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Show that

a) if $ \frac{b_{n}}{b_{n+1}}=1+\beta_{n}, n=1,2, \dots $, and the series $ \sum\limits_{n=1}^{\infty} \beta_{n} $ converges absolutely, then the limit $\lim\limits_{n\to\infty}{b_n}=b\in\mathbb{R} $ exists;

b) if $ \frac{a_{n}}{a_{n+1}}=1+\frac{p}{n}+\alpha_{n}, n=1,2, \dots $, and the series $ \sum\limits_{n=1}^{\infty} \alpha_{n} $ converges absolutely, then $a_n \sim \frac{c}{n^p}$ as $n\to\infty$

c)if the series $\sum\limits_{n=1}^{\infty}{a_n}$ is such that $\frac{a_{n}}{a_{n+1}}=1+\frac{p}{n}+\alpha_{n}$ and the series $ \sum\limits_{n=1}^{\infty} \alpha_{n} $ converges absolutely, then $ \sum\limits_{n=1}^{\infty} a_{n} $ converges absolutely for $p > 1$ and diverges for $p\leq1$ (Gauss' test for absolute convergence of a series).

Obviously,we can get c) from b).My question is how to prove b)

I found this problem in "Mathematical Analysis I" of Zorich, just Google Books and turn to page 149 of the preview. I believe this problem is somehow incorrectly stated (since I can't solve it :) ) , so I asked for sure.

Thanks.

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    $\begingroup$ These are three problems in one. We usually prefer posts to focus on a single problem. Also, what have youtried? Where are you stuck? What theorems have you tried, and what tools do you have at your disposal? $\endgroup$ – Arthur Feb 19 at 14:03
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For (a): first we have $$ \sum_n |\frac{b_n}{b_{n+1}}-1| = \sum_n |\beta_n| < \infty $$ hence $\frac{b_n}{b_{n+1}} \to 1$. If $(b_n)$ is bounded, we are done: if $|b_n|\le M$ for all $n$, then $$ \sum_n |\frac{b_n}{b_{n+1}}-1| \ge \frac1M \sum_n |b_n - b_{n+1}| . $$ To prove boundedness, we take the inverse of the defining equation and multiply. Since $\beta_n\to 0$ there is $K$ such that $\beta_n\ge-\frac12$ for all $n\ge K$. Then we get $$ \frac{b_{N+1}}{b_K} = \prod_{n=K}^N \frac{b_{n+1}}{b_n} =\prod_{n=K}^N (1-\frac{\beta_n}{1+\beta_n}). $$ Taking absolute values $$ |\frac{b_{N+1}}{b_K}| = \prod_{n=K}^N |1- \frac{\beta_n}{1+\beta_n}| \le \prod_{n=K}^N (1+ \frac{|\beta_n|}{1+\beta_n}) \le \prod_{n=K}^N (1+ 2 |\beta_n|). $$ As $\sum|\beta_n|<\infty$, the product on the right-hand side is bounded for all $N$, and so is $(b_n)$.

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  • $\begingroup$ But how to prove b)? @daw $\endgroup$ – LiTaichi Feb 20 at 1:42
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Since your series $\{\alpha _n \}$ converges then $\forall \epsilon > 0 \exists n_0 \in \N$ such that for $n > n_0$, $|\alpha _n|<\epsilon$. Now note that

$\tag{1}\frac{c/n^p}{c/(n+1)^p} = \left( 1 + \frac{1}{n}\right)^p = \sum_{k=0}^{p}{p \choose k}n^{-k}$

implying

$\tag{2}\left|\frac{a_n}{a_{n + 1}}- \sum_{k=0}^{p}{p \choose k}n^{-k}\right| = \left|\frac{a_n}{a_{n + 1}}- \left( 1+\frac{p}{n}\right) - \sum_{k=2}^{p}{p \choose k}n^{-k}\right| < \left|\frac{a_n}{a_{n + 1}}- \left( 1+\frac{p}{n}\right)\right| < \epsilon$

Thus

$\tag{3}\lim \left[\frac{a_n}{a_{n + 1}}- \left( 1+\frac{p}{n}\right)\right] = \lim \left[\frac{a_n}{a_{n + 1}}- \frac{c/n^p}{c/(n+1)^p}\right] = 0$

which is your desired result.

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  • $\begingroup$ I typed from a phone and left some details. As soon as I get hand on my computer I add those, but I think you can clean up the proof. $\endgroup$ – Lucas Aurélio Feb 20 at 4:25
  • $\begingroup$ How to get $\left|\frac{a_n}{a_{n + 1}}- \left( 1+\frac{p}{n}\right) - \sum_{k=2}^{p}{p \choose k}n^{-k}\right| < \left|\frac{a_n}{a_{n + 1}}- \left( 1+\frac{p}{n}\right)\right| < \epsilon$, maybe $\frac{a_n}{a_{n + 1}}- \left( 1+\frac{p}{n}\right)<0$?@Lucas Aurélio $\endgroup$ – LiTaichi Feb 20 at 9:35

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