I need to prove that the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am having some trouble:


The sequence seems to be monotone and increasing. This can be proved by induction: Claim that $a_n\leq a_{n+1}$ $$a_1=1\leq 1+\frac{1}{2^2}=a_2$$

Need to show that $a_{n+1}\leq a_{n+2}$ $$a_{n+1}=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}\leq 1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}=a_{n+2}$$ Thus the sequence is monotone and increasing.


Since the sequence is increasing it is bounded below by $a_1=1$. Upper bound is where I am having trouble. All the examples I have dealt with in class have to do with decreasing functions, but I don't know what my thinking process should be to find an upper bound.

Can anyone enlighten me as to how I should approach this, and can anyone confirm my work thus far? Also, although I prove this using monotonicity and boundedness, could I have approached this by showing the sequence was a Cauchy sequence?

Thanks so much in advance!

  • 2
    $\begingroup$ Hint: compare it to another sequence with only powers of two in the denominator. $\endgroup$ – user2055 Mar 18 '13 at 2:58
  • $\begingroup$ @Jason: Could you please elaborate on your hint? I don't understand it. $\endgroup$ – Jonas Meyer Mar 18 '13 at 3:08
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    $\begingroup$ @JonasMeyer: The sequence is less than $1 + 1/2^2 + 1/2^2 + 1/4^2 + 1/4^2 + 1/4^2 + 1/4^2 + ... \leq 2$ $\endgroup$ – user2055 Mar 18 '13 at 3:13
  • $\begingroup$ @ShuXiaoLi my objection was to the way the comment was presented, i.e., like a solution to the question. A comment like "by the way, the sum is $\pi^2/6$" would have been fine. $\endgroup$ – Ittay Weiss Mar 18 '13 at 3:17
  • $\begingroup$ A similar question: Does $\sum\limits_{k=1}^n 1 / k ^ 2$ converge when $n\rightarrow\infty$? $\endgroup$ – Martin Sleziak Mar 18 '13 at 9:21

Your work looks good so far. Here is a hint: $$ \frac{1}{n^2} \le \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n} $$

To elaborate, apply the hint to get: $$ \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \le \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n-1} - \frac{1}{n}\right) $$

Notice that we had to omit the term $1$ because the inequality in the hint is only applicable when $n > 1$. No problem; we will add it later.

Also notice that all terms on the right-hand side cancel out except for the first and last one. Thus: $$ \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \le 1 - \frac{1}{n} $$

Add $1$ to both sides to get: $$ a_n \le 2 - \frac{1}{n} \le 2 $$

It follows that $a_n$ is bounded from above and hence convergent.

It is worth noting that canceling behavior we saw here is called telescoping. Check out the wikipedia article for more examples.

  • $\begingroup$ Sorry, I'm afraid I'm going to need a bit more of a push. I can't think of how/where to use your hint. Am I supposed to say that $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+{…}+\frac{1}{n^2}\leq 1+\frac{1}{2^2}+\frac{1}{3^2}+{…}+\frac{1}{n-1}-\frac{1}{n}$ and do something further with that? $\endgroup$ – user66807 Mar 18 '13 at 4:02
  • $\begingroup$ try to apply it more than once. $\endgroup$ – chx Mar 18 '13 at 4:17
  • $\begingroup$ @chx Try to apply it more than once in the "thing" I supposed? $\endgroup$ – user66807 Mar 18 '13 at 4:22
  • $\begingroup$ @user66807, we have $a_n=\sum_{k=2}^{n}\frac{1}{k^2}\le b_n=\sum_{k=2}^{n}\frac{1}{k-1}-\frac{1}{k}$, try to find the first few terms of $b_n$ you will see a pattern (things will start cancelling) then you can find an upper bound of $a_n$. Notice i started the sum from 2 to be able to use Ayman's hint. $\endgroup$ – i.a.m Mar 18 '13 at 4:32
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    $\begingroup$ @user66807, try to sum a few terms of the telescopic series $\,\sum\left(\frac{1}{n-1}-\frac{1}{n}\right)\,$ and see what happens... $\endgroup$ – DonAntonio Mar 18 '13 at 4:33

Besides to Ayman's neat answer, you may take $f(x)=\frac{1}{x^2}$ over $[1,+\infty)$ and see that $f'(x)=-2x^{-3}$ and then is decreasing over $[1,+\infty)$. $f(x)$ is also positive and continuous so you can use the integral test for $$\sum_{k=1}^{+\infty}\frac{1}{n^2}$$ to see the series is convergent. Now your $a_n$ is the $n-$th summation of this series.

  • $\begingroup$ The problem is we haven't gone over series in class yet, but thanks anyways. $\endgroup$ – user66807 Mar 18 '13 at 3:41
  • $\begingroup$ @user66807: Sorry, I didn't know that. So the Ayman's is more preferable nice and simple approach for you. :-) $\endgroup$ – mrs Mar 18 '13 at 3:43
  • $\begingroup$ I hope so, let's just see if I can catch his hint. And thanks. $\endgroup$ – user66807 Mar 18 '13 at 3:46
  • $\begingroup$ +1 This is a good answer, though the OP can't use it yet. $\endgroup$ – DonAntonio Mar 18 '13 at 4:34
  • $\begingroup$ I agree with DonAntonio that this is a good answer! $+1 \;\ddot\smile\;\;$ (I also like the new gravatar!) Good day, to you! $\endgroup$ – amWhy Mar 18 '13 at 15:08

You can show this geometrically too.

