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It is pretty well-known that $\sum_{k = 1}^{\infty} \frac {1} {k^2} = \frac {\pi^2} {6}$. I am interested in evaluating the partial sum $\sum_{k = 1}^{N} \frac {1} {k^2}$. Here is what I have done so far. Since we have

$$\sum_{k = 1}^{N} \frac {1} {k^2} = 1 + \sum_{k = 2}^{N} \frac {1} {k^2} < 1 + \sum_{k = 2}^{N} \frac {1} {k^2 - 1} =\\= 1 + \frac {1} {2} \left ( \sum_{k = 2}^{N} \frac {1} {k - 1} - \sum_{k = 2}^{N} \frac {1} {k + 1} \right ) = 1 + \frac {1} {2} \left ( \frac {3} {2} - \frac {2N + 1} {N^2 + N} \right ) = \frac {7} {4} - \frac {2N + 1} {2 N^2 + 2 N}$$ and

$$\sum_{k = 2}^{N} \frac {1} {k^2 - 1} - \sum_{k = 2}^{N} \frac {1} {k^2} < \int_{2}^{N} \left ( \frac {1} {x^2 - 1} - \frac {1} {x^2} \right ) \textrm {d}x = \int_{2}^{N} \frac {\textrm {d}x} {x^2 - 1} - \int_{2}^{N} \frac {\textrm {d}x} {x^2} < \frac {1} {N^2 - N} - \frac {1} {2} + \log \sqrt {3},$$

we have

$$\frac {7} {4} + \frac {1} {2} - \log \sqrt {3} - \frac {2 N^2 + N + 1} {2N^3 - 2N} < \sum_{k = 1}^{N} \frac {1} {k^2} < \frac {7} {4} - \frac {2N + 1} {2 N^2 + 2 N}$$

But how to obtain a more precise expression without using analytic methods such as Euler summation?

Masacroso
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    Did you try the Euler–Maclaurin summation formula ? –  Nov 24 '15 at 16:11
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    But you have used already an integral yourself. Euler's summation formula also has not more analytic tools than an integral. – Dietrich Burde Nov 24 '15 at 17:35
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    Using the same trick $$\frac1k-\frac1{k+1}=\frac1{k(k+1)}\leqslant\frac1{k^2}\leqslant\frac1{k(k-1)}=\frac1{k-1}-\frac1k$$ one gets $$\frac1{n+1}\leqslant\sum_{k=n+1}^\infty\frac1{k^2}\leqslant\frac1n$$ hence $$\frac{\pi^2}6-\frac1{n}\leqslant\sum_{k=1}^n\frac1{k^2}\leqslant\frac{\pi^2}6-\frac1{n+1}$$ – Did Oct 23 '16 at 18:35

2 Answers2

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A quick fix on the upper bound is possible.

Just change everything from $\frac{1}{k^2-1}$ to $\frac{1}{k^2-\frac{1}{4}}$.

Then, we have $$ \sum_{k=1}^N \frac{1}{k^2} = 1+ \sum_{k=2}^N \frac{1}{k^2} < 1+\sum_{k=2}^N \frac{1}{k^2-\frac{1}{4}} = 1+\sum_{k=2}^N \frac{1}{k-\frac{1}{2}} - \sum_{k=2}^N \frac{1}{k+\frac{1}{2}} = 1+\frac{2}{3}-\frac{1}{N+\frac{1}{2}} = \frac{5}{3}-\frac{2}{2N+1}$$

Therefore, we have $$\sum_{k=1}^N \frac{1}{k^2} < \frac{5}{3}-\frac{2}{2N+1}$$

The upper bound should be more tight since $$\frac{1}{k^2}<\frac{1}{k^2-\frac{1}{4}}< \frac{1}{k^2-1}$$

rkm0959
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The Euler summation formula gives, for all $x\ge 1$, $$ \sum_{n\le x}\frac{1}{n^2}=\frac{\pi^2}{6}-\frac{1}{x}+O(x^{-2}), $$ by taking $f(t)=t^{-2},\, f'(t)=-2t^{-3}$.

Dietrich Burde
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