Accuracy from approximating $\zeta(2)$ with a partial sum This is for an introductory numerical analysis class. The answer shouldn't be too complicated, but if you have one, feel free to post it.
Figure out what $n$ should be such that
$$\sum_{k=n+1}^\infty {1\over k^2}<10^{-8}.$$

My Simple Algebraic Attempt
We know that 
$$\sum_{k=1}^\infty {1\over k^2} = \sum_{k=1}^n {1\over k^2} + \sum_{k=n+1}^\infty {1 \over k^2} \implies \sum_{k=n+1}^\infty {1 \over k^2} = \sum_{k=1}^\infty {1\over k^2} - \sum_{k=1}^n {1\over k^2}$$
And
$$\sum_{k=1}^\infty {1\over k^2} = \zeta(2)={\pi^2 \over6}$$
So,
$$\sum_{k=n+1}^\infty {1 \over k^2} = {\pi^2 \over6} - \sum_{k=1}^n {1\over k^2}$$
Then, 
$$\sum_{k=n+1}^\infty {1 \over k^2} < 10^{-8} \implies {\pi^2 \over6} - \sum_{k=1}^n {1\over k^2} < 10^{-8} \implies \sum_{k=1}^n {1\over k^2} > {\pi^2 \over6} - 10^{-8}$$
I'm currently trying to brute force an answer to the last expression, but is there a better way to do this?
 A: Since $\displaystyle \frac{1}{k^2} < \frac{1}{(k-1)k } = \frac{1}{k-1} - \frac{1}{k}$ we can bound your term by a telescoping sum: $$\sum_{k=n+1}^{\infty} \frac{1}{k^2} < \left(\frac{1}{n} - \frac{1}{n+1} \right)+\left(\frac{1}{n+1} - \frac{1}{n+2} \right)+\left(\frac{1}{n+2} - \frac{1}{n+3} \right) + \cdots = \frac{1}{n}$$ so $n=10^8$ works. The estimate we used isn't too weak, and this $n$ shouldn't be much worse than the minimal $n.$
A: Hint:
$$
\sum_{k=n+1}^\infty\frac1{k^2}\le\int_n^\infty\frac1{x^2}\mathrm{d}x=\frac1n
$$
A: A simple idea is to replace the remainder of the series by the corresponding integral :
$$\sum_{k=N+1}^\infty \frac 1{k^2}\sim \int_{N+1}^\infty \frac {dk}{k^2}$$
or get the inequality :
$$\sum_{k=N+1}^\infty \frac 1{k^2}<  \int_N^\infty \frac {dk}{k^2}=\frac 1N$$
If you wish more precision you may use Euler Maclaurin formula and get (for $x=2$ in your case) :
$$\zeta(x)-\sum_{k=1}^N \frac 1{k^x} \sim   
               \frac 1{(x-1)N^{x-1}}-\frac 1{2\,N^x}
               + \frac x{12\,N^{x+1}}
               - \frac {x\,(x+1)(x+2)}{720\,N^{x+3}}
               +\frac {x\,(x+1)(x+2)(x+3)(x+4)}{30240\,N^{x+5}}
               +...$$
(the constants $\ \frac 1{2},\ \frac 1{12},\ \frac 1{720}$ are $\frac {|B_n|}{n!}$ with $B_n$ a Bernoulli number)
This is an asymptotic expansion and the error made corresponds to the first term omitted.
