# What is the minimum value of $x$ for which this “reciprocal pronic” series, $\sum_{n=1}^\infty \frac{1}{n^x(n+1)^x}$, converges?

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.

• Think integral test. Your numerator is a constant, so your denominator should be of degree more than 1. (Why?) – imranfat Oct 13 at 21:21
• I am aware that convergence is possible for values 0.5<x<1. – user714524 Oct 13 at 21:23
• 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 $s=(1-2m)/2,m\ge 0$. – reuns Oct 14 at 0:13
• Very clever - thanks – user714524 Oct 16 at 21:20

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.
• @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. – Travis Willse Oct 13 at 21:44