Are there sequence $(u_n)\subset \mathbb R^+$ s.t. $\sum_{n=1}^\infty u_n<\infty $ but $nu_n\not\to 0$?
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.
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.