$\sum a_n$ converges $\implies\ \sum a_n^2$ converges? [duplicate]

If $\sum a_n$ with $a_n>0$ is convergent, then is $\sum {a_n}^2$ always convergent? Either prove it or give a counter example.

Im trying in this way, Suppose $a_n \in [0,1] \ \forall\ n.\ $ Then ${a_n}^2\leq a_n\ \forall\ n.$ Therefore by comparison test $\sum {a_n}^2$ converges.

So If $a_n$ has certain restrictions then the result is true. what about the general case?

How to proceed further? Hints will be greatly appreciated.

Hint: Since, $a_i>0\forall i\ge 1$, $$\sum_{n=1}^N a_i^2\le \left(\sum_{i=1}^N a_i\right)^2\ \forall N\ge 1$$

If $\sum_{n=1}^\infty a_n$ is convergent, then $\lim_{n\to\infty}a_n=0$, hence from somewhere upward we have $0<a_n<1$, now use comparison test...

Looks like there has been some significant editing but to answer the question as given, as I interpret it, does $\sum a_n$ convergent imply $\sum a_n^2$ convergent for $a_n$ not necessarily non-negative?

The answer is no. Take $a_{2n+1} = 1/\sqrt{n}$ and $a_{2n} = -1/\sqrt{n}$. Then the partial sums will be $s_{2n} = 0$ and $s_{2n+1} = 1/\sqrt{n}$. This means $s_n \to 0$ and the series converges.

However, $a_{2n}^2 = 1/n$ and $a_{2n+1}^2 = 1/n$, so $s_{2n} = 2 \sum_{k=1}^n 1/k$ in this case. This is the harmonic series and it diverges.