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If for any $n\ge n_0, n(\frac{a_n}{a_{n+1}}-1) \gt r \gt 1,$Prove $\sum_{n=1}^\infty a_n$ converges.
(Request: Apply the following theorem: If for any $n\ge n_0, \frac{a_{n+1}}{a_n}\le \frac{b_{n+1}}{b_n},\sum_{n=1}^\infty b_n$ converges, then $\sum_{n=1}^\infty a_n$ converges.)
(Hint: take $s\in (1,r)$ and make use of $\lim\limits_{n \to \infty}{\frac{(1+\frac{1}{n})^s-1}{\frac{1}{n}}}=s$)
Giving some hint on how to constuct {$b_n$} would be great, thanks.

My Attempt:
Since $n(\frac{a_n}{a_{n+1}}-1) \gt r \gt 1$ and $s \in (1,r)$, we can get $\frac{a_n}{a_{n+1}}>1+\frac{r}{n}>1+\frac{s}{n}$ and I lost my way.

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Let $b_n=\frac 1 {n^{s}}$. Then $\sum b_n <\infty$. Let us show that $n$ sufficiently large $\frac {a_{n+1}} {a_n} < \frac {b_{n+1}} {a_n}$ for $n$ sufficiently large. (This would finish the proof).

We have $\frac {(1+/n)^{s}-1} {1/n} \to s$ as $n \to \infty$. Hence $\frac {(1+/n)^{s}-1} {1/n} <r$ for $n$ sufficiently large. I will leave it to you to get $\frac {a_{n+1}} {a_n} < \frac {b_{n+1}} {a_n}$ from this by a simple algebraic manipulation (using the fact that $\frac {a_{n+1}} {a_n} < \frac 1 {1+ \frac r n}$ which you already know).

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  • $\begingroup$ Thank you for your brilliant answer. I tried this construction but failed to find a way to go to the end. However, I see the gate towards it in your answer by using algebraic way and make use of $r$. $\endgroup$ – 秦彬皓 Oct 12 '19 at 12:15

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