Decay of a decreasing summable sequence Let $(x_n)$ be a decreasing sequence of non-negative real numbers, such that $\sum_n x_n\leq C<\infty$ for some $C>0$. 
What is the optimal upper bound I can have about $x_n$'s decay?
I think I remember something like $x_n=o(1/n)$ but I am not sure, and I would more like something quantitative like $x_n\leq C/n$. Do you have any idea?
EDIT: I wrote $o(n)$ instead of $o(1/n)$ in the first place, i apologize for this. Many answers state that $o(1)$ is the best bound, but I'm quite sure that the fact that $a_n$ is non-increasing yields $a_n=o(1/n)$.
 A: The question seems now to be the following:

Let $(x_n)$ denote a positive and decreasing sequence such that the series $\sum\limits_nx_n$ converges, show that $x_n=o(1/n)$.

A proof goes as follows: the series $\sum\limits_nx_n$ converges hence $\sum\limits_{k\gt n}x_k\to0$ when $n\to\infty$.
In particular, $\sum\limits_{k=n+1}^{2n}x_k\to0$. The sequence $(x_k)$ is decreasing hence $x_k\geqslant x_{2n}$ for every $n+1\leqslant k\leqslant2n$ and $\sum\limits_{k=n+1}^{2n}x_k\geqslant nx_{2n}$, which proves that $2nx_{2n}\to0$. 
Likewise (but other arguments are possible), $\sum\limits_{k=n}^{2n+1}x_k\to0$ hence $\sum\limits_{k=n}^{2n}x_k\geqslant (n+1)x_{2n+1}\geqslant (n+\frac12)x_{2n+1}$, which proves that $(2n+1)x_{2n+1}\to0$. Finally, $nx_n\to0$, that is, $x_n=o(1/n)$.
Edit: The same elementary approach yields $x_n\leqslant C/n$ for every $n\geqslant1$, with $C=\sum\limits_{k=1}^\infty x_k$.
A: User @Did's answer is the real deal, but I'm posting this argument year because it is really elementary.

Because the sequence is nonnegative decreasing the $n$ term is less than or equal to the average of the first $n$ terms, i.e $x_n \le S_n/n$, where $S_n$ is the $n$th partial sum. But $S_n \le \sum_{k=1}^\infty x_k \le C \lt \infty$ since $(x_k)_k \in \ell_1$ by assumption. We deduce that
$$
x_n \le C/n\text{ for all } n\ge 1.
$$
