Finding the value of a tricky limit I have some difficulties with my homework in mathematical analysis and I don't really have any good ideas, how to get off the mark. If you, guys, could  give me any ideas, tips or solutions for the following task, I would be really thankful. 
The task is as follows:  
Let $a_{n}$ be a sequence with positive members so that
$$ \lim_{n} \frac{a_{n}}{n} = 0 ,$$
$$\limsup\limits_{n}\frac{a_1 + a_2 + ... + a_n }{n} \in \mathbb{R}.$$
Find the value of:
$$ \lim_{n} \frac{a_1^2 + a_2^2 + ... + a_n^2}{n^2}  $$

Trying the problem with $a_n=c$ suggests the answer is $0$.

Since $\limsup\limits_{n}\frac{a_1 + a_2 + ... + a_n }{n} <\infty$, there exists some $M>0$ such that for all $n$, $\displaystyle \frac{a_1 + a_2 + ... + a_n }{n}\leq M$.
Let $\epsilon>0$. Since $\lim_n \frac{a_n}n=0$, there is some $N$ such that $n\geq N\implies a_n\leq \frac{\epsilon}{2M} n\implies a_n^2\leq \frac{\epsilon}{2M} na_n$.
Then for $n\geq N$, $$\begin{align} \frac{a_1^2 + a_2^2 + ... + a_n^2 }{n^2}
&\leq
 \frac{a_1^2 + a_2^2 + ... + a_N^2 }{n^2} + \frac{a_N^2 + a_{N+1}^2 + ... + a_n^2 }{n^2}\\&\leq \frac{a_1^2 + a_2^2 + ... + a_N^2 }{n^2} +\frac{\epsilon}{2M}\frac{a_N + a_{N+1} + ... + a_n }{n}  \\
&\leq \frac{a_1^2 + a_2^2 + ... + a_N^2 }{n^2} +\frac{\epsilon}{2M}\frac{a_1 + a_2 + ... + a_n }{n} \\
&\leq  \frac{a_1^2 + a_2^2 + ... + a_N^2 }{n^2} + \frac{\epsilon}{2}
\end{align}$$
Since $\displaystyle \lim_n \frac{a_1^2 + a_2^2 + ... + a_N^2 }{n^2} = 0$, there is some $N'$ such that $n\geq N'\implies \frac{a_1^2 + a_2^2 + ... + a_N^2 }{n^2}\leq \frac{\epsilon}{2}$
For $n\geq \max(N,N')$, $$\frac{a_1^2 + a_2^2 + ... + a_n^2 }{n^2}\leq \epsilon$$
Hence convergence to $0$.
