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Let $(a_n)_n$ be a real, convergent, monotonic sequence. Prove that if the limit $$\lim_{n \to \infty}n(a_{n+1}-a_n)$$ exists, then it equals $0$.

I tried to apply the Stolz-Cesaro theorem reciprocal: $$\lim_{n\to \infty}n(a_{n+1}-a_n)=\lim_{n \to \infty} \frac{a_n}{1+\frac{1}{2}+\dots+\frac{1}{n-1}}=0$$ but I can't apply it since for $b_n=1+\frac{1}{2}+\dots+\frac{1}{n-1}$ we have $\lim_\limits{n \to \infty} \frac{b_{n+1}}{b_n}=1$. I also attempted the $\epsilon$ proof but my calculations didn't lead to anything useful.

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Hint. You are on the right track. Note that $a_n$ is convergent to a finite limit and the harmonic series at the denominator is divergent. Therefore $$0=\lim_{n \to \infty} \frac{a_n}{1+\frac{1}{2}+\dots+\frac{1}{n-1}}\stackrel{\text{SC}}{=}\lim_{n\to \infty}\frac{a_{n+1}-a_n}{\frac{1}{n}}=\lim_{n\to \infty}n(a_{n+1}-a_n).$$

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  • $\begingroup$ Oooh right, so since $\lim_{n \to \infty}n(a_{n+1}-a_n)$ exists and the two sequences in the fraction verify the lemma's conditions, I can directly apply Cesaro-Stolz $\endgroup$ – Shroud Nov 7 '17 at 17:50
  • $\begingroup$ Yes, the existence of the limit allow to apply ths Stolz-Cesaro Theorem. $\endgroup$ – Robert Z Nov 7 '17 at 17:54
  • $\begingroup$ Nothing wrong with this. My A reveals a little more about the Q.......+1 $\endgroup$ – DanielWainfleet Nov 7 '17 at 19:13
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We can also solve this without the Stoltz-Cesaro theorem as follows: By contradiction, suppose $\lim_{n\to \infty}n(a_{n+1}-a_n)=r>0.$ (If $r<0$ we can, WLOG, replace each $a_n$ by $-a_n).$

Then $\exists m\;(n\geq m\implies n(a_{n+1}-a_n)>r/2)$. So $\exists m\;(n\geq m\implies a_{n+1}>a_n+(r/2n)\;).$

Then $\exists m (n> m\implies a_n>a_m+\sum_{j=m}^{n-1}(r/2j)\;)$ which implies that $a_n\to \infty$ as $n\to \infty.$

Note that the conditions that $(a_n)_n$ is monotonic and convergent are unnecessary. It suffices that $(a_n)_n $ is a bounded sequence (and that $\lim_{n\to \infty}n(a_{n+1}-a_n)$ exists).

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