# 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.

• no ,m increases with $\epsilon$ so unclear that their product is small Commented Jan 23, 2018 at 10:51
• @StuartMN is the proof now correct? Commented Jan 23, 2018 at 10:58
• Yes ,the proof is now correct . Good job ! Commented Jan 24, 2018 at 10:25

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.

• Does the edit make sense? Commented Jan 23, 2018 at 10:36
• @SuperMario Well now I see what you did, but it also highlights where your argument fails. Sure, you have proven that $nu_n<\frac{m+1}{2}\epsilon$, but that doesn't prove yet that $nu_n\to 0$, since you have no bound on the value of $m$.
– 5xum
Commented Jan 23, 2018 at 10:40
• $m$ can be fixed to $N$ and my answer does the rest. Commented Jan 23, 2018 at 10:44
• @ArnaudMortier True, but the answer currently does not fix $m$, so you can hardly call the proof "perfect", can you?
– 5xum
Commented Jan 23, 2018 at 10:45
• But you really need to write "there exists N such that for all..." Commented Jan 23, 2018 at 10:46

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!

• That's not just an edit and the proof is not perfect. The proof is incorrect.
– 5xum
Commented Jan 23, 2018 at 10:43

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.

• Again, you want epsilon over 2m... Commented Jan 23, 2018 at 11:01