Closed form evaluation of series in terms of fractional zeta function values? It is easy to verify using the binomial theorem that
$$\sum_{m=1}^\infty \bigg(1+\frac{1}{m}\bigg)^n-1-\frac{n}{m}=\sum_{k=2}^n \binom{n}{k}\zeta(k)$$
for positive integer $n\ge 2$. But what about for fractional $n$? For example, can one evaluate the series
$$\sum_{m=1}^\infty \bigg(1+\frac{1}{m}\bigg)^{3/2}-1-\frac{3}{2m}$$
in terms of the zeta function at its fractional values?
The series converges, but Wolfram yields no value.
 A: We have
$$\left(1+\frac{1}{m}\right)^{3/2}-1-\frac{3}{2m}=\frac{3}{2}\sum_{n\geq 0}\frac{\binom{2n}{n}(-1)^n}{(n+2)(2n+2)4^n m^{n+2}} \tag{1}$$
hence
$$\sum_{m\geq 1}\left[\left(1+\frac{1}{m}\right)^{3/2}-1-\frac{3}{2m}\right]=\color{red}{\frac{3}{4}\sum_{n\geq 0}\frac{\binom{2n}{n}(-1)^n\,\zeta(n+2)}{(n+1)(n+2)\,4^n}}.\tag{2} $$
The series in the RHS of $(2)$ is clearly convergent due to the alternating signs and the fact that the absolute value of the main term behaves like $\frac{C}{n^{5/2}}$ for large values of $n$. By exploiting the integral representation for the $\zeta$ function, the RHS of $(2)$ can also be written as
$$ \int_{0}^{+\infty}\frac{(3+2z)\,I_0(z/2)\,e^{-z/2}+(1+2z)\,I_1(z/2)\,e^{-z/2}-3}{2(e^z-1)}\,dz \tag{3}$$
where $I_0$ and $I_1$ are modified Bessel functions of the first kind. The integrand function approximately behaves like $\frac{3}{8}e^{-7z/12}$ in a right neighbourhood of the origin, giving us that the value of the wanted series is not that far from $\frac{9}{14}$. On the other hand, by interpolating the partial sums of the RHS of $(2)$ we already have the accurate approximation $\color{red}{\frac{33}{59}}$.
A: I honestly do not see any problem with doing the same kind of derivation for:
$$\sum_{m=1}^\infty \bigg[ \bigg(1+\frac{1}{m}\bigg)^r-1-\frac{r}{m} \bigg]=\sum_{k=2}^\infty \binom{r}{k}\zeta(k)$$
Where $r \in \mathbb{R}$.
We have for $|m| >1$ the binomial series:
$$\bigg(1+\frac{1}{m}\bigg)^r=\sum_{k=0}^\infty \binom{r}{k} \frac{1}{m^k}$$
The result above obviously follows. No fractional Zeta functions are needed, since $k$ is an integer.
If I missed something here, please comment.
