Trying to show that $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$ converges by using the direct comparison test I know that if $\forall n \in \mathbb{N}$, $ 0 \le a_n \le b_n$. Then if $\sum b_n$ converges then $\sum a_n$ converges.

I am having trouble finding a convergent $\sum b_n$ for which
\begin{align*}
 \sum_{n=1}^{\infty}{\frac{1}{n^2}} \le \sum b_n
\end{align*}
Is it possible to do it with comparison test, or do I need another method?
You can compare, as said in the comments, by
$$
\frac1{n^2}&lt;\frac1{n(n-1)}=\frac1{n-1}-\frac1n
\text{ so that } ~~
\sum_{n=1}^\infty\frac1{n^2}\le1+\sum_{n=2}^\infty \frac1{n-1}-\frac1n=2$$
by telescoping the upper bound. A closer bound is
$$
\frac1{n^2}&lt;\frac1{(n+1)(n-1)}=\frac1{2(n-1)}-\frac1{2(n+1)}
\text{ so that } ~~
\sum_{n=1}^\infty\frac1{n^2}\le1+\frac12\sum_{n=2}^\infty \frac1{n-1}-\frac1{n+1}=\frac74
$$
where the telescoping "leap-frogs".

You could also use the construction of the Cauchy condensation test where
$$
\frac1{n^2}&lt;\frac1{4^k}~~\text{ for }~~ 2^k\le n &lt; 2^{k+1} ~~
\text{ so that } ~~
\sum_{n=1}^\infty\frac1{n^2}\le\sum_{k=0}^\infty2^k\cdot \frac1{4^k}=\sum_{k=0}^\infty2^{-k}=2$$
We know $$n^2 &gt; (n - 1)^2$$ 

Inverting the fractions the sign changes and we get 

$$n^2 &gt; (n - 1)^2 \implies \frac{1}{n^2} &lt; \frac{1}{(n - 1)^2}$$

Take $b_n = \frac{1}{(n - 1)^2}$

Hence $a_n &lt; b_n$ is satisfied.

P-series are of the form $$f(x) = \frac{1}{x^p}$$

And when $p &gt; 1$ the series converges; $p \le 1$ the series diverges.

Since $b_n$ is a p-series with $p &gt; 1 (= 2)$, $b_n$ converges.

Since $b_n$ converges and $b_n &gt; a_n$; $a_n$ converges as well.