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.
 A: 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 $\sum_{k=0}^\infty \frac{1}{2^n}=2$.
A: Another way of proving it that generalizes to sums of the type $\sum f(n)$ with $f$ decreasing:
$$
\sum_{n=1}^N\frac{1}{n^2}\le1+\int_1^N\frac{dx}{x^2}.
$$
A: Maybe the simplest way of proving this is to use :
$ \frac{1}{n^2} \leq \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n} $ for n $\geq 2$
Introducing partial sums you get : 
$ S_n = \sum_{k=2}^n \frac{1}{k^2} \leq \sum_{k=2}^n \frac{1}{k-1} - \frac{1}{k}  =1 - \frac{1}{n} \leq 1 $
So $(S_n)$ is an increasing sequence, upper-bounded hence...
Edit: I'll write down the part about fractions since you're not very familiar with it :) .
You can deal with this in two ways:
1) Write that: $\frac{1}{n(n-1)} = \frac{A}{n} +\frac{B}{n-1} $, then reduce to a common denominator ,i.e :
$\frac{A}{n} +\frac{B}{n-1} = \frac{A(n-1) + Bn}{n(n-1)}$ , then identify with the numerator: $-1=A ; A+B=0 \implies A=-1; B=1$
2) Directly verifying: $\frac{1}{n-1} - \frac{1}{n} = \frac{n}{n(n-1)} - \frac{n-1}{(n-1)n} =\frac{1}{n(n-1)} $
Here the first one sounds silly because the result can be easily obtained by finding the A and B directly, that is reducing $\frac{1}{n-1} - \frac{1}{n}$ to a common denominator and verify the equality. However this method of expanding a fraction or reducing it, depending on what we are supposed to prove, is very usual. For further information about that, I suggest you take a look a this page:
http://en.wikipedia.org/wiki/Partial_fraction_decomposition
You'll find everything you need to know. It explains why I can write the decomposition with A and B and expect to find something that has a meaning. It's not always the case. 
As for bounding the sum, the latter introduces a telescoping with the sequence:
$V_n = \frac{1}{n}$
$ \frac{1}{k^2} \leq \frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k} = V_{k-1} -V_k $
$ \implies \sum_{k=2}^n \frac{1}{k^2} \leq \sum_{k=2}^n (V_{k-1} -V_k) = (V_1-V_2)+(V_2-V_3)+...+(V_{n-2}-V_{n-1})+(V_{n-1}-V_n) = V_1 - V_n = 1-\frac{1}{n} $
Only the first and last terms remain.
