<s>[INST] Which of the two sums gives a better approximation to $\pi^2/6$? Which of the following two sums 
$$\sum_{n=1}^{1000000}\frac1{n^2} \quad \text{or} \quad 1+\sum_{n=1}^{1000}\frac1{n^2(n+1)}$$
gives a better approximation to $\pi^2/6?$
I tested this on MATLAB and surprisingly obtained as a result the second sum. However, I do not know how to prove this rigorously.
 [/INST] $$\eqalign{\sum_{k+1}^\infty\frac{1}{n^2(n+1)}&<\sum_{k+1}^\infty\frac{1}{(n-1)n(n+1)}\cr
&=\frac12\sum_{k+1}^\infty\left(\frac{1}{(n-1)n}-\frac{1}{n(n+1)}\right)\cr
&=\frac{1}{2k(k+1)}}$$
$$\eqalign{\sum_{k^2+1}^\infty\frac{1}{n^2}&>\sum_{k^2+1}^\infty\frac{1}{n(n+1)}\cr
&=\sum_{k^2+1}^\infty\left(\frac{1}{n}-\frac{1}{n+1}\right)\cr
&=\frac{1}{k^2+1}}$$
So, clearly the error in the second is smaller than half the error in the first, that is
$$\frac{\pi^2}{6}-\left(1+\sum_{n=1}^k\frac{1}{n^2(n+1)}\right)<\frac{1}{2}\left(\frac{\pi^2}{6}-\sum_{n=1}^{k^2}\frac{1}{n^2}\right)$$
The problem corresponds to $k=1000$.
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