Can I use limits to calculate summation of a progression I had a question 
$$\sum_{a=1}^n \frac{1}{a^2}=?$$
I had learned newton's method of undetermined method 
,but that doesn't work here because of negative power
than I saw another question 
$$\sum_{a=1}^n \frac{1}{a(a+1)}$$
here we transformed it as
$$\sum_{a=1}^n (\frac{1}{a}-\frac{1}{a+1})$$
which on expanding gave 
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{n}-\frac{1}{n+1}$$
where each element cancels out except
$$1-\frac{1}{n+1}$$
which gives the formula
$$\sum_{a=1}^n \frac{1}{n(n+1)}=1-\frac{1}{n+1}=\frac{n}{n+1}$$
but I can't express $a^{-2}$ as two diffrent factors so I thought to use limits(calculus) as 
$$\frac{1}{a^2}=\lim_{h\to 0}\frac{1}{a(a+h)}$$
but when I tried to break it then I got $h$ in denominator.
I may be wrong, so please help to calculate the summation.
 A: There isn't such a beautiful solution as the one you've presented, since it is the case that as $n\to\infty$, you get the Basel problem.
Notice that
$$\frac1{a(a+h)}=\frac1h\left(\frac1a-\frac1{a+h}\right)$$
but since $h$ is not a whole number, we will not be seeing any cancellations like with your telescoping series.

But, if you could recall the geometric series:
$$\frac{1-r^{n+1}}{1-r}=\sum_{k=0}^nr^k$$
Integrate both sides and divide by $x:$
$$\frac1x\int_0^x\frac{1-r^{n+1}}{1-r}\ dr=\frac1x\int_0^x\sum_{k=0}^nr^k\ dr=\sum_{k=0}^n\frac1{k+1}x^k$$
Repeating the process,
$$\int_0^y\frac1x\int_0^x\frac{1-r^{n+1}}{1-r}\ dr\ dx=\int_0^y\sum_{k=0}^n\frac1{k+1}x^k\ dx=\sum_{k=0}^n\frac1{(k+1)^2}y^{k+1}$$
And as $y\to1$,
$$\sum_{k=1}^nk^{-2}=\int_0^1\frac1x\int_0^x\frac{1-r^n}{1-r}\ dr\ dx$$
I believe I've read that this is how Euler approximated the Basel problem.
A: I am not aware of any formulas for the finite partial sum... however we can compute the limit ! I think the easiest way to compute it is through trigonometric sums/estimations. Here is a sketch of the proof : 
By using De Moivre's Formula, you can easily show that, for all real $t$ :
\begin{equation*}
\sin(2n+1)t = \sum_{k=1}^{n} \binom{2n+1}{2k+1}(-1)^k \sin^{2k+1}(t) \cos^{2(n-k)}(t)
\end{equation*}
Then dividing by $\sin^{2n+1}(t)$, when $t \notin \pi\mathbb{Z}$, you get 
\begin{equation*}
\frac{\sin(2n+1)t}{\sin^{2n+1}(t)} = P_n({cotan}^{2}(t)). \; \text{where} \: P_n=\sum_{k=1}^{n} \binom{2n+1}{2k+1}(-1)^k X^{n-k}
\end{equation*}
For $t \in (0, \frac{\pi}{2})$, the formula above holds and thus $P_n$ vanishes at points $x_k=cotan(t_k)$ where $t_k= \frac{k\pi}{2n+1}$ for $k \in \{1, \cdots, n\}$. Since $P_n$ is of degree $n$, we found all the roots of $P_n$. And by the coefficient/roots relationship, we can assert that :
\begin{equation*}
\sum_{k=1}^n x_k = -\frac{-\binom{2n+1}{3}}{\binom{2n+1}{1}} = \frac{n(2n-1)}{3}
\end{equation*}
Then by squaring and inverting the inequalities for $ t \in(0,\frac{\pi}{2})$ : $0<sin(t)\leq t \leq \tan(t)$, we get :
\begin{equation*}
{cotan}^2(t) \leq \frac{1}{t^2} \leq 1 + {cotan}^2(t)
\end{equation*}
which yields by taking $t=t_k$ and then summing over $k$ :
\begin{equation*}
\frac{n(2n-1)}{3} \leq \frac{(2n+1)^2}{\pi^2} \sum_{k=1}^n \frac{1}{k^2} \leq n + \frac{n(2n-1)}{3}
\end{equation*}
Thanks to the squeeze Theorem, you may conclude that :
\begin{equation*}
\sum_{k=1}^{+ \infty} \frac{1}{k^2} = \frac{\pi^2}{6}
\end{equation*}