If you take a square, and divide the height into $\frac 12$, $\frac 14$, $\frac 18$, and so forth, doubling the denominator on each deal.

Now take the squares $\frac 12$ and $\frac 13$: these go onto the top shelf.

On the second shelf go $\frac 14$ to $\frac 17$. These fractions are all less than $\frac 14$, so fit onto the second shelf. Likewise, squares from $8$ to $15$ go onto the third shelf, each $\frac 1n$ is smaller than $\frac 18$, and so forth.

Therefore $\sum_{n=2}^{\infty}\frac 1n < 1$, and therefore the whole lot is less than two squares.

  • $\begingroup$ This answer also provides a picture for this geometric argument. $\endgroup$ – Martin Sleziak Sep 28 '15 at 9:51

by the integral test :

$\int^\infty_1 \frac{1}{n^2} dn \le \sum_{i=1}^\infty \frac{1}{n^2}\le 1+\int^\infty_1 \frac{1}{n^2}dn$ .

you can compute the integral so the answer is :

$1 \le \sum_{i=1}^\infty \frac{1}{n^2}\le 2$ .

because : $\int^\infty_1 \frac{1}{n^2} dn =1 $


Hint: Prove the the following holds for all $n$ by induction.

$$\sum_1^n \frac{1}{k^2} \le 2 - \frac{1}{n}.$$

Is it not uncommon that when proving some inequality by induction, you will first need to strengthen the hypothesis to get the induction to work.

You can use the same technique to bound other values of the zeta function. For example, try showing $\zeta(3)$ is bounded above by $\frac{3}{2}$.

(I am copying my answer from a duplicate question that was closed as a copy of this one since the induction approach is not available here.)


Notice that $ 2k^2 \geq k(k+1) \implies \frac{1}{k^2} \leq \frac{2}{k(k+1)}$.

$$ \sum_{k=1}^{\infty} \frac{2}{k(k+1)} = \frac{2}{1 \times 2} + \frac{2}{2 \times 3} + \frac{2}{3 \times 4} + \ldots $$

$$ \sum_{k=1}^{\infty} \frac{2}{k(k+1)} = 2\Big(\, \Big(1 - \frac{1}{2}\Big) + \Big(\frac{1}{2} - \frac{1}{3} \Big) + \Big(\frac{1}{3} - \frac{1}{4} \Big) + \ldots \Big)$$

$$ \sum_{k=1}^{\infty} \frac{2}{k(k+1)} = 2 (1) = 2 $$.

Therefore $ \sum_{k=1}^{\infty} \frac{1}{k^2} \leq 2$.


$a_n=\frac{1}{n^2}$ and because $a_n>0$ we have $|a_n|=a_n$
First: check the necessary condition
$$\lim_{n\to +\infty} na_n=\lim_{n\to +\infty}\frac{1}{n}=0$$
Second: check D'Alembert's ratio test
$$\lim_{n\to +\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to +\infty}(\frac{n}{n+1})^2=1$$
Third: Because the answer of D'Alembert's test is $1$, you should use Raabe's test:
$$\lim_{n\to +\infty}n(1-|\frac{a_{n+1}}{a_n}|)=\lim_{n\to +\infty}n(1-\frac{a_{n+1}}{a_n})=\lim_{n\to +\infty}n(\frac{2n+1}{n^2+2n+1})=2\gt1$$
so the series is convergent


This here should work with $n \geq 1 $ :

$$s_n = \sum\limits_{n=1}^\infty \frac{1}{n^2}= \frac{1}{1} + \frac{1}{4} +\frac{1}{9} + \frac{1}{16}+ ...+ \frac{1}{n²} $$$$ b_n = \sum\limits_{n=1}^\infty \frac{1}{2^{n-1}} = \frac{1}{1} + \frac{1}{2} +\frac{1}{4}+\frac{1}{8} + ...+ \frac{1}{2^{n-1}} $$

$b_n$ is directly compared greater than $s_n$ : $$s_n < b_n$$ and $b_n$ converges, because of its ratio test : $$\frac{1}{2^{n-1+1}} / \frac{1}{2^{n-1}} = \frac{2^{n-1}}{2^{n}} = \frac{1}{2} < 1$$

  • $\begingroup$ That proof is false. $2^{n-1}>n^2$ once $n\geq 7$ and thus $s_n>b_n$. $\endgroup$ – user35953 Jan 20 '18 at 13:00
  • $\begingroup$ $\frac{1}{s_n}$ < $s_n$ $\endgroup$ – user1511417 Jan 22 '18 at 10:54
  • $\begingroup$ It's true that $s_n < b_n$ for $n > 1$, but the direct comparison test would require that eventually $\frac{1}{n^2} < \frac{1}{2^{n-1}}$, which doesn't hold. Proving that indeed $s_n \leqslant b_n$, or just $s_n < 2$ amounts to proving the convergence of the series in question. $\endgroup$ – Daniel Fischer Jan 22 '18 at 11:21

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