I came across the sum:
$$I(m)=\sum_{n=0}^\infty \frac{(-1)^n}{mn+1} $$
where $m>0$.
It's easy to see that this sum is equal to:
$$\int_0^1 \frac{1}{1+x^m}dx$$
for $m>0$
So I tried my hardest to find a way of evaluating it. First I thought using Euler's Beta Function but to no avail. Then I decided to brute force a partial fraction decomposition, and I got this ugly mess (which is more than like incorrect):
$$I(m)=\sum_{n=0}^\infty \frac{(-1)^n}{mn+1} = \sum_{k=0}^\infty A_k\ln|1-e^{-(i(2k+1))/m}| $$Where $A_k$ equals
$$A_k= \frac{1}{\prod_{n=0}^k(e^{i(2k+1)}-e^{i\pi(2n+1)/m})\prod_{n=k}^\infty(e^{ik\pi/m}-e^{i(2n+3)}) } $$
I would check my work, but it's really tricky to check so I'm just hoping it's right. Is there a way of evaluating the integral that's less ugly than what I did?
Best Answer
Using the digamma function $$\int_0^1 \frac{1}{1+x^m}dx=\frac{\psi ^{(0)}\left(\frac{m+1}{2 m}\right)-\psi ^{(0)}\left(\frac{1}{2 m}\right)}{2 m}$$
Using the Hurwitz-Lerch transcendent function $$\int_0^1 \frac{1}{1+x^m}dx=\frac{\Phi \left(-1,1,\frac{1}{m}\right)}{m}$$