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If we define a function as: $$P(x)=\sum_{n=1}^\infty \frac{1}{n^x(n+1)^x}$$ For $x=1$, we have a standard telescoping series that sums to $1$. For $x=2$, the series sums to $\frac{\pi^2}{3}-3$. For $x=3$, the series sums to $10-\pi^2$, ... and so on.

My question is, what is the minimum value for which x allows this "Reciprocal Pronic" series to converge. I think it is something above $x>0.5$ but I cannot prove this.

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    $\begingroup$ Think integral test. Your numerator is a constant, so your denominator should be of degree more than 1. (Why?) $\endgroup$ – imranfat Oct 13 at 21:21
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    $\begingroup$ I am aware that convergence is possible for values 0.5<x<1. $\endgroup$ – user714524 Oct 13 at 21:23
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    $\begingroup$ Note the binomial series $(1+n^{-1})^{-s}=\sum_{k\ge 0} {-s \choose k} n^{-k}$ means $$\sum_{n\ge 1} n^{-s}(n+1)^{-s} = \sum_{n\ge 1} n^{-2s}\sum_{k\ge 0} {-s \choose k} n^{-k}=\sum_{k\ge 0} {-s \choose k} \sum_{n\ge 1}n^{-2s-k}=\sum_{k\ge 0} {-s \choose k} \zeta(2s+k)$$ which gives the analytic continuation to the whole complex plane, with simple poles at $s=(1-2m)/2,m\ge 0$. $\endgroup$ – reuns Oct 14 at 0:13
  • $\begingroup$ Very clever - thanks $\endgroup$ – user714524 Oct 16 at 21:20
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Hint Use inequalities to write bounds for the summand that involve simpler expressions: $$\frac{1}{(n + 1)^{2x}} \leq \frac{1}{n^x (n + 1)^x} \leq \frac{1}{n^{2x}}.$$

It follows from doing so that there is no minimal $x$ for which the series converges, but one can instead for the infimum of the set of $x$ for which it does.

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  • $\begingroup$ That is helpful but not conclusive. This only places an upper limit which I suspect is the correct limit but I want to go further. $\endgroup$ – user714524 Oct 13 at 21:36
  • $\begingroup$ Actually it places both an upper and lower limit $\endgroup$ – vujazzman Oct 13 at 21:42
  • $\begingroup$ @user714524 The inequalities bound the summand both above and below---essentially by $p$-series, which suggests using the $p$-series Test and the Comparison Test. $\endgroup$ – Travis Willse Oct 13 at 21:44

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