# Let $a_n \to 0$, then $\lim_{n \to \infty} (a_{n+1}-a_n)n$ equals to ??

Let $$a_n \to 0.$$

Then $$\lim_{n \to \infty} (a_{n+1}-a_n)n$$ equals to ??

I have taken a few examples and got that the limit equals to zero. It seems that the limit is zero, but how to prove it in general ?

or my guess is wrong...

Please provide some hint. Thank you.

• Interestingly, this limit is true under the stronger assumption that $a_n$ is summable and monotonic. – Theo Bendit May 6 '19 at 4:12
• @Zhanxiong with your example limit is zero. – Eklavya May 6 '19 at 4:14
• @Zhanxiong Your counter example is apparently wrong, the limit is $0$ in your case. – StAKmod May 6 '19 at 4:14

Even if your sequence is monotone, this may not be true. If you consider $$a_{k}=\frac{1}{n}$$ for $$n^3\leq k<(n+1)^3$$, then $$\lim_{n\to\infty}(a_{n^3}-a_{n^3-1})(n^3-1)=\lim_{n\to\infty}\left(\frac{1}{n}-\frac{1}{n-1}\right)(n^3-1)=\lim_{n\to\infty}\frac{n^3-1}{n-n^2}=-\infty$$ So, $$\{(a_{n^3}-a_{n^3-1})(n^3-1)\}$$is a divergent subsequence of $$\{(a_{n+1}-a_n)n\}$$.
No general conclusion here. For example let $$a_n$$ be the sequence $$0,1/2,0,1/4,0,1/6,\cdots$$ You have $$(a_{2k+1}-a_{2k})(2k)=-\frac{2k}{2k}=-1$$ and $$(a_{2k}-a_{2k-1})(2k-1)=\frac{2k-1}{2k}\to1$$
Your guess is wrong. Consider $$a_n=(-1)^n/\sqrt{n}, n\geq 1$$ Then $$n(a_{n+1}-a_n)=(-1)^{n+1}n(1/\sqrt{n+1})+1/\sqrt{n}).$$
However, $$n(1/\sqrt{n+1}+1/\sqrt{n})>2\sqrt{n}\to +\infty$$, which excludes $$(-1)^{n+1}n(1/\sqrt{n+2}+1/\sqrt{n+1})\to 0$$.