Ever since I came over this method of finding a limit:
Example: Find the limit of $f(n)=\cfrac{1+2+3+...+n}{n^2}$ as $n\rightarrow \infty$, which turns out to be as follows
Since the limit exists, let the limit be $L$, then $$n^2f(n)-(n-1)^2f(n-1)=n$$ $$\lim_{n\rightarrow \infty}\left(n^2f(n)-(n-1)^2f(n-1)\right)=\lim_{n\rightarrow \infty}n$$ $$\lim_{n\rightarrow \infty}\left(n^2L-(n-1)^2L\right)=\lim_{n\rightarrow \infty}n$$ $$\lim_{n\rightarrow \infty}\left((2n-1)L\right)=\lim_{n\rightarrow \infty}n$$ Which can be shown that $L=1/2$.
I'm wondering if the Basel Problem can be solved this way, namely $f(n)=\sum_{k=1}^n\frac{1}{k^2}$, $$f(n+1)-f(n)=\frac{1}{(n+1)^2}$$