Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$n$ is a positive integer, then


please don't refer to the famous $1+\frac1{2^2}+\frac1{3^2}+\dotsb=\frac{\pi^2}6$.

I want to find a better proof.

My stupid method:

$$1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt (1+\frac1{2^2}+\dotsb+\frac1{10^2})+\frac1{10\cdot11}+\dotsb+\frac1{n(n-1)}\\<1.549768...+\frac1{10}\lt\frac53$$

share|improve this question

3 Answers 3

up vote 74 down vote accepted

I suggest

$$\sum_{k=1}^n \frac{1}{k^2} \leqslant 1 + \sum_{k=2}^n \frac{1}{k^2 - \frac14} = 1 + \sum_{k=2}^n \left(\frac{1}{k-\frac12} - \frac{1}{k+\frac12}\right) = 1 + \frac23 - \frac{1}{n+\frac12}.$$

share|improve this answer

We show by induction that if $n\gt 1$ then $$1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{n^2}\lt \frac{5}{3}-\frac{2}{2n+1}.$$ The result is true at $n=2$. Suppose that the result holds at $n=k$. We show it holds at $n=k+1$.

By the induction assumption, $$1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{k^2}+\frac{1}{(k+1)^2}\lt \frac{5}{3}-\frac{2}{2k+1}+\frac{1}{(k+1)^2}.\tag{1}$$ The right-hand side of (1) is equal to $$\frac{5}{3}-\left(\frac{2}{2k+1}-\frac{1}{(k+1)^2}\right).$$ Now we need to show that $\frac{2}{2k+1}-\frac{1}{(k+1)^2}\gt \frac{1}{2k+3}$ or equivalently that $\frac{4}{(2k+1)(2k+3)}\gt \frac{1}{(k+1)^2}$. So we show that $4(k+1)^2\gt (2k+1)(2k+3)$. This is straightforward.

share|improve this answer
@Daniel Fischer: Thanks for the correction! –  André Nicolas Dec 31 '13 at 16:42
Can I ask how that term ($\frac{2}{2n+1}$) reached to your mind in order to strength your Induction hypothesis ? –  Fardad Pouran Jun 7 '14 at 13:01
Nothing as nice as the (identical) choice by Daniel Fischer. A glance at the series shows how the error term behaves, so I looked for a suitable $\frac{1}{2n+ a}$. I think! It was a while ago. –  André Nicolas Jun 7 '14 at 14:39
Thank you so much –  Fardad Pouran Jun 7 '14 at 15:29

I don't know if this is "better," but since $1/x^2$ is strictly decreasing, we have

$$\sum_{n=N+1}^\infty{1\over n^2}\lt\int_N^\infty{1\over x^2}\,dx={1\over N}$$

after which a little systematic trial and error gives

$$\sum_{n=1}^\infty{1\over n^2}\lt 1+{1\over4}+{1\over9}+{1\over16}+{1\over25}+{1\over5}={5989\over3600}\lt{6000\over3600}={5\over3}$$

Added later: It occurs to me that my approach is not inherently different from the OP's "stupid" method. We each basically argue that

$$\sum_{n=1}^\infty{1\over n^2}\lt1+{1\over4}+\cdots+{1\over N^2}+{1\over N}$$

for any $N$. If there's anything stupid in the OP's solution (which there isn't, really), it's just that it's not necessary to go all the way out to $N=10$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.