Convergence of series given a monotonic sequence Let $(x_n)_{n \in \Bbb N}$ be a decreasing sequence such that its series converges, want to show that $\displaystyle \lim_{n \to \infty} n x_n = 0$.
Ok I don't even know where to start.
I need a direction please! 
Thankyou!
 A: The sequence $(x_n)$ is nonincreasing and the series $\sum\limits_nx_n$ converges, hence $x_n\geqslant0$ for every $n$. Otherwise, there exists $n$ such that $x_n\lt0$, and $x_k\leqslant x_n$ for every $k\geqslant n$, hence
$$
\sum\limits_{k=n+1}^mx_k\leqslant(m-n)x_n\to-\infty
$$ 
when $m\to\infty$, which is absurd.
Furthermore, the series $\sum\limits_kx_k$ converges hence $\sum\limits_{k\gt n}x_k\to0$ when $n\to\infty$. But 
$$
\sum\limits_{k\gt n}x_k\geqslant\sum\limits_{k=n+1}^{2n}x_k\geqslant nx_{2n}\geqslant0,
$$ 
hence $nx_{2n}\to0$ hence $nx_n\to0$.
A: Just another approach. Since $\{x_n\}_{n\in\mathbb{N}}$ id decreasing and its associated series converges, we have $x_n\geq 0$ for every $n\in\mathbb{N}$ (otherwise, $\lim_{n\to +\infty}x_n < 0$ and the series cannot converge).
Assume now the existence of a positive real number $\alpha$ such that
$$ n\, x_n \geq \alpha $$
for an infinite number of positive natural numbers $n$; let $A=\{n_1,n_2,\ldots\}$ be the set of such natural numbers. Let now $a_0=0,a_1=n_1$, $a_2$ be the minimum element of $A$ greater than $2a_1$, $a_3$ be the minimum element of $A$ greater than $2a_2$ and so on. We have:
$$\sum_{n=1}^{+\infty}x_n \geq \sum_{k=1}^{+\infty}(a_k-a_{k-1})x_{a_k} \geq\sum_{k=1}^{+\infty}\frac{\alpha}{2}=+\infty,$$
that is clearly a contradiction, so
$$\lim_{n\in\mathbb{N}} (n\,x_n) = 0$$
must hold.
