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This sum $\sum_{n=2}^{+\infty}\frac{(-1)^n}{\zeta(n)}$ gives from $n=2$ to odd integer negative value which is close to $-0.27,..$ and gives $0.72 ...$ to even integer, This analysis mixed me to predict the exact value of that sum, probably that is not exists , Now my question here is : What's this :$$\sum_{n=2}^{+\infty}\frac{(-1)^n}{\zeta(n)}$$ equal ?

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  • $\begingroup$ Why would you expect it to have a closed or particularly nice form? The even terms might have a convenient form, at least. $\endgroup$ – anomaly Jul 13 '18 at 2:56
  • $\begingroup$ Man lives with his dreams! $\endgroup$ – Nosrati Jul 13 '18 at 2:59
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    $\begingroup$ @anomaly They say they expect it doesn't exist. (Though they could certainly have been clearer on this point in how they phrased the question.) $\endgroup$ – spaceisdarkgreen Jul 13 '18 at 3:07
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    $\begingroup$ The series is divergent but may be regularized (say by searching the limit as $\,x\to 1,\;x<1\;$ of $\displaystyle\;f(x):=\sum_{n=2}^{+\infty}\frac{(-x)^n}{\zeta(n)}\;$) to get approximatively $0.22408327940177299230992503989177309601$. $\endgroup$ – Raymond Manzoni Jul 14 '18 at 7:35
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Since $\lim_{n\to\infty}\zeta(n) = 1$, we have that the general term $$ \frac{(-1)^n}{\zeta(n)} $$ does not converge to zero (and does not converge at all, period). Thus the series $\sum_n \frac{(-1)^n}{\zeta(n)}$ diverges by the term test.

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  • $\begingroup$ It might be more interesting to look at $\lim \sup$ and $\lim\inf$. $\endgroup$ – Szeto Jul 13 '18 at 9:07

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