3
votes
1answer
238 views

Series of the inverse quadratic trinomial

Maybe it's a very simple question, but I have a problem with the following series $$\sum\limits_{n=1}^{\infty}\frac{1}{n^2+pn+q},$$ where $p, q \in \mathbb{R}$. I know about five ways how to calculate ...
1
vote
0answers
267 views

Do these infinite series expressing $\zeta(s)$ only (partially) converge at $\Re(s)=\frac12$?

The following analytic continuation for $\zeta(s)$ towards $\Re(s)>-1$ was derived here: $$\displaystyle \zeta(s) = \frac{1}{2\,(s-1)} \left(\sum _{n=1}^{\infty } {\frac {s-1-2\,n}{{n}^{s}}} + ...
1
vote
0answers
170 views

Zeta sum $\sum_{n=2}^\infty \frac{\zeta(n)}{a^n}$

Probably this is known, but mathworld and wolfram alpha don't recognize this potential identities. Numerical evidence suggests: $$ \sum_{n=2}^\infty \frac{\zeta(n)}{a^n} =? \sum_{n=1}^\infty ...
3
votes
0answers
199 views

What is the (fractional) half-derivative of zeta at $s=0$ (and how to compute it)?

(I asked this in MSE before but there was only a general reference which did not help for my specific question) I think I understood the concept of fractional derivatives applied to ...
13
votes
0answers
762 views

Regularizing the divergent sum $1^k + 2^k + \cdots$

EDIT: Under this "regularization", the harmonic series can be interpreted as $s_{-1}$ and assigned a value $$s_{-1} = \lim_{k \rightarrow 1} \zeta(k) (2-2^k) = -2 \log 2$$ I was looking at ...