Proving the sum of the reciprocals squared converges I'm investigating the Basel Problem, and the sum to consider is:
$\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{n^2}$
How can I show this converges? Using graphs/computer software is also fine, but how would I do it? Is there a way using calculus?
I've looked on Google for a while and I found a few proofs, but I didn't understand any of them.

Note that
$$
\begin{alignat}{8}
&\frac{1}{1^2}+&&\frac{1}{2^2}+&&\frac{1}{3^2}+&&\frac{1}{4^2}+&&\frac{1}{5^2}+&&\frac{1}{6^2}+&&\frac{1}{7^2}+&&\dots\\
\leq &\frac{1}{1^2}+&&\frac{1}{2^2}+&&\frac{1}{2^2}+&&\frac{1}{4^2}+&&\frac{1}{4^2}+&&\frac{1}{4^2}+&&\frac{1}{4^2}+&&\dots
\end{alignat}
$$
where there are $2^n$ copies of $\frac{1}{(2^n)^2}$.
But $\frac{1}{(2^n)^2}=\frac{1}{4^n}$, so $2^n$ copies of it is $\frac{2^n}{4^n}=\frac{1}{2^n}$. So the sum is bounded above by $\sum_{k=0}^\infty \frac{1}{2^n}=2$.
