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?

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: $$0<\sum_{n\mathop=2}^{2014}\zeta(n)-1<1$$ $$2013<\sum_{n\mathop=2}^{2014}\zeta(n)<2014$$ $${\sum_{n\mathop=2}^{2014}\zeta(n)}\approx 2014$$

$$\eqalign{\sum_{j=2}^N \zeta(j) &=\sum_{n=1}^\infty \sum_{j=2}^N \dfrac{1}{n^j}\cr &= N-1 + \sum_{n=2}^\infty \dfrac{n^{-1} - n^{-N}}{n-1}\cr &= N - 1 + \sum_{n=2}^\infty \dfrac{1}{n(n-1)} - \sum_{n=2}^\infty \dfrac{n^{-N}}{n-1}} $$ 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

$$N > \sum_{j=2}^N \zeta(j) > N - 2^{1-N} $$