# Can this infinite sum be simplified

I have the following sum that I would like to explore: $$d(S) = \sum_{n_1, n_2, \ldots, n_S = 0}^\infty \left( \left[\prod_{j=1}^S\frac{\alpha^{n_j}}{n_j!}\right]\sum_{k=1}^Nn_k^2\mu_k^2\right),$$

where $\alpha$ is a positive real number, $S$ is a positive integer, and

$$\mu_k = \xi\left(k - \frac{S+1}{2}\right).$$ I am using Mathematica to numerically find $d(S)$ for different values of $S$, which of course takes a long time (even for the truncated infinite sum). Is there some identity that I could make use of to simplify the summation in any way.

• Does pulling the summation outside the product and takes its $S^{\mathrm{th}}$ power speed things up? (since it is independent of $j$) – Alex Vong Dec 13 '17 at 14:50
• @AlexVong, thank you. This has been made clear in the notation – Sid Dec 13 '17 at 15:02

The way you wrote the round bracket was initially slightly ambiguous. I assume you mean the following $$d(S) = \sum_{n_1, n_2, \ldots, n_S = 0}^\infty \left[\left(\prod_{j=1}^S\frac{\alpha^{n_j}}{n_j!}\right) \sum_{k=1}^Nn_k^2\mu_k^2\right]= \sum_{n_1=0}^\infty\frac{\alpha^{n_1}}{n_1!}\sum_{n_2=0}^\infty\frac{\alpha^{n_2}}{n_2!}\cdots \sum_{n_S=0}^\infty\frac{\alpha^{n_S}}{n_S!}\left(n_1^2\mu_1^2+\ldots +n_N^2\mu_N^2\right)\ .$$ If this is what you meant, and assuming $N\leq S$, then your sum is not that difficult. It is given by $$d(S)=\sum_{n_1=0}^\infty\frac{\alpha^{n_1}}{n_1!}\sum_{n_2=0}^\infty\frac{\alpha^{n_2}}{n_2!}\cdots \sum_{n_S=0}^\infty\frac{\alpha^{n_S}}{n_S!}\left(n_1^2\mu_1^2+\ldots +n_N^2\mu_N^2\right)=\left[\sum_{n_1=0}^\infty\frac{\alpha^{n_1}}{n_1!}\sum_{n_2=0}^\infty\frac{\alpha^{n_2}}{n_2!}\cdots \sum_{n_S=0}^\infty\frac{\alpha^{n_S}}{n_S!}n_1^2\mu_1^2\right]+$$ $$+\left[\sum_{n_1=0}^\infty\frac{\alpha^{n_1}}{n_1!}\sum_{n_2=0}^\infty\frac{\alpha^{n_2}}{n_2!}\cdots \sum_{n_S=0}^\infty\frac{\alpha^{n_S}}{n_S!}n_2^2\mu_2^2\right]+\cdots+\left[\sum_{n_1=0}^\infty\frac{\alpha^{n_1}}{n_1!}\sum_{n_2=0}^\infty\frac{\alpha^{n_2}}{n_2!}\cdots \sum_{n_S=0}^\infty\frac{\alpha^{n_S}}{n_S!}n_N^2\mu_N^2\right]\ .$$ Hence $$d(S)=\sum_{k=1}^N\underbrace{\sum_{n_1=0}^\infty\frac{\alpha^{n_1}}{n_1!}\cdots\sum_{n_{k-1}=0}^\infty\frac{\alpha^{n_{k-1}}}{n_{k-1}!}}_{(k-1)-terms}\sum_{n_k=0}^\infty\frac{\alpha^{n_k}}{n_k!}n_k^2\mu_k^2 \underbrace{\sum_{n_{k+1}=0}^\infty\frac{\alpha^{n_{k+1}}}{n_{k+1}!}\cdots\sum_{n_{S}=0}^\infty\frac{\alpha^{n_{S}}}{n_{S}!}}_{(S-k)-terms}\ .$$ All the sums except the $k$th give the same result $=e^\alpha$, while the $k$th sum gives $=e^{\alpha } \left(\alpha ^2+\alpha \right) \xi ^2 \left(k+\frac{1}{2} (-S-1)\right)^2$. Therefore the final result is $$d(S)=\sum_{k=1}^N e^{\alpha (S-1)}e^{\alpha } \left(\alpha ^2+\alpha \right) \xi ^2 \left(k+\frac{1}{2} (-S-1)\right)^2=e^{\alpha S}\left(\alpha ^2+\alpha \right) \xi ^2 \frac{1}{12} \left(4 N^3-6 N^2 S+3 N S^2-N\right)\ .$$
• sorry about the ambiguity - I have changed the question. Your first assumption is correct. How did you manage to rewrite $d(S)$ in that case? – Sid Dec 13 '17 at 15:18
• uhm...I just distribute the product, $\sum_{n_1,n_2}f_1(n_1)f_2(n_2)=\sum_{n_1}f_1(n_1)\sum_{n_2}f_2(n_2)$... – Pierpaolo Vivo Dec 13 '17 at 15:20