# Product of 2 elements in $\ell^2$ is in $\ell^1$

If $$a,b \in \ell^2$$ then $$a\bar b \in \ell^1$$

($$\ell^1$$ and $$\ell^2$$ on $$\mathbb C$$ and $$\bar b$$ is complex conjugate of $$b$$)

What did I write :

Since $$a,b \in \ell^2$$, $$a=(a_n)$$ and $$b=(b_n)$$ such that $$\sum_{n=1}^{\infty}|a_n|^2 \lt \infty$$ and $$\sum_{n=1}^{\infty}|b_n|^2 \lt \infty$$

But I cannot realize that how can I show $$\sum_{n=1}^{\infty}|a_n\bar {b_n}| \lt \infty$$ from above?