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?

 A: Hint. One may observe that
$$
\frac1{n^2}< \frac1{n(n-1)}=\frac1{n-1}-\frac1{n},\quad n\ge2,
$$ then one may use a telescoping sum.
A: HINT:
SIMPLE METHODOLOGY:
Note that 
$$\sum_{i=1}^N\frac{1}{i^2}\le 1+\int_1^N \frac{1}{x^2}\,dx$$

MORE ADVANCED METHODOLOGY:
Using $\int_0^1 x^{i-1}\,dx=\frac{1}{i}$, we can show that
$$\sum_{i=1}^\infty \frac{1}{i^2}=\int_0^1\int_0^1 \frac{1}{1+xy}\,dx\,dy$$
In THIS ANSWER, I evaluated the double integral by using the coordinate transformation $x=s+t$ and $y=s-t$ and showed that the series is equal to $\pi^2/6$.
A: We can actually apply the comparison test to your series, though not in the manner you were trying to do:
\begin{align} \sum_{n=1}^\infty\frac1{n^2}&=\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\frac1{5^2}+\frac1{6^2}+\frac1{7^2}+\dots\\ &< \color{red}{\frac1{1^2}}+\color{#44aa77}{\frac1{2^2}+\frac1{2^2}}+\color{#4477aa}{\frac1{4^2}+\frac1{4^2}+\frac1{4^2}+\frac1{4^2}}+\dots\\ &=\sum_{n=0}^\infty\frac{2^n}{(2^n)^2}\\&=\sum_{n=0}^\infty\frac1{2^n}\end{align}
A: 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*}
A: It is interesting to remark that Stirling's inequality can be proved by approximating similar series (related with $\psi'(z)$) through creative telescoping. We may notice that:
$$ \sum_{n\geq 2}\frac{1}{n^2-\frac{1}{4}}=2\sum_{n\geq 2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{2}{3}\tag{1} $$
hence
$$ \zeta(2)=\sum_{n\geq 1}\frac{1}{n^2}\color{red}{\leq} 1+\sum_{n\geq 2}\frac{1}{n^2-\frac{1}{4}} = \frac{5}{3}\tag{2} $$
leads to a not-so-bad approximation, equivalent to (the Babylonian?) $\pi\approx \sqrt{10}$.
By iterating the same approach we also get the interesting identity
$$ \sum_{n\geq 1}\frac{1}{n^2}=\sum_{n\geq 1}\frac{3}{n^2\binom{2n}{n}}\tag{3} $$
where the RHS converges much faster than the LHS.
