How to show that $\limsup_{n\to \infty}\frac{a_n}{n}\leq \limsup_{n\to\infty}(a_{n}-a_{n-1})$ for any sequence $(a_n)$? 
Let $(a_n)$ be a sequence of real numbers. Prove that
  $$\limsup_{n\to \infty}\frac{a_n}{n}\leq \limsup_{n\to\infty}\,
 (a_{n}-a_{n-1}).
$$

If the sequence $(a_n)$ converges, then this is immediate: both sides are $0$. If it is not convergent, how should I approach this problem? 
 A: First suppose $\limsup\limits_{n\rightarrow\infty} (a_n - a_{n-1}) < \infty$ (otherwise the statement is certainly true), and say it is equal to $M$. Then, intuitively speaking, as we look further along the sequence, each successive term is at most approximately $M$ more than the previous one. Looking further and further along, the worst case scenario is that approximately $M$ is continued to be added to each term. As more terms are added, any aberrant earlier behaviour will be nullified by the increasing $n$ in the denominator, which will make the inequality true.
Let's prove it. First, fix $\varepsilon > 0$. It is my aim to prove that,
$$\limsup_{n\rightarrow \infty} \frac{a_n}{n} \le M + \varepsilon.$$
By the definition of $M$, it follows that we may find some $N \in \mathbb{N}$ such that,
$$n \ge N \implies a_{n+1} \le a_n + M + \varepsilon.$$
It therefore follows by induction that,
\begin{align*}
& a_{N+k} \le a_N + k(M + \varepsilon) \\
\implies& \frac{a_{N+k}}{N+k} \le \frac{a_N}{N + k} + \frac{k}{N + k}(M + \varepsilon).
\end{align*}
Take the limit superior of both sides, as $k \rightarrow \infty$. The left side will simplify to $\limsup\limits_{n\rightarrow\infty} \frac{a_n}{n}$, as we are considering a tail of the full sequence. The right side is convergent as $k \rightarrow \infty$, and converges to $M + \varepsilon$. Thus,
$$\limsup_{n\rightarrow\infty} \frac{a_n}{n} \le M + \varepsilon,$$
as required.
A: Thanks to Theo's answer and Steven's comment, I have a slightly different interpretation of the proof. Let $b_1=a_1$ and $b_n=a_n-a_{n-1}$ for $n>1$. Then it suffices to prove that
$$
\limsup_{n\to\infty}\frac{1}{n}\sum_{k=1}^nb_k\leq \limsup_{n\to\infty}b_n=:M<\infty
$$
Given $\epsilon>0$, there exists $N>0$ such that for all $n\geq N$, 
$$
b_n< M+\epsilon
$$
and thus for all $n\geq N$, 
$$
\frac1n\sum_{k=1}^n b_k\leq\frac{1}{n}(b_1+\cdots+b_N)+\frac1n\big[M(n-N)+\epsilon(n-N)\big]
$$
It follows that
$$
\limsup_{n\to\infty}\frac1n\sum_{k=1}^n b_k\leq M+\epsilon.
$$
Since $\epsilon$ is arbitrary, we are done. 
