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I would like to ask a simplified version of this question on MO: question about infinite series

Assume $x \in \mathbb{R}$.

I like to conjecture that the sum of the following two infinite series:

$$\displaystyle Z(x) = \frac{1}{2\,(x-1)} \sum _{n=1}^{\infty } {\frac {x-1-2\,n}{{n}^{x}}} + \frac{1}{2\,(-x)} \sum _{n=0}^{\infty } {\frac {1-x+1+2\,n}{\left( n+1 \right) ^{1-x}}}$$

only converges for $x=\frac12$.

Is there a possible approach to prove this or am I touching on something extremely complex?

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    $\begingroup$ The first converges only if $x>2$, the second if $x<-2$, so there is no $x$ for which both converge $\endgroup$ – Hagen von Eitzen Dec 27 '13 at 22:32
  • $\begingroup$ That is correct, however the sum/difference of two divergent series could also converge, right? My conjecture is that the latter only happens at $x=\frac12$ and then it is equal to $\zeta(\frac12)$. $\endgroup$ – Agno Dec 27 '13 at 22:44
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Well, let us take $\;x=\frac12\;$:

$$Z\left(\frac12\right)=-\sum_{n=1}^\infty\frac{-\frac12-2n}{\sqrt n}-\sum_{n=0}^\infty\frac{\frac12+1+2n}{\sqrt{n+1}}=$$

$$=\frac12\left[\sum_{n=1}^\infty\frac{4n+1}{\sqrt n}-\sum_{n=0}^\infty\frac{4n+3}{\sqrt{n+1}}\right]=\frac12\left[\sum_{n=0}^\infty\frac{4n+5}{\sqrt{n+1}}-\sum_{n=0}^\infty\frac{4n+3}{\sqrt{n+1}}\right]=$$

$$=\frac12\sum_{n=0}^\infty\frac2{\sqrt{n+1}}=\sum_{n=1}^\infty\frac1{\sqrt n}$$

and the last series is, of course, divergent.

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  • $\begingroup$ This proves that it is convergent for $x=1/2$. I think the OP was struggling with the "only" part. $\endgroup$ – John Dec 27 '13 at 23:22
  • $\begingroup$ Well, I think the above proves it is not convergent for $\;x=\frac12\;$ unless, as many other times, I missed something, @John ...Besides this, what "only" part?? $\endgroup$ – DonAntonio Dec 27 '13 at 23:24
  • $\begingroup$ Sorry, I judged the idea before looking at the final result. I blame the day--exhausted. :( $\endgroup$ – John Dec 27 '13 at 23:25
  • $\begingroup$ Welcome to the "Rats, I could have sworn over my holy cousin's girlfriend's great-greatmother's grave that I saw something else!" club, @John: I'm emerit founder, president and emperor of it for life. $\endgroup$ – DonAntonio Dec 27 '13 at 23:28
  • $\begingroup$ To simplify his question, he is asking for which values of $x$ does the series converge, and in this case its seems the proposed value is wrong :\. $\endgroup$ – John Dec 27 '13 at 23:28

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