Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges

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:

Monotonic:

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.

Boundedness:

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!

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.

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.

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.

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.)