# Find $a$ Such That $\sum_{n=1}^{\infty}\frac{1}{n}(\frac{2a}{a+1})^n$ Diverges/Converges Absolutely/Conditionally

Let $$\sum_{n=1}^\infty \frac{1}{n}\left(\frac{2a}{a+1}\right)^n$$

Find $$a$$ for which the series diverges/converges absolutely/converges conditionally

diverges:

if $$\sum_{n=1}^\infty \frac{1}{n}\big(\frac{2a}{a+1}\big)^n = \sum_{n=1}^\infty \frac{1}{n}$$ it will diverge:

$$\left(\frac{2a}{a+1}\right)^n = 1\iff2a=a+1\iff a=1$$

Converges absolutely:

We have to show that $$\sum_{n=1}^\infty \frac{1}{n}\left|{\frac{2a}{a+1}}\right|^n$$ converges:

It will be sufficient to show that it pass the root-test

$$\lim_{n\to \infty} \sqrt[n]{\frac{1}{n}\big|{\frac{2a}{a+1}}\big|^n}<1\Rightarrow \lim_{n\to \infty} {\frac{1}{\sqrt[n]{n}}\big|{\frac{2a}{a+1}}\big|}<1\Rightarrow \big|{\frac{2a}{a+1}}\big|\lim_{n\to \infty} {\frac{1}{\sqrt[n]{n}}}<1\Rightarrow \big|{\frac{2a}{a+1}}\big|<1\Rightarrow \\|a|<1$$

How find $$a$$ such it will converges conditionally, the test for non negative series does not give a number, they are most bound $${\frac{2a}{a+1}}$$?

• Consider $a= - 1/3$. Commented Feb 16, 2020 at 18:04

You made a mistake on absolute convergence when you solved the inequality $$\frac{|2a|}{|a+1|}<1$$. When $$a\geq0$$, this reduces to $$\frac{2a}{a+1}<1 \implies a<1.$$ When $$-1, this reduces to $$\frac{-2a}{a+1}<1 \implies a>-\frac{1}{3}$$. When $$a<-1$$, this reduces to $$\frac{-2a}{-(a+1)}<1 \implies -2a<-a-1\implies a>1$$, a contradiction. Thus, absolute convergence occurs when $$a\in (-\frac{1}{3},1)$$. Obviously, divergence occurs when $$a=1$$.
Plugging in $$a=-\frac{1}{3}$$, the series reduces to $$\sum_n \frac{1}{n}(-1)^n$$, which converges as an alternating series since $$\frac{1}{n}$$ is decreasing.
If $$\frac{2a}{a+1} = -1$$ then the series converges conditionally because it is $$\sum_{n=1}^\infty \frac{(-1)^n} n. \vphantom{\dfrac{\displaystyle\sum}{}}$$