Let $\sum^\infty_{n=1}x_n$ where all the terms are positive. Prove that if this series is convergent, so is $\sum^\infty_{n=1}x_n^2.$ Let $$\sum^\infty_{n=1}x_n$$
where all the terms are positive.


*

*Prove that if this series is convergent, so is 
$$\sum^\infty_{n=1}x_n^2.$$

*Prove that the previous statement does not necessarily hold the other way around.


I know how to approach this when the terms can also be negative but when the terms are now all positive, I don't even know how to start.
 A: Hint:
Since $\sum_{n=1}^\infty x_n$ converges, there exists $N \in \Bbb{N}$ such that for all $n \ge N$ we have $x_n \le 1$. Hence $x_n^2 \le x_n$ so
$$\sum_{n=1}^\infty x_n^2 \le \sum_{n=1}^N x_n^2 + \sum_{n=N+1}^\infty x_n < +\infty.$$
To show that the converse doesn't hold, consider $x_n = \frac1n$.
A: Because $x_n$ is positive for all $n$
$$X_N=\sum_{n=1}^{N}x_n^{2}\leq \left ( \sum_{n=1}^{N}x_n \right )^{2} \leq \left ( \sum_{n=1}^{\infty}x_n \right )^{2}$$
So $X_N$ is a monotonically increasing and bounded sequence, therefore convergent.
A: If $S_1 = \sum^\infty_{n=1}x_n$ is convergent, then $\lim_{n \to \infty} = 0$ and there exists a $k$ such that $0<x_n<1 \ \forall n \geq k$. So you can split the sum into  $\sum^{k-1}_{n=1}x_n +  \sum^{\infty}_{n=k}x_n$. The first sum is clearly finite, so for the sum to be convergent, the second has to be as well.
Now consider $S_2 = \sum^\infty_{n=1}x^2_n = \sum^{k-1}_{n=1}x^2_n +  \sum^{\infty}_{n=k}x^2_n$. The first sum is clearly again finite. The second sum is smaller term for term than $\sum^{\infty}_{n=k}x_n$, so it clearly converges as well. Hence $S_2$ is also convergent.
For the other part, it is sufficient to provide a counter-example. An easy one is $\zeta(2)$ vs the harmonic series.
A: Since $\sum_{n\ge 1}x_{n}$ is a series of positive terms and it converges we have $$\lim_{n \to \infty}x_n=0\implies \lim_{n\to \infty}\frac{x_n^2}{x_n}=0$$
Now lets choose $\epsilon=1$, then $\exists N_0\in \mathbb N$ such that $$\frac{x_n^2}{x_n}\leq 1;\forall n\ge N_0\implies x_n^2\leq x_n, \forall n\ge N_0$$
(notice that all terms are positive here)
Then by this comparison we have
$$\lim_{M\to \infty}\sum_{n\ge N_0}^{M}x_{n}^2\leq \lim_{M\to \infty}\sum_{n\ge N_0}^{M}x_{n}<\infty$$
Hence 
$\sum^\infty_{n=1}x_n^2< \infty$. And for the second just take $x_n=\frac{1}{n}$
