How can we prove that $\displaystyle\sum_{i\in\mathbb N^+}1/i^2$ is convergence.(Elementary or advanced) I've tried to apply compare test with  $\displaystyle\sum_{i\in\mathbb N^+}1/2^i$ like this:
$f(x)=\dfrac{\ln x}{x}$, and if we derivative this we would find that its extrem maximum point is $x=e$, therefore, for very big numbers we can say this;
$$n>>l\\ \Rightarrow \\ l^n>n^l$$ if we choose $l=2$
$$2^n>n^2\quad\rightarrow\quad \dfrac1{2^n}<\dfrac1{n^2}$$
But it is not what we wanted.

How can we prove this, and can we put a upperbound better than 2?


Besides the comparison with the geometric series, you can also use a telescoping one:
\begin{align*}
\sum_{n=1}^\infty\frac{1}{n^2}=1+\sum_{n=2}^\infty\frac{1}{n\cdot n}\leq 1+\sum_{n=2}^\infty\frac{1}{(n-1)n}=1+\sum_{n=2}^\infty\left(\frac{1}{n-1}-\frac{1}{n}\right)=1+1=2.
\end{align*}
In order to get a better upper bound than 2, just start the telescoping series a little later. For instance:
\begin{align*}
\sum_{n=1}^\infty\frac{1}{n^2}=1+\frac{1}{4}+\sum_{n=3}^\infty\frac{1}{n\cdot n}\leq \frac{5}{4}+\sum_{n=3}^\infty\frac{1}{(n-1)n}=\frac{5}{4}+\sum_{n=3}^\infty\left(\frac{1}{n-1}-\frac{1}{n}\right)=\frac{5}{4}+\frac{1}{2}=\frac{7}{4}.
\end{align*}
