Given $(a_{n+1} - a_n) \rightarrow g$ show $\frac{a_n}{n} \rightarrow g$ So yea, basically this is the problem. 
$(a_{n+1} - a_n) \rightarrow g$ show $\frac{a_n}{n} \rightarrow g$
It looks like Cauchy's sequence, but I'm not sure. Can we say that if 
$(a_{n+1} - a_n) \rightarrow g$ then $|a_{n+1}-a_n| \rightarrow g$ ? 
So, for large enough $n$ we can say that $$|a_n-g|<\frac{\epsilon}{2}$$ for $\epsilon > 0$. So in this case we have $|a_{n+1}-g| + |a_n-g| <\epsilon$ 
So using triangle's inequality we get 
$$|a_{n+1}-a_n| = |a_{n+1}-g+g-a_n| \leq |a_n-g| + |a_{n+1}-g| < \epsilon$$
So... from that we know that $a_n$ is bounded by $g$.
Is it so far true? If no, how to correct it? If it is true, how to show second part of it? I would appreciate some help. Thanks in advance
 A: Use the following fact:
If $x_n\to x$, then also $\dfrac{x_1+x_2+\cdots+x_n}{n}\to x$.
In our case
$$
\frac{1}{n}\big((a_1-a_0)+(a_3-a_2)+\cdots+(a_n-a_{n-1})\big)=\frac{a_n-a_1}{n}\to g,
$$
but as $\,\dfrac{a_1}{n}\to 0$, then we conclude that $\,\dfrac{a_n}{n}\to g$.
A: Choose $N_1$ so that $|a_{n+1} - a_n - g| < \epsilon / 2$ whenever $n \geq N_1$. Then choose $N_2$ so that $|a_{N_1}| / n < \epsilon / 2$ whenever $n \geq N_2$. Then for $n \geq \max (N_1, N_2)$ you will have $a_n / n = ((a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + \ldots + a_{N_1})/n$ which will be within $\epsilon$ of $(n - N_1)g/n$ which will approach $g$ as $n \to \infty$.
A: You can use Stolz–Cesàro theorem 

Let $(a_n)_{n \geq 1}, (b_n)_{n \geq 1}$ be two sequences of real numbers. Assume that $b_n$ is strictly increasing and approaches infinity and the following limit exists:
  $$\lim_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\ell.$$ 
  Then, the limit
  $$\lim_{n \to \infty} \frac{a_n}{b_n}$$ 
  also exists and it is equal to ℓ.

Now, in your case $b_n=n$.
A: Using Stolz–Cesàro theorem:
$$\lim_{n\to\infty}\frac{a_n}{n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{(n+1)-n}
=\frac g 1 = g.$$
