Showing a series converges absolutely

Question

The goal is to prove that if $|\frac{c_{n+1}}{c_n}|\leq1+\frac{a}{n}$, where $a<-1$ and $a$ does not depend on $n$, then the series $\sum_{n=1}^\infty c_n$ converges absolutely.

My idea: to have the series converge absolutely, then we want to show that there is some $N\in\mathbb{N}$ such that $|\frac{c_{n+1}}{c_n}|\rightarrow n>N$ uniformly. I was wanting to play with the idea that $n=1$ would work, but then no such $N$ could exist. I was thinking about maybe using the Ratio test, but I wasn't able to get it to go anywhere. Moreover, I am assuming the whole "$a<-1$ and $a$ doesn't depend on $n$" bit is pretty important (why I was thinking about playing with $n=1$, but I'm not quite sure how that fits in. Any thoughts or help would be greatly appreciated :)

Answer 1

Let $c_n$ be a sequence such that

$$\left|\frac{c_{n+1}}{c_n}\right|\le \left(1+\frac an\right)\tag1$$

for some number $a<-1$ and $a$ does not depend on $n$.

Let $k$ be a positive integer $k\ge2 $ such that $-k<a<-1$. Then, we see from $(1)$ that for $n>k$

$$\begin{align} |c_{n+1}|&\le \left(1+\frac an\right)|c_n|\\\\ &\le \left(1+\frac an\right)\left(1+\frac a{n-1}\right)|c_{n-1}|\\\\ &\vdots\\\\ &\le \prod_{m=0}^{n-k} \left(1+\frac{1}{n-m}\right)|c_k|\\\\ &=|c_k|\exp\left(\sum_{p=k}^{n}\log\left(1+\frac{a}{p}\right)\right)\\\\ &\le |c_k|\exp\left(\sum_{p=k}^{n}\frac{a}{p}\right)\\\\ &\le |c_k|e^{\left(a \log(n/k)\right)}\\\\ &=|c_k|k^{|a|}\frac1{n^{|a|}} \end{align}$$

Inasmuch as the series $\sum_{n=1}^\infty \frac1{n^{|a|}}$ converges for $|a|>1$, the series $\sum_{n=1}^\infty |c_n|$ converges and hence the series $\sum_{n=1}^\infty c_n$ converges absolutely.

Answer 2

One method is to use Raabe's Test, a refinement of the Ratio Test. Let $S=\sum_n c_n$ be a series of positive terms. If it exists, let $$ \rho =\lim_{n\to\infty} n\left(\frac{c_n}{c_{n+1}}-1\right) $$Then $$ \begin{cases} \rho >0 & S \text{ converges}\\ \rho <0 & S \text{ diverges}\\ \rho =0 & \text{ (inconclusive)}\\ \end{cases} $$Since we know something about the ratios, we have $$ \lim_{n\to\infty} n\left(\frac{c_n}{c_{n+1}}-1\right) \geq \lim_{n\to\infty} n\left((1+a/n)^{-1}-1\right)=\lim_{n\to\infty}-\frac{a n}{a+n} = -a>1 $$ So your series converges.

• https://mathworld.wolfram.com/RaabesTest.html