TITLE: Series with $\zeta$
QUESTION [10 upvotes]: How do I calculate the following series:
$$ \zeta(2)+\zeta(3)+\zeta(4)+ \dots + \zeta(2013) + \zeta(2014) $$
All I know is that $\zeta(2)=\pi^2/6$ and $\zeta(4)=\pi^4/90$. But this is not enough to solve this problem. How do I do this?

REPLY [6 votes]: $$\eqalign{\sum_{j=2}^N \zeta(j) &=\sum_{n=1}^\infty \sum_{j=2}^N \dfrac{1}{n^j}\cr  &= N-1 + \sum_{n=2}^\infty  \dfrac{n^{-1} - n^{-N}}{n-1}\cr
&= N - 1 + \sum_{n=2}^\infty \dfrac{1}{n(n-1)} - \sum_{n=2}^\infty \dfrac{n^{-N}}{n-1}} $$
Now $$\sum_{n=2}^\infty \dfrac{1}{n(n-1)} = \sum_{n=2}^\infty \left(\dfrac{1}{n-1} - \dfrac{1}{n}\right) = 1$$
while
$$0 < \sum_{n=2}^\infty \dfrac{n^{-N}}{n-1} = \sum_{n=2}^\infty \dfrac{n^{1-N}}{n(n-1)} < 2^{1-N} \sum_{n=2}^\infty \dfrac{1}{n(n-1)} = 2^{1-N}$$
so that
$$N > \sum_{j=2}^N \zeta(j) > N - 2^{1-N} $$