$0\leq u_n\leq \frac {1}{n^2}\sum_{k=1}^nu_k\implies $ $\sum u_k$ converges. 
Let $(u_n)_{n\geq 1}$ be a sequence of real numbers such that $\displaystyle \forall n\geq 1, \;\;\;\;0\leq u_n\leq \frac {1}{n^2}\sum_{k=1}^nu_k$.
Prove that $\sum u_k$ converges.

Let $S_n=\sum_{k=1}^nu_k$ and $A_n=\sum_{k=1}^n \frac 1{k^2}$.
It suffices to prove that $S_n$ converges.
Summing the inequalities $0\leq u_n\leq \frac {1}{n^2}\sum_{k=1}^nu_k$ from $n=1$ to $n=N$ yields $S_N\leq \sum_{k=1}^N \frac{S_k}{k^2}$
Summation by part gives $\displaystyle \sum_{k=1}^N \frac{S_k}{k^2}= S_NA_N-\sum_{k=1}^{N-1}u_{k+1}A_k\leq S_NA_N$
Hence the estimate  $S_N\leq \sum_{k=1}^N \frac{S_k}{k^2}\leq S_NA_N$
As a consequence, $S_N$ converges if and only if the series $\displaystyle \sum_{k}\frac{S_k}{k^2}$ is convergent.
This is quite interesting, but it doesn't answer the problem...

You can prove by induction that : $0 \leqslant u_n \leqslant \frac{2}{n(n+1)}u_1$.
Base case for $n=1$, that is okay
Inductive step Assume it is true for all $k < n$. Then :
$u_n \leqslant \frac{u_n}{n^2} + \frac{1}{n^2}\sum \limits_{k=1}^{n-1} u_k$, so $(1 - \frac{1}{n^2})u_n \leqslant \frac{1}{n^2}\sum \limits_{k=1}^{n-1} \frac{2}{k(k+1)}u_1 = \frac{1}{n^2}\sum \limits_{k=1}^{n-1} \Big( \frac{2}{k} - \frac{2}{k+1} \Big)u_1$.
Hence $\frac{n^2 - 1}{n^2} u_n \leqslant \Big( 2 - \frac{2}{n} \Big)u_1 = \frac{2(n-1)}{n} u_1$. We end up with $u_n \leqslant \frac{2}{n(n+1)}u_1$.
Finally you deduce that $\sum \limits_{k=1}^n u_k \leqslant 2 u_1$ (and all the $u_k$ are non-negative), so $\sum u_k$ converges.
