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?

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 $n.$
