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!

$a_n=\frac{1}{n^2}$ 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
