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In this answer it is stated that $$ \sum_{n\geq1}\frac{1}{n^2}=\sum_{n\leq x}\frac1{n^2}+\mathcal O(1/x). $$ Is this statement true as $x\to\infty$?

What I've done: If $x$ is fixed, then I think the answer is almost trivial, because we may set $C=\pi^2x/6$, so $$ \sum_{n=x}^\infty\frac1{n^2}\leq\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}{6}=\frac{C}{x}, $$ therefore $$ \sum_{n\geq1}\frac1{n^2}=\sum_{n\leq x}\frac{1}{n^2}+\sum_{n=x}^\infty\frac{1}{n^2}\leq\sum_{n\leq x}\frac{1}{n^2}+C/x=\sum_{n\leq x}\frac{1}{n^2}+\mathcal O(1/x). $$ But is there a constant independent of $x$ that makes this true?

Bill Dubuque
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nfhplrpd
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1 Answers1

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Euler-Maclaurin Sum Formula

What is being referred to here is that $$ \sum_{k=n+1}^\infty\frac1{k^2}=\frac1n-\frac1{2n^2}+\frac1{6n^3}+O\!\left(\frac1{n^5}\right) $$ This can be gotten using the Euler-Maclaurin Sum Formula. Thus, the constant is $1$.

Integral Estimate for the Tail of the Series

We also have the basic integral estimate for the tail of a monotonically decreasing series. This is based on the fact that $$ \frac1{k^2}\le\int_{k-1}^k\frac1{x^2}\,\mathrm{d}x $$ Summing this inequality yields $$ \begin{align} \sum_{k=n+1}^\infty\frac1{k^2} &\le\int_n^\infty\frac1{x^2}\,\mathrm{d}x\\ &=\frac1n \end{align} $$

Telescoping Series

Using the simple inequality $$ \frac1{k^2}\le\frac1{k-1}-\frac1k $$ we can sum, using the telescoping series, to get $$ \sum_{k=n+1}^\infty\frac1{k^2}\le\frac1n $$

Example $$ \sum_{k=1}^{1000}\frac1{k^2}=1.6439345667 $$ whereas $$ \frac{\pi^2}6=1.6449340668 $$ The error is $0.0009995001\lt\frac1{1000}$.

robjohn
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  • Thanks for the answer -- is there a more elementary way to show that $\sum^\infty1/n^2=\sum_{n\leq x}1/n^2+\mathcal O(1/x)$ without using the Euler-Maclaurin Sum Formula? – nfhplrpd Mar 19 '21 at 07:39
  • The Euler-Maclaurin Sum Formula is essentially based in the integral approximation for the tail of the sum. I will add that to my answer. – robjohn Mar 19 '21 at 07:46
  • I have added a third, and simpler, telescoping approach. – robjohn Mar 19 '21 at 07:58
  • The second and third approaches are very clean and illuminating. Especially, the inequality in the third approach looks very clever. Thank you! – nfhplrpd Mar 19 '21 at 07:59