# Limit of $b_n$ when $b_n=\frac{a_n}{n}$ and $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=l$ [duplicate]

I have the following problem.

I have to find the limit of $b_n = \frac{a_n}{n}$, where $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=l$

My approach:

I express $a_n$ in terms of $b_n$, i.e. $a_n=nb_n$ and $a_{n+1}=(n+1)b_n$

We look at the difference: $a_{n+1}-a_n=(n+1)b_{n+1}-nb_n$

Assuming that $b_n$ converges to a real number m, we see that:

$l=(n+1)m-nm$, from where I conclude that $m=l$.

What I'm left with is proving that $b_n$ is convergent which I'm not sure how to do.

## marked as duplicate by Robert Wolfe, Lord Shark the Unknown, Chris Custer, Namaste, Frank W.Jan 8 at 15:52

• The answer section does not make sense. Take for instance $a_n = n$ so that $l = 1$, and $b_n = 1 \ne e$. – angryavian Sep 21 '17 at 16:24
• You still have to prove that $(b_n)_n$ is convergent. – mechanodroid Sep 21 '17 at 16:34
As @ParamanandSingh suggested, from Stolz–Cesàro theorem $$a_{n+1}-a_n=\frac{a_{n+1}-a_n}{(n+1)-n} \rightarrow l, n \rightarrow \infty$$ where $\{n\}_{n \in \mathbb{N}}$ is monotone and divergent, then $$\frac{a_n}{n} \rightarrow l, n \rightarrow \infty$$