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

Question

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 ?

Answer 1

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.