Are there sequence $(u_n)\subset \mathbb R^+$ s.t. $\sum_{n=1}^\infty u_n<\infty $ but $nu_n\not\to 0$? I have an exercise that ask me to prove that if $(u_n)$ is decreasing, $u_n\geq 0$ for all $n$ and $\sum_{n=1}^\infty u_n$ converge, then $nu_n\to 0$. I proved it, but for me this result in fact correct even if we remove $(u_n)$ decreasing. Because I have in mind that $1/n$ is the quickest speed of non convergence, i.e. if $\sum_{n}v_n$ converge, it will have $\alpha $ s.t. $v_n\leq \frac{1}{n^\alpha }<\frac{1}{n}$ for all $n$ for a certain $n$. But if we impose $u_n$ decreasing, maybe my imagination is wrong. So is there a positive sequence s.t. $\sum_{n=1}^\infty u_n$ converge but $nu_n\not\to 0$ ?
 A: You can prove that if $a_n$ is decreasing and $na_n\not\to 0$, then $\sum a_n$ diverges. 
Proof: If $na_n\not\to0$, then there is an $\epsilon>0$ and infinitely many indices $n(1)<n(2)<\dots$ for which $a_{n(i)}\ge \epsilon/n(i)$. Since $a_n$ is decreasing, we can group terms based on which of the intervals $[1,n(1)),[n(1),n(2)),\dots$ they lie in, and you get
$$
\sum_n a_n \ge \sum_i (n(i+1)-n(i))\cdot \frac{\epsilon}{n(i)}\stackrel{*}=-\sum_i n(i+1)\Big(\frac{\epsilon}{n(i+1)}-\frac{\epsilon}{n(i)}\Big)=\epsilon \Big(\sum_i \frac{n(i+1)}{n(i)}-1\Big)\tag{1}
$$
where $\stackrel{*}=$ follows by summation by parts (check this!).
Now, let $r(i) = \frac{n(i+1)}{n(i)}$, with the convention $r(0)=n(1)$. It is a fairly standard result that 
$$
\sum_i r(i)-1<\infty \iff \prod_i r(i)<\infty 
$$
which follows from $\log x\sim x-1$ as $x\to1$. However, since $\prod_{i=0}^k r(i)=n(k+1)\to\infty$, we must have that the  final sum in $(1)$ diverges. 
A: To answer the question in your title. For each positive integer $t$ set $u_{t^2} = \frac{1}{t^2}$. And for all nonsquare positive integers $n$ set $u_n = \frac{1}{n^3}$. Then 


*

*On the one hand $\sum_{j=1}^{\infty} u_j$ converges (as $\sum_{t=1}^{\infty} \frac{1}{t^2}$ converges as does $\sum_{n=1}^{\infty} \frac{1}{n^3}$). 

*On the other hand, $t^2u_{t^2} = 1$ for all positive integers $t$ so $nu_n$ does not converge to 0 as $n$ goes to infinity. i.e., No matter how large $n$ is, there is a positive integer $n'$ s.t. $n'u_{n'} =1$, namely $n'$ is any perfect square larger than $n$.
The $u_n$s are not nonincreasing though. 
