Series with $\zeta$ 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?
 A: Note that $\zeta(n)-1>0 $ for $n>1$.
First we need to show that:
$$\sum_{n\mathop=2}^{\infty} \zeta(n)-1=1$$

Proof for $ \sum_{n=2}^{\infty} (\zeta(n)-1) = 1$:
Use $ \displaystyle\zeta(n) - 1 =  \sum_{s=1}^{\infty} \frac{1}{s^n} - 1 = \sum_{s=2}^{\infty} \frac{1}{s^n}$ to get
$ \displaystyle\sum_{n=2}^{\infty} (\zeta(n)-1) = \displaystyle \sum_{n=2}^{\infty} \sum_{s=2}^{\infty} \frac{1}{s^n} $
$$ =  \sum_{s=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{s^n}= \sum_{s=2}^{\infty} \frac{1}{s(s-1)}= \sum_{s=2}^{\infty} \bigg(\frac{1}{s-1}-\frac{1}{s}\bigg)=1$$

Therefore:
$$0<\sum_{n\mathop=2}^{2014}\zeta(n)-1<1$$
$$2013<\sum_{n\mathop=2}^{2014}\zeta(n)<2014$$
$${\sum_{n\mathop=2}^{2014}\zeta(n)}\approx 2014$$
A: $$\eqalign{\sum_{j=2}^N \zeta(j) &=\sum_{n=1}^\infty \sum_{j=2}^N \dfrac{1}{n^j}\cr  &= N-1 + \sum_{n=2}^\infty  \dfrac{n^{-1} - n^{-N}}{n-1}\cr
&= N - 1 + \sum_{n=2}^\infty \dfrac{1}{n(n-1)} - \sum_{n=2}^\infty \dfrac{n^{-N}}{n-1}} $$
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
$$N > \sum_{j=2}^N \zeta(j) > N - 2^{1-N} $$
