Limit with a summation I'm getting crazy with the next limit:
$$\lim_{n\rightarrow\infty}\frac{\sum \limits_{k=1}^n(k·a_k)}{n^2}$$
The exercise also says that is known that: $$\lim_{n\rightarrow\infty}a_n = L$$
I suppose that I have to answer in function of "$L$", but I still can't see anything. Any ideas?
 A: There are a few ways of doing this problem - ranging from first principles to using the Stolz-Cesàro theorem as a canned theorem. I will explain both the proofs here; however, note that if we unroll the proof of Stolz-Cesàro, the resulting proof is essentially the same as the second approach. 
Method 1: Using the Stolz-Cesàro theorem: The Stolz-Cesàro theorem can be seen as a discrete version of l'Hôpital's rule. Verify that the denominator $n^2$ satisfies the hypotheses of the theorem; i.e., is strictly increasing and unbounded. Now 
$$
\frac{\sum \limits_{k=1}^n k a_k - \sum \limits_{k=1}^{n-1} k a_k}{n^2 - (n-1)^2} = \frac{n a_n}{2n-1} = \frac{a_n}{2 - \frac{1}{n}} \to \frac{L}{2}
$$
as $n \to \infty$. Therefore the given sequence also converges to the same limit $\frac{L}{2}$.  $\quad  \diamond$

Method 2: First principles approach. This is a fleshed out version of Davide's hint. Fix $\varepsilon > 0$. Since $a_n \to L$ as $n \to \infty$, there exists $q$ such that $|a_n - L| \leqslant \varepsilon$ for all $n \geqslant q$. Therefore,
$$
\begin{align*}
\left| \frac{1}{n^2} \sum_{k=1}^n k a_k  - \frac{L}{2} \right| 
&= \left| \frac{1}{n^2} \sum_{k=1}^n k (a_k - L) + \frac{1}{n^2} \sum_{k=1}^n k L - \frac{L}{2} \right|
\\ &= \left| \frac{1}{n^2} \sum_{k=1}^n k \cdot (a_k - L) + \frac{L}{2n}  \right|
\\ &\leqslant \frac{1}{n^2} \sum_{k=1}^n k \cdot |a_k - L| + \frac{L}{2n} 
\\ &\leqslant \frac{1}{n^2} \sum_{k< q} k \cdot |a_k - L| + \frac{1}{n^2} \sum_{k=q}^n k \cdot \varepsilon + \frac{L}{2n} 
\\ &\leqslant \frac{1}{n^2} \sum_{k< q} k \cdot |a_k - L| +  \varepsilon + \frac{L}{2n} .
\end{align*}
$$
Now for fixing $\varepsilon$ and $q$, for large enough $n$, we can bound the above by $2 \varepsilon$. Since this holds for all $\varepsilon > 0$, it follows that the sequence approaches $\frac{L}{2}$. $\quad \diamond$

Note: Connection to a generalised Cesàro theorem. The following theorem generalises the standard theorem on Cesàro means. 

Suppose $\sum_{n} b_n$ is a positive series that diverges to infinity. Then if $(a_n) \to L$, then the generalised Cesàro mean
  $$
\frac{\sum_{k=1}^{n} b_k \cdot a_k}{\sum_{k=1}^{n} b_k}
$$
  also converges to $L$. 

(The term “generalised Cesàro mean” is something I made up just now; I do not know if it is standard or not.) This theorem can be proved by mimicking the above two approaches, so I will skip the proof. 
Applying the above theorem to $b_n = n$, we conclude that 
$$
\frac{\sum_{k=1}^{n} k \cdot a_k}{\sum_{k=1}^{n} k} = 2 \frac{\sum_{k=1}^{n} k \cdot a_k}{n(n+1)} 
$$
converges to $L$. From this, it follows that $\frac{\sum_{k=1}^{n} k \cdot a_k}{n^2}$
 converges to $\frac{L}{2}$. 
A: Hint: It will be $$\lim_{n\rightarrow\infty}\frac{\sum_{k=1}^n(kL)}{n^2}$$ which is $$L \lim_{n\rightarrow\infty}\frac{\sum_{k=1}^n k}{n^2}$$ which you should be able to work out.
