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I'm asking because numerical tests seem to give nonsensical answers, and I thought I would check if there was an analytic way of checking for divergence, but I couldn't think of one offhand.

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up vote 14 down vote accepted

$(s-1)\zeta(s) = 1 + a(s-1) + b(s-1)^2 + \dots$ is analytic near $1$ (in fact entire, but we don't need that for this problem).

For $s=1+1/n$ this gives $\zeta(1+1/n)=n + a + b/n + \dots =n + O(1)$, so the sum diverges.

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You could use an integral comparison to get a bound on $\zeta(1+1/n)$:

$$\zeta(1+1/n)\lt 1+\int_1^\infty x^{-1-\frac{1}{n}}=1+n.$$

More generally, if $0\lt a\lt 1$, then

$$\frac{1}{a}=\int_1^\infty x^{-1-a}\lt\zeta(1+a)\lt1+\int_1^\infty x^{-1-a}=1+\frac{1}{a}<\frac{2}{a}.$$

Thus if $a_1,a_2,\ldots$ is a sequence of positive numbers converging to $0$, then $\sum_{n=1}^{\infty} \frac{1}{\zeta(1+a_n)}$ converges if and only if $\sum_{n=1}^\infty a_n$ does.

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