# Show that $\lim\limits_{n\to\infty}{b_n}=b\in\mathbb{R}$ exists

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.

• 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? – Arthur Feb 19 at 14:03

## 2 Answers

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 $$(b_n)$$.

• But how to prove b)? @daw – LiTaichi Feb 20 at 1:42

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.

• 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. – Lucas Aurélio Feb 20 at 4:25
• 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 – LiTaichi Feb 20 at 9:35