How to find the infinite sum $\sum\limits_{m=0}^\infty \frac{1}{(4m+1)^n}$ I know that $$\sum\limits_{m=0}^\infty \frac{1}{(2m+1)^n}=(1-2^{-n})\zeta(n),$$ where $\zeta(n)$ is the Riemann zeta value. But how to find the infinite sum $$\sum\limits_{m=0}^\infty \frac{1}{(4m+1)^n},$$ where $n$ is an integer with $n\geq 2$. I have tried that
$$\sum\frac{1}{(4m+1)^n}+\sum\frac{1}{(4m+2)^n}+\sum\frac{1}{(4m+3)^n}+\sum\frac{1}{(4m+4)^n}=\zeta(n),$$
and $$\sum\frac{1}{(4m+2)^n}=2^{-n}(1-2^{-n})\zeta(n),$$
$$\sum\frac{1}{(4m+4)^n}=4^{-n}\zeta(n).$$ But the two left sums can not be well represented.
 A: I think that the simplest is to use the Hurwitz zeta function
$$\sum\limits_{m=0}^\infty \frac{1}{(a\,m+b)^n}=a^{-n} \zeta \left(n,\frac{b}{a}\right)$$and
$$a^{-n} \zeta \left(n,\frac{a}{a}\right)=a^{-n} \zeta(n)$$
A: For $n\ge 2$ integer there is no closed-form for $$\sum_{m= 0}^\infty (4m+1)^{-n}$$ There is one only for $$\sum_{m=-\infty}^\infty (4m+1)^{-n}$$
Similarly there is a closed-form for $\zeta(2n)$ and $\sum_{m=-\infty, m\ne 0}^\infty m^{-2n+1} = 0$ but there isn't a closed-form for $\zeta(2n+1)$.
A: Maple does the case $n=2$ in terms of Catalan's constant $G$.
$$
\sum_{m=0}^\infty\frac{1}{(4m+1)^2} = \frac{\pi^2}{16}+\frac{G}{2}
$$
But of course there is no known simple evaluaton of Catalan's constant.
Maple does all other cases in terms of the digamma function $\psi(x) = \Gamma'(x)/\Gamma(x)$.
$$
\sum_{m=0}^\infty \frac{1}{(4m+1)^n} = 
\frac{(-1)^n}{4^n(n-1)!}\;\psi^{(n-1)}\left(\frac14\right)
$$
Here $\psi^{(n-1)}$ is the derivative of order $n-1$.  These derivatives of the digamma function are also called polygamma functions.
