg)^k = \sum_{k=1}^\infty\frac{2^k\lambda_k}{(2 + \nu)^k}s^k, \]

which is summable.

Summarizing, in both cases the order of summation in (34) can be interchanged, and doing so we obtain $f_Y(s) = \exp\big\{{-}A_0 + \sum_{n=1}^\infty A_n s^n\big\}$ , where

(36) \begin{equation} \begin{split} A_0 & = \sum_{k=1}^{\infty}\frac{2^k\lambda_k}{\nu^k} \Bigg(\sum_{i=1}^{k-1}({-}1)^{i}\frac{\nu^{i}}{i2^{i}} + \log\bigg(1+\frac{\nu}{2}\bigg)\Bigg), \\ A_n & = \sum_{k=1}^{\infty}\frac{2^k\lambda_k}{\nu^k} \Bigg(\bigg(\frac{\nu}{2+\nu}\bigg)^n n^{-1} - \sum_{i=n}^{k-1}({-}1)^{i+n}\binom{i}{n}\frac{\nu^{i}}{i2^{i}}\Bigg). \end{split} \end{equation}

Recall the notation from the proof of Lemma 8. By (30), using the inversion formula for Stirling numbers of the first and second kind (see, e.g., [Reference Graham, Knuth and Patashnik10, Exercise 12, p. 310]), we have

\[ k! \lambda_k = \sum_{i=0}^k ({-}1)^{k+i} \genfrac[]{0pt}{0}{k}{i} \mu_k, \]

where represents the Stirling number of the first kind. Substituting (33), and using

\[ \sum_{i=0}^k ({-}1)^{k+i} \genfrac[]{0pt}{0}{k}{i} j^k = j (j-1) \cdots (j-k+1) \]

(see, e.g., [Reference Graham, Knuth and Patashnik10, p. 263, (6.13)]), we obtain the formula

(37) \begin{equation} \lambda_k = \sum_{j=k}^\infty \binom{j}{k} q_j. \end{equation}

We also see that $\lambda_k = 0$ implies $q_j = 0$ for all $j \geq k$ . Substituting (37) back into (36) we claim that the order of summation with respect to k and j can be interchanged. This is again clear under (i), while under (ii), by (35),

\begin{align*} \sum_{k=n+1}^\infty\sum_{j=k}^\infty\frac{2^k}{\nu^k}q_j\binom{j}{k}\Bigg|\sum_{i=n}^{k-1}({-}1)^{i+n} & \binom{i}{n} \frac{\nu^i}{i 2^i}- n^{-1} \frac{\nu^n}{(2+\nu)^n} \Bigg| \\ & \leq \sum_{k=n+1}^\infty\sum_{j=k}^\infty\frac{2^k}{\nu^k}q_j\binom{j}{k}\frac{\nu^n}{2^n}n^{-1} \binom{k-1}{k-n}\frac{\nu^{k-n}}{2^{k-n}} \\ & = \sum_{k=n+1}^\infty\lambda_k n^{-1}\binom{k-1}{k-n} < \infty, \end{align*}

since

\[ n^{-1} \binom{k-1}{k-n} = k^{-1} \binom{k}{n} \leq k^{n}. \]

Thus, the order of summation is indeed interchangeable, and we obtain

(38) \begin{align} A_n & = \sum_{j=1}^\infty q_j\Bigg\{\bigg[\bigg(1 + \frac{2}{\nu}\bigg)^j - 1\bigg] \bigg(\frac{\nu}{2 + \nu}\bigg)^n n^{-1} - \sum_{k=n+1}^j\frac{2^k}{\nu^k}\binom{j}{k}\sum_{i=n}^{k-1}({-}1)^{i+n}\binom{i}{n}\frac{\nu^i}{i 2^i}\Bigg\} \nonumber \\ & \;=\!:\; \sum_{j=1}^\infty q_j B(n,j). \end{align}

We claim that

(39) \begin{equation} B(n,j) = \frac{\nu^n}{(2+\nu)^n n}\bigg[\bigg(1 + \frac{2}{\nu}\bigg)^{n \wedge j}-1\bigg]. \end{equation}

This is clear for $j \leq n$ . Writing $\ell = k -i$ in the summation and using Lemma 9 with $L = j-n$ we obtain

\begin{align*} \sum_{k=n+1}^j\sum_{i=n}^{k-1}({-}1)^i\binom{j}{k}\binom{i-1}{n-1}x^{k-i} & = \sum_{k=n+1}^j\sum_{\ell=1}^{k-n}({-}1)^{k+\ell}\binom{j}{k}\binom{k-\ell-1}{n-1}x^\ell \\ & = ({-}1)^n[(1+x)^{j-n}-1]. \end{align*}

Using that $n \binom{i}{n} = i \binom{i-1}{n-1}$ , and substituting back into (38) with $x= 2/\nu$ , we have

\[ B(n,j) = n^{-1}\bigg[1 - \bigg(1 + \frac{2}{\nu}\bigg)^{j-n}\bigg] + \bigg[\bigg(1 + \frac{2}{\nu}\bigg)^j - 1\bigg]\bigg(\frac{\nu}{2 + \nu}\bigg)^n n^{-1}, \]

which after simplification gives (39).

Now let $\nu = 0$ . In this case the calculations are again simpler. Expanding $(s-1)^i$ and changing the order of summation,

\[ \log f_Y(s) = \sum_{k=1}^\infty({-}1)^k\frac{\lambda_k}{k} + \sum_{n=1}^\infty s^n\sum_{k=n}^\infty({-}1)^{n+k}\frac{\lambda_k}{k}\binom{k}{n} \;=\!:\; -A_0 + \sum_{n=1}^\infty s^n A_n. \]

Substituting (37) and changing the order of summation again we get

(40) \begin{equation} A_n = \sum_{j=n}^\infty q_j\sum_{k=n}^j\binom{k}{n}\binom{j}{k}({-}1)^{k+n}\frac{1}{k}. \end{equation}

The changes of the order of summation can be justified as in the previous case. For the coefficient of $q_j$ we have

\[ \sum_{k=n}^j\binom{k}{n}\binom{j}{k}({-}1)^{k+n}\frac{1}{k} = \binom{j}{n}\sum_{k=n}^j\frac{1}{k}\binom{j-n}{k-n}({-}1)^{k+n} = \frac{1}{n}, \]

where we used that for $j \geq n$ ,

\[ \sum_{k=0}^{j-n}({-}1)^k\frac{j}{k+n}\binom{j-n}{k} = \binom{j-1}{n-1}^{-1}. \]

The latter formula follows by induction on j. Substituting back into (40), the proof is complete.