I know there are lots of proofs about this, however I wanted to come up with my own proof using what I learnt in class and in this website.


It is easy to see $A_n$ is an increasing function, since $A_{n+1}-A_n=\frac{1}{n}\geq 0, \forall n$

Now I prove $A_n$ is bounded, $A_{2n}-A_{n}=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+\frac{1}{(2n)^2}\leq \frac{1}{n^2}+\frac{1}{n^2}+\dots+\frac{1}{n^2}=\frac{1}{n}$.

Supose it $\exists M$ so that $\forall n, A_n\leq M$. Then, when $A_n$ tends to $M$,

$M-M\leq 1/n \implies 0\leq \frac{1}{n}$, which is true, and therefore $M$ exists and $A_n$ is Cauchy.

Is this correct?

  • $\begingroup$ $\frac{1}{n^2} \ne \frac{1}{n}$ $\endgroup$ – Travis Jan 30 '17 at 13:38
  • $\begingroup$ You assume it is bounded, then show that under this assumption this leads to the conclusion it is bounded. Even if there were no errors how does this prove anything? $\endgroup$ – Ahmed S. Attaalla Jan 31 '17 at 0:46

This part of your proof is flawed:

Now I prove $A_n$ is bounded, $A_{2n}-A_{n}=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+\frac{1}{(2n)^2}\leq \frac{1}{n^2}+\frac{1}{n^2}+\dots+\frac{1}{n^2}=\frac{1}{n}$.

Supose it $\exists M$ so that $\forall n, A_n\leq M$. Then, when $A_n$ tends to $M$,

$M-M\leq 1/n \implies 0\leq \frac{1}{n}$, which is true, and therefore $M$ exists.

First you assume that the sum is bounded, which means exactly that such $M$ exists. Using this assumaption, you deduce a true statement. However this does not allow you to conclude that the assumption is correct.
For example, if we assume that $\sqrt{2}$ is rational, then we can conclude that $\sqrt{2}*\sqrt{2}=2$ is also rational (which is true), but the assumption is still wrong.

You probably confused this with a proof by contradiction. A proof by contradiction would be of the following form:
Assume $M$ is not bounded. Using this assumption we deduce a statement which contradicts some facts from which you certainly know that they are true. As you showed before that the assumtion ($M$ not bounded) cannot hold simultaneously with the fact, but the fact is certainly true, the assumption must be false. But then the negation of the assumption has to hold (excluded middle), hence $M$ is bounded.

A possible proof could be this:

We want to show that the sequence $(x_k)_{k\in \mathbb N}$ defined by $x_n=\sum_{k=1}^n \frac 1 {k^2}$ is bounded.

For any $k\in \mathbb N$ we have $$\left \vert x_k\right \vert = \left \vert \sum_{n=1}^k \frac 1 {n^2} \right \vert = \sum_{n=1}^k \frac 1 {n^2}.$$ As for any $n \ge 2$, $$\frac 1 {n^2} < \frac 1 {n (n-1)}=\frac 1 {n-1} - \frac 1 n ,$$ $|x_k|$ can be estimated by $$|x_k| =1+\sum_{n=2}^k \frac 1 {n^2} <1+\sum_{n=2}^k \frac 1 {n-1} -\sum_{n=2}^k \frac 1 {n}= 1+\sum_{n=1}^{k-1} \frac 1 {n} -\sum_{n=2}^k \frac 1 {n}= 2 - \frac 1 k<2. $$ Hence if we set $S=2$, then for any $k\in \mathbb N$ the member $x_k<S$, i.e. $(x_k)_{k\in \mathbb N}$ is bounded.


Sorry, it's wrong in many ways.

You write

Suppose there exists $M$ such that, for all $n$, $A_n\le M$

This is exactly assuming the sequence is bounded, which you have not yet proved.

What does “when $A_n$ tends to $M$” mean?

  • $\begingroup$ What I am trying to say here is that since $A_n$ is an increasing sequence and it is bounded by $M$ there will be a point where all $A_n$ will be in the interval $(\epsilon-M,M)$ $\endgroup$ – J doeoeo Jan 30 '17 at 13:41
  • $\begingroup$ @Jdoeoeo Not at all. The sequence $(1+1/n)^n$ is bounded by $1000$, but no way its terms go near $1000$ by less than $500$, for instance. Anyway, if you assume the sequence is bounded, then you can certainly say it's bounded, but you have proved nothing. $\endgroup$ – egreg Jan 30 '17 at 13:44

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