# What's this $\sum_{n=2}^{+\infty}\frac{(-1)^n}{\zeta(n)}$ equal?

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 ?

• Why would you expect it to have a closed or particularly nice form? The even terms might have a convenient form, at least. – anomaly Jul 13 '18 at 2:56
• Man lives with his dreams! – Nosrati Jul 13 '18 at 2:59
• @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.) – spaceisdarkgreen Jul 13 '18 at 3:07
• 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 $0.22408327940177299230992503989177309601$. – Raymond Manzoni Jul 14 '18 at 7:35

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