TITLE: If $x_{n}$ is decreasing and $\sum x_{n}$ converges, prove that $\lim nx_{n} = 0$
QUESTION [14 upvotes]: I need a little hint in a proof on convergence of a sequence. The problem I have is: Suppose $(x_n)$ is a monotone decreasing sequence of real numbers such that $\sum x_n$ converges, prove that:
$$\lim_{n \to \infty} nx_n=0$$
My idea is: proving this fact aims to prove that given $\epsilon >0$ there's some natural $n_0 \in \mathbb{N}$ such that if $n >n_0$ we have $|nx_n|<\epsilon$.
Now, I know that $\sum x_n$ converges, so that given $\epsilon'>0$ there's some $k_0 \in \mathbb{N}$ such that if $k > k_0$ we have:
$$\left|\sum_{i=1}^{k}x_i - S\right|<\epsilon'$$
Where $S = \sum x_n$. Now, since $(x_n)$ is monotone decreasing we know that we must have $x_1 > \cdots > x_k$ so that the sum of all $x_i$ should be less or equal to $k x_k$. So I know that I have:
$$\left|\sum_{i=1}^{k}x_i - S\right|\leq\left|\sum_{i=1}^{k}x_i\right|+|S|\leq|kx_k|+|S|$$
I feel that the proof will come from this, however I'm stuck at this point. Can someone give just a little hint on how to proceed from here?
Thanks very much in advance for your help.

REPLY [5 votes]: Hint
By index change prove:
$$\sum_{k=1}^n x_k=\sum_{k=1}^n k(x_k-x_{k+1})+nx_{n+1}$$
then prove that the series 
$$\sum_{n=1}^\infty n(x_n-x_{n+1})$$
is convergent and deduce that the sequence $(nx_{n+1})$ is convergent to some $\lambda$.
Can $\lambda$ be different to zero? Why? and finaly how to prove that the sequence $(nx_n)$ converges also to zero?