1
$\begingroup$

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 ${\frac{2a}{a+1}}$?

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

2 Answers 2

1
$\begingroup$

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 $-1<a<0$, 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.

$\endgroup$
1
$\begingroup$

If $\frac{2a}{a+1} = -1$ then the series converges conditionally because it is $\sum_{n=1}^\infty \frac{(-1)^n} n. \vphantom{\dfrac{\displaystyle\sum}{}}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .