Convergence of an infinite sum at $s=\frac12$? The following infinite, alternating sum:
$$\displaystyle A(s):= \sum _{n=1}^{\infty } \left( {\frac {s+(-1)^n(1+2\,n)}{{n}^{s}}} \right)$$
converges for $\Re(s)>1$.
In the domain $\Re(s) \le 1$ the sum seems to only converge EDIT: for finite odd upper bounded sums, when $s=\frac12$ and it then becomes $2.095285523...$, EDIT: but does diverge for even upper bounded finite sums. I am now curious to understand why that is and/or whether it can be proven.
Thanks!
 A: Any convergent series must have a general term which converges against 0. (Easy to prove. consider $\lim_{n\to \infty} S(n)-S(n-1)$.) This is not the case for your series, so it must diverge.
You are right about the convergence if you pair every two consecutive values.
$a(n):=\frac{(-1)^n (2 n+1)+\frac{1}{2}}{\sqrt{n}}$,
Then
$\sum_{n=1}^{\infty}a(2n)+a(2n+1)$ converges to $\frac{\pi  \left(\left(2 \sqrt{2}-1\right) \zeta \left(\frac{1}{2}\right)+5\right)+\left(1-2 \sqrt{2}\right) \zeta \left(\frac{3}{2}\right)}{2 \pi }$
That is what Mathematica said. The calculation was a bit complicated, so I didn't do it myself. Sorry.
A: Consider $a_n(s)=(s+(-1)^n(1+2\,n))\,n^{-s}$ and $b_n(s)=a_{2n}(s)+a_{2n+1}(s)$, then the convergence you observe and the values you give for $A(s)$ concern the series $$
B(s)=a_1(s)+\sum\limits_{n\geqslant1}b_n(s).
$$
Now,
$$
(2n+1)^sb_n(s)=\left(1+\frac1{2n}\right)^s(s+1+4n)+s-3-4n.
$$
Expanding the power as
$$
\left(1+\frac1{2n}\right)^s=1+\frac{s}{2n}+\frac{s(s-1)}{8n^2}+O\left(\frac1{n^3}\right),
$$
one gets
$$
(2n+1)^sb_n(s)=4s-2+\frac{s^2}n+O\left(\frac1{n^2}\right).
$$
Two cases appear:


*

*If $s\ne1/2$, then $b_n(s)=\Theta(1/n^s)$ hence $B(s)$ converges if and only if $\Re(s)\gt1$.

*If $s=1/2$, then the term $4s-2$ disappears and $b_n(1/2)=\Theta(1/n^{3/2})$, in particular $B(1/2)$ converges.


This yields the convergence/divergence of the series $A(s)$ since the series $A(s)$ converges if and only if the series $B(s)$ converges and $a_n(s)\to0$:


*

*The series $A(s)$ converges absolutely if and only if $\Re(s)\gt2$.

*The series $A(s)$ converges if and only if $\Re(s)\gt1$ (and, obviously, not if $s=1/2$).

