# Convergence or divergence of series

Does this series $$\sum_{k=2}^{\infty}\dfrac{\zeta{(k+1)}}{\zeta{(k)}}$$ converge or diverge?

Here $$\zeta{(k)}=\dfrac{1}{1^k}+\dfrac{1}{2^k}+\dfrac{1}{3^k}+\cdots+\dfrac{1}{n^k}+\cdots$$ If it converges, how to find the value?

If it diverges, how to prove it?

Thank you

Discussion for a while, and I have edited to a new problem.

How to find the result of this sum $$\sum_{k=2}^{\infty}\left(\dfrac{\zeta{(k+1)}}{\zeta{(k)}}-1\right)$$

-
Did you think about it one minute? Hint: when $k\to\infty$, $\zeta(k)\to$ $____$, hence... – Did Aug 4 '13 at 10:15
Already with $\,k=1\,$ you have big problems since $\,\zeta(1)\;$ diverges...so what is the first summand in your series?! – DonAntonio Aug 4 '13 at 10:18
The $k=1$ summand can be saved by setting $\zeta(2)/\zeta(1)=0$ since $\zeta(2)$ is finite and $\zeta(1)$ diverges. – Did Aug 4 '13 at 10:22
Perhaps a more interesting question would be $$\sum\left({\zeta(k+1)\over\zeta(k)}-1\right)$$ – Gerry Myerson Aug 4 '13 at 10:41
You're asking me, how do I prove that my question is more interesting than yours? Well, it can hardly be any less interesting. – Gerry Myerson Aug 4 '13 at 10:44

Consider the sum $$\sum_{n=2}^\infty\sum_{k=2}^\infty\frac1{n^k}=\sum_{n=2}^\infty\frac{n^{-2}}{1-n^{-1}}=\sum_{n=2}^\infty\frac{1}{n(n-1)}=1.$$ As the above sum is absolutely convergent,$$1=\sum_{n=2}^\infty\sum_{k=2}^\infty\frac1{n^k}=\sum_{k=2}^\infty\sum_{n=2}^\infty\frac1{n^k}=\sum_{k=2}^\infty(\zeta(k)-1).$$ Hence,$$0<\sum_{k=2}^{\infty}\left(\dfrac{\zeta{(k)}}{\zeta{(k+1)}}-1\right)<\sum_{k=2}^\infty(\zeta(k)-1)=1.$$