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.
 A: $\sum_{n=1}^{\infty}\alpha_{n}$ converges absolutely, means $\lim_{n\to\infty}\alpha_{n}=0$ and there for
\begin{align*}
 &\lim_{n\to\infty}\left(\frac{p}{n}+\alpha_{n}\right)=0\\
 \Rightarrow&\ln\left(1+\frac{p}{n}+\alpha_{n}\right)\sim\frac{p}{n}+\alpha_{n}\text{ (over $n\to\infty$)}\\
 \Rightarrow&1+\frac{p}{n}+\alpha_{n}\sim e^{\left(\frac{p}{n}+\alpha_{n}\right)}
\end{align*}
Then
\begin{align*}
 &\frac{a_{1}}{a_{2}}\cdots\frac{a_{n-1}}{a_{n}}=\frac{a_{1}}{a_{n}}=\left(1+\frac{p}{1}+\alpha_{1}\right)\cdots\left(1+\frac{p}{n}+\alpha_{n}\right)\\
 \sim&e^{\left(\frac{p}{1}+\alpha_{1}\right)}\cdots e^{\left(\frac{p}{n}+\alpha_{n}\right)}=e^{\sum_{k=1}^{n}\left(\frac{p}{n}+\alpha_{n}\right)}\\
 \Rightarrow&a_{n}\sim a_{1}e^{-\sum_{k=1}^{k}\left(\frac{p}{k}+\alpha_{k}\right)}\\
 =&a_{1}e^{-\sum_{k=1}^{k}\frac{p}{k}}\cdot e^{-\sum_{k=1}^{k}\alpha_{k}}
\end{align*}
Since $\sum_{k=1}^{\infty}\alpha_{n}$ converges absolutely, so $e^{-\sum_{k=1}^{\infty}\alpha_{k}}$ also converges. Let it be $M$, then we can write
$$a_{n}\sim a_{1}Me^{-\sum_{k=1}^{n}\frac{p}{k}}=a_{1}Me^{-p\sum_{k=1}^{n}\frac{1}{k}}$$
Then notice that
$$\sum_{k=1}^{n}\frac{1}{k}-\ln(n)$$
is positive and bounded: as $n\to\infty$, it tends to Euler-Mascheroni constant $\gamma=0.57721\ldots$. Therefore,
\begin{align*}
 &\sum_{k=1}^{n}\frac{1}{k}\sim\ln(n)+\gamma\\
 \Rightarrow&a_{n}\sim a_{1}Me^{-p\sum_{k=1}^{n}\frac{1}{k}}\\
 \sim&a_{1}Me^{-p\left(\ln(n)+\gamma\right)}=a_{1}Me^{-p\ln n}\cdot e^{-p\gamma}\\
 =&a_{1}Me^{-p\gamma}\cdot e^{\ln n^{-p}}=a_{1}Me^{-p\gamma}\cdot n^{-p}
\end{align*}
Thus we combine all constants $a_{1}Me^{-p\gamma}=c$, then
\begin{align*}
 a_{n}\sim a_{1}Me^{-p\gamma}\cdot n^{-p}=cn^{-p}=\frac{c}{n^{p}}
\end{align*}
A: 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 $(b_n)$.
A: 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.
