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$$\sum\limits_{y=1}^{Y}\sum\limits_{z=1}^{y} a^{y-1} b^y \binom{y-1}{z-1} (c + 2z)^d $$
Does this series converge when $Y=∞$? If the series converges, what does it converge to? If the series does not converge, how can I evaluate it for a finite $Y$?


$0 \lt a \lt 1$

$0 \lt b \lt 1$

$0 \lt d \lt 1$

$0 \lt c$

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What have you tried? Where does this problem come from? You have a much better chance to get good answers if you provide some context. – mrf Jan 15 '13 at 21:34
And more precisely? – Did Jan 15 '13 at 21:39
Hint: Sum over $y$ first. The result is a simple function of $z$, and the convergence of their serie can be studied. It might be the case that the double series converges if and only if $ab\lt\frac12$. – Did Jan 15 '13 at 22:56
There is an event that can occur in each period with the probability $b$. If the event happens in a period, the value in that period is $(c+2z)^d$, where $z$ is the number of times the event has happened in total. If the event $b$ does not happen, the value for the period is $0$. The value of future periods is discounted by the factor $a$. – LKL Jan 19 '13 at 0:34
Is $c>0$ or $0<c<1$? – Ian Mateus Jan 19 '13 at 0:56

Here is a solution for your first question. Take $Y=\infty$. As a pedagogical device, you can draw all the pairs $(y,z)$ appearing in the sum: they form a region bounded by the $y$ axis starting from $y=1$ and the line $z=y$. Now put your right ear on your right shoulder (or turn the sheet, either way works), and you will see that the possible values of $z$ are the positive integers, and for fixed $z$, the admissible values of $y$ are precisely those with $y\geq z$. Since your double series consists entirely of nonnegative summands, we can reorder it as we want and so your sum becomes

$$\sum_{z=1}^\infty\sum_{y=z}^\infty a^{y-1}b^y\binom{y-1}{z-1}(c+2z)^d\,.$$

Now we make the change of variable $k=z-1, j=y-1$ (I feel uncomfortable with indices $z$ and $y$), and the series becomes

$$\sum_{k=0}^\infty\sum_{j=k}^\infty a^jb^{j+1}\binom jk(2k+2+c)^d=b\sum_{k=0}^\infty(2k+2+c)^d\sum_{j=k}^\infty\binom jk(ab)^j\,.$$

Let $f(x)=\sum_{j=k}^\infty\binom jkx^j$. Then


Therefore we have (modulo convergence issues) $g^{(j)}(0)=(j+k)!$. You can see by induction that the function $h(x)=k!(1-x)^{-k-1}$ satisfies $h^{(j)}(x)=(j+k)!(1-x)^{-j-k-1}$ for all $j\geq0$; in particular $h^{(j)}(0)=(j+k)!$, and so


(any shorter deduction of this fact is welcomed). Using this result, our double series becomes

$$b\sum_{k=0}^\infty(2k+2+c)^df(ab)=\frac b{1-ab}\sum_{k=0}^\infty(2k+2+c)^d\biggl(\frac{ab}{1-ab}\biggr)^k\,.$$

Now we apply the root test: we have


which implies that your series converge whenever $\frac{ab}{1-ab}<1$, that is $ab<1/2$, and diverges whenever $\frac{ab}{1-ab}>1$, that is $ab>1/2$. When $ab=1/2$ you have $\frac{ab}{1-ab}=1$, and your series becomes


which obviously diverges.

I don't think that a explicit formula exists for the case $ab<1/2\ldots$ but I would be grateful if someone disappoint me.

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The fact that $f(x) = x^k (1-x)^{-k-1}$ follows directly from Newton's generalized binomial theorem. It's one of the standard generating functions for the binomial coefficients, too. See, for example, p. 335 of Concrete Mathematics. – Mike Spivey Jan 20 '13 at 3:31
Your final sum can be written in terms of the Lerch zeta function. A "more closed" expression may not exist. (+1) – user26872 Jan 21 '13 at 7:08
@MikeSpivey You are right. At first I want to use the geometric series for $(1-x)^{-1}$, and multiply by itself $k+1$ times. Of course, this is more complicated than desired, so finally I resorted to heuristics ... next time I will carry Newton's GBN on my toolbox ;-) @ oen Thanks, I hope that the OP finally become convinced that a explicit formula probably does not exist. – Matemáticos Chibchas Jan 21 '13 at 15:09

From the comments, I think the question may be based on the following situation:

Let $X_1$, $X_2$, $\dots$ be independent, identically distributed Bernoulli random variables such that ${\Bbb P}(X_i=1)=b$, ${\Bbb P}(X_i=0)=1-b$. The payoff at time $y=1$, $2$, $\dots$ is $$V_y:=\left\{\begin{array}{ll}(c+2(X_1+\dots+X_y))^d,& X_y=1, \\ 0,& X_y=0.\end{array}\right.$$ The problem is to find the expected discounted payoff from times $1$ through $Y$, $$P_Y:=\sum_{1\le y\le Y} a^{y-1}\, {\Bbb E}V_y,$$ where $Y$ may be $\infty$, in which case the sum is over all $y\ge 1$.

Since, if $1\le z\le y$, $${\Bbb P}(X_1+\dots+X_y=z \hbox{ and } X_y=1)=\binom{y-1}{z-1} b^{z} (1-b)^{y-z},$$ $P_Y$ can be computed by the double sum on $y$ and $z$, $$ P_Y=\sum_{1\le y\le Y,\ 1\le z\le y} \binom{y-1}{z-1} a^{y-1} b^{z} (1-b)^{y-z} (c+2z)^d. $$ This is a slightly different sum from that given by the questioner.

This formula would be simpler if $d$ were integral. If $0<a<1$, $0<b<1$, and $c$ and $d$ are positive, it is still possible to say that $P_{\infty}$ converges, because $P_{\infty}$ can be overestimated by its value in the case where each $X_i$ always occurs, i.e., $b=1$. In this case, \begin{eqnarray*} P_{\infty}&=&\sum_{y\ge 1} a^{y-1} (c+2y)^d \\ &\le&(c+2)^d \sum_{y\ge 1} a^{y-1} y^d, \end{eqnarray*} and $\sum_{y\ge 1} a^{y-1} y^d$ converges whenever $|a|<1$. It is not necessary to assume that $d<1$.

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