If $f(a_n)\rightarrow L$ then $a_{n}\rightarrow L$ Let $\{a_{k}\}$ be a non-negative, increasing sequence. Consider

$$S_{n} = \frac{1}{n}\sum_{k=1}^{n}a_{k}.$$

I would like to show that if $$\lim_{n\rightarrow\infty}S_{n} = L,$$ then $$\lim_{k\rightarrow\infty}a_{k} = L.$$

I don't really have any clue how I could solve this problem. By definition, there exists $N$ such that

$$\lvert S_{n} -L\rvert &lt;\epsilon$$

or

$$\frac{1}{n}\left\lvert \sum_{k=1}^{n}a_{k} - nL\right\rvert&lt; \epsilon$$

for all $\epsilon&gt;0$ whenever $n&gt;N$ and  

$$\frac{1}{n}\left\lvert \sum_{k=1}^{n}a_{k} - nL\right\rvert \leq \frac{1}{n}\sum_{k=1}^{n}\lvert a_{k}-L \rvert.$$

Now, if we define a new sequence $\{\alpha_{k}\}$ by

$$\alpha_{k} := \lvert a_{k} -L\rvert,$$

then I am pretty sure there is a theorem from harmonic analysis that states something like $\sum_{k=1}^{n}{\alpha_{k}}$ converges if and only if $(1/n)\sum_{k=1}^{n}{\alpha_{k}}$ converges too.

From here, by the Divergence Theorem, I would get that necessarily $\alpha_{k} \rightarrow L$ for $\sum{\alpha_{k}}$ to converge and hence $(1/n)\sum \alpha_{k}$ would also require $\alpha_{k}\rightarrow L$ to converge.

However, I can neither find a reference to this theorem I only vaguely remember and this still doesn't get me anywhere because I don't think the convergence of $(1/n)\sum \alpha_{k}$ is helpful anyway.

I have also tried proving this by contradiction too, but I would really prefer a direct proof.
Since the sequence is increasing $a_n \rightarrow \sup \{a_k : k \in \mathbb {N}\}$

Let prove that $L = \sup  \{a_k : k \in \mathbb {N}\}$. First for every $n$, $a_n \le L$. Indeed if there is $N$ that $a_N &gt; L+\epsilon$ for $\epsilon &gt; 0$. Since the sequence is increasing then for every $n \ge N$, $a_n &gt; L+\epsilon$. Since the sequence is non negative, then $S_n &gt; \frac{n - N}{n} (L+\epsilon)$ after tending $n$ to $\infty$ we have $L \ge L + \epsilon$ which is absurde. 

On the other hand if for all $n$, $a_n &lt; L - \epsilon$ for $\epsilon &gt; 0$ then $S_n &lt; L-\epsilon$. Tending $n$ to infinity $L \le L- \epsilon$ which is also absurde.
If $a_n$ tends to infinity them $L$ is infinity and if $a_n$ tends to a finite limit a then $L=a$. This is standard and the proof is elementary. The result follows immediately by considering these two cases.