Does the sum of the Zeta function taken on natural numbers converge?

Question

Does this series \begin{equation*} \sum_{n\ge2} \zeta(n) \end{equation*} converge? If yes is it easy to prove ?

Answer 1

$\zeta(n)>1$ for all $n>1$, so no.

Answer 2

$\zeta(n)=1+\frac{1}{2^n}+\frac{1}{3^n}\cdots>1$, so the sum of zeta diverges, in particular because of the first $1$ term. On the other hand there's a very cute modification:

$$\sum_{n=2}^\infty(\zeta(n)-1)=1,$$

and generalizations, I believe due to Borwein, Bradley and Crandall.

Answer 3

As $Z(1)$ is infinite, the sum will be infinite as well.

Taking the sum from $n=2$, consider $$ \lim_{n\to\infty}\frac{Z(n)}{1}. $$ We have $$ \lim_{n\to\infty}\frac{Z(n)}{1}=\lim_{n\to\infty}\frac{1+\frac{1}{2^n}+\dots}{1}=1. $$ By the limit comparison test, http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx, this means that the Zeta series converges if and only if the sum $\sum_{n=1}^\infty 1$ converges. Hence the series diverges.