Does this double series converge? $$\sum\limits_{y=1}^{Y}\sum\limits_{z=1}^{y} a^{y-1} b^y \binom{y-1}{z-1} (c + 2z)^d $$
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 $Y$? 
where:  
$0 \lt a \lt 1$  
$0 \lt b \lt 1$  
$0 \lt d \lt 1$  
$0 \lt c$  
 A: 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
$$f(x)=\sum_{j=0}^\infty\binom{j+k}kx^{j+k}=\frac{x^k}{k!}\sum_{j=0}^\infty\frac{(j+k)!}{j!}\,x^j=\frac{x^k}{k!}g(x)\,.$$
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
$$f(x)=x^k(1-x)^{-k-1}$$
(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
$$\lim_{k\to\infty}\sqrt[k]{(2k+2+c)^d\biggl(\frac{ab}{1-ab}\biggr)^k}=\frac{ab}{1-ab}\,,$$
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
$$2b\sum_{k=0}^\infty(2k+2+c)^d\,,$$
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.
A: 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 $d<1$.
