Let $u_n$ be a positive and decreasing sequence of real numbers such that $\sum u_n$ converges. Show that $\lim_{n\to \infty}nu_n= 0$. Let $u_n$ be a positive and decreasing sequence of real numbers such that $\sum u_n$ converges. Show that $\lim_{n\to \infty}nu_n= 0$.
My Attempt: Since $\sum u_n$ is convergent we know that the sequence $S_n=\sum_{i=1}^nu_i $ is convergent and in particular given $\epsilon/2>0$ there exists $n\geq m\geq N$ such that $$|u_{m+1}+u_{m+2}+...+u_n|<\epsilon/2.$$ Since $\{u_n\}$ is decreasing we have that $$(n-m)u_n<|u_{m+1}+u_{m+2}+...+u_n|<\epsilon/2.$$ For $m=N$ we have that $$(n-N)u_n<\epsilon/2.$$Now we know that $u_n\to 0$ and so for $n\geq N'$ $$u_n<\epsilon/2N.$$ Thus for $n\geq \max\{N,N'\}$ we have that $$nu_n<\epsilon/2+Nu_N<\epsilon.$$ This shows that $nu_n\to 0.$ 
Is this argument correct? I am asking this because the solution in the book is quite different from mine. 
 A: I think you need to explain that last part better. How you get from


*

*$(n-m)u_n<\frac\epsilon2$

*$u_n<\frac\epsilon2$


to
$$nu_n\to 0$$
is unclear.
A: What you really want is $ u_n\leq \frac {\epsilon}{2m} $. Otherwise it is perfect,  although I would recommend that you take my edit in consideration. Edit if that was unclear: and of course,  fix $ m $ first! 
A: I'm going to finish your proof but there are much cleaner ways of doing this. 
So given $\epsilon>0$ you found an integer $N$ such that for all $n\geq m\geq N$, 
$$|u_{m+1}+\cdots+u_{n}|<\epsilon/2$$
Also, since $u_n$ is decreasing,
$$nu_n\leq |u_{m+1}+\cdots+u_{n}|+mu_n.$$
Fix $m$. Now since $u_n\to 0$, there is an $N'$ such that for all $n\geq N'$,
$$u_n<\epsilon/(2m).$$
Letting $M=\max(N,N')$ we see that for all $n\geq M$
$$nu_n\leq|u_{m+1}+\cdots+u_{n}|+mu_n<\epsilon,$$
as desired.

Easier Way
Suppose $nu_n\not\to 0$. WLOG, suppose $nu_n\to a>0$ then there is $N$ such that for $n\ge N$, $nu_n > a/2$ (fill in the details here). So $u_n > a/2n$ for $n\geq N$ and hence
$$\sum_{n=N}^\infty u_n\geq\sum_{n=N}^\infty \frac{a}{2n}=\frac{a}{2}\sum_{n=N}^\infty\frac{1}{n}=\infty\implies \sum_{n=1}^\infty a_n=\infty.$$ 
which is the desired conclusion.
