Proving that $\lim_{n\rightarrow\infty}n(a_{n+1}-a_{n})=1 \implies a_n$ diverges to $\infty$ I'm trying to prove that given a sequence $a_{n}$ such as $\displaystyle\lim_{n\rightarrow\infty}n(a_{n+1}-a_{n})=1,$ then $a_{n}$ diverges to $\infty.$
I'm lost searching a path to prove it. I believe the fact of $n(a_{n+1}-a_{n})$ is bound, because of the convergence, guides me to conclude that $(a_{n+1}-a_{n})$ converges to zero. In this point I'm searching the "connection" between de convergence of $(a_{n+1}-a_{n})$ and the way to find an "epsilon" such as $|a_{n}|>\epsilon.$
Any idea to solve this it will be very thankful.   
 A: Hint:
Define $b_n = n(a_{n+1} - a_n)$, then $a_n = \sum_{k=1}^{n-1}\frac{b_k}{k} + a_1$
So we can conclude by remarking that $\sum_{k=1}^{n}\frac{1}{k} \to +\infty$ and $b_n \to 1$
A: Define recursively $b_{n+1}=b_n+\frac{1}{n}$, and prove that $\{b_n\}_n$ increases to $+\infty$. Then apply the Stolz-Césaro Theorem to conclude that $$\lim_{n \to +\infty} \frac{a_n}{b_n}=1.$$ Therefore $\lim_{n \to +\infty} a_n=+\infty$.
A: I think you can show the sequence is not Cauchy, so it diverges: If $a_n$ was Cauchy, then there would be some N so that $|a_n -a_{k+n}|< \frac {1}{n^2}$ for $k>N$. Then $\lim_{n \rightarrow \infty} (n (a_{n+1} -a_n))<\lim_{n \rightarrow\infty} n(\frac{1}{n^2})=\lim_{n \rightarrow \infty}\frac {1}{n}=0 $. Then the sequence is not Cauchy, and so it does not converge.
A: Let $b_k=a_{k+1}-a_k$. Since 
$$\lim_n \frac{b_n}{\frac{1}{n}}=1$$
and $\sum_n\frac{1}{n}=\infty$ it follows that $\sum_n b_n$ is also diverging to $+ \infty$.
Therefor, the partial sum
$$S_n=\sum_{k=1}^n b_k=a_{n+1}-a_1$$
diverges to $+ \infty$.
A: The simplest thing is to consider that $a_n$ is (up to $a_0$) the sum of the first terms of some series (that of differences). Then, for $n>0$ 
$$
a_n=a_0+\sum_{0\leq j\leq n-1} (a_{j+1}-a_j)
$$
and apply the criterion about equivalences (see the limit comparison test here), you get $lim_{n\rightarrow +\infty}a_n=+\infty$.
