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!

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$.
