# Positive Term Series

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.

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 $$\frac {(1+/n)^{s}-1} {1/n} for $$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).
• 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 $r$. – 秦彬皓 Oct 12 '19 at 12:15