Prove that $\lim\limits_{n\to\infty}nx_n=0$

Question

For a sequence $\left\{x_n\right\}$, prove that $\lim\limits_{n\to\infty}nx_n=0$, given that $\left\{x_n\right\}$ is decreasing and $\sum_{n=0}^\infty x_n$ converges.

My given hint is to prove: $$\sum_{n=k}^\infty x_n \ge \sum_{n=k}^m x_n \ge \left(m-k\right)x_m$$ Then, I am meant to deduce: $$\limsup\limits_{m\to\infty}mx_m \le \sum_{n=k}^\infty x_n$$

How do I prove the latter deduction? I tried to write $m$ as $\left(m-k\right)+k$ to make $kx_m$ disappear but then I failed to make the $\limsup$ appear.

Answer 1

I imagine, that $\sum_0^\infty x_n<\infty$ is also part of the assumption.

In that case, $$ \sum_{n=0}^\infty\ge \sum_{k=\lfloor n/2\rfloor}^nx_k\ge (n-\lfloor n/2\rfloor+1)x_n\ge \frac{nx_n}{2.} $$ It hence suffices to show that $$ y_n=\sum_{k=\lfloor n/2\rfloor}^nx_k\to 0. $$ If not, then as $y_n\ge 0$, there would be a $\varepsilon>0$ and subsequence $y_{k_n}$, such that $y_{k_n}>\varepsilon$. Take a sub-subsequence $y_{j_n}$, such that, $$ j_{n+1}>2j_n $$ Then $$ \sum_{n=1}^\infty\ge\sum_{k=1}^{j_n}x_k\ge \sum_{\ell=1}^n y_{j_\ell}\ge n\varepsilon\to\infty. $$ Contradiction.