Find a number $x$ such that $\sum_{n=1}^\infty\frac{n^x}{2^n n!} = \frac{1539}{64}e^{1/2}$

Question

I need to find a number $x$ such that $$\sum_{n=1}^\infty\frac{n^x}{2^n n!} = \frac{1539}{64}e^{1/2}.$$ What is the best approach to this problem?

Answer 1

One may note that $$\left(z\frac{d}{dz}\right)^a\;\underbrace{\sum_{n=1}^{\infty}\frac{z^n}{n!}}_{e^{z}-1}=\sum_{n=1}^{\infty}\frac{n^a z^n}{n!}.$$ Now indeed setting $a=6$, calculating the derivatives of $e^z-1$ and setting $z=\frac12$ in the final expression, we get $\frac{1539}{64}\sqrt{e}$.

Answer 2

According to Wikipedia, $$\sum_{n = 0}^\infty \frac{n^k z^n}{n!} = e^z T_k(z),$$ where $T_k(z)$ is the $k^\text{th}$ Touchard polynomial. Using $z = \tfrac{1}{2}$ and $k = 6$, we find $$\sum_{n = 0}^\infty \frac{n^6}{2^n n!} = e^{1/2} T_6(1/2) = \frac{1539}{64}e^{1/2}.$$

(Hopefully someone can find a more satisfying solution.)

Answer 3

The idea is that this sum can be computed if $x \in \mathbb{N}$.

Let $f_0(x) = \sum_{n\geq0} \frac{x^n}{2^n n!}$. This is a power series; its radius of convergence is infinite, and we have $f_0(x) = \exp\left(\frac x2\right)$. For all $k \in \mathbb{N}$, let $f_{k+1}(x) = xf_k'(x)$. By induction, it is straightforward to show that $f_k(x) = \sum_{n\geq0} \frac{n^k x^n}{2^n n!}$. Let $P_k(x) = 2^k \exp\left(-\frac x2\right)$, so that $$f_k(x) = \sum_{n\geq0} \frac{n^k x^n}{2^n n!} = 2^{-k} \exp\left(\frac x2\right) P_k(x)$$ We have $P_0(x) = 1$, and the relation $f_{n+1}(x)=xf_n'(x)$ yields immediatly $$P_{n+1}(x)=x\bigl( P_n(x) + 2P_n'(x) \bigr).$$ One can compute $P_1(x)=x$, $P_2(x)=x^2+2x$, ..., $P_6(x) = x^6+30x^5+260x^4+720x^3+496x^2+32x$.

For $k \in \mathbb{N}^*$, we have $$\sum_{n\geq1} \frac{n^k}{2^n n!} = f_k(1) = \exp\left(\frac 12\right) \frac{P_k(1)}{2^k}. $$

Now is time for a bit of luck: the denominator is $64=2^6$ and $P_6(1)=1539$.

Hence $x=6$ is solution.