Find all real number $x$ such that the series:
$\sum_{n=1}^\infty {n x^n\over 2n^2+1}$
is absolutely convergent?
Find all real number $x$ such that the series:
$\sum_{n=1}^\infty {n x^n\over 2n^2+1}$
is absolutely convergent?
Use the limit comparison test with $\sum_{n=1}^{\infty}\frac{x^n}{n}$
Which you already know its radius of convergence(This is $\log(1-x)$).
The radius of convergence of the power series is : $R=\lim_{n\to +\infty}\frac{\frac{n}{2n^2+1}}{\frac{n+1}{2n^2+4n+3}}= \lim_{n\to +\infty}\frac{2n^3+4n^2+3n}{2n^3+2n^2+n+1}=\lim_{n\to +\infty}\frac{2+\frac{4}{n}+\frac{3}{n^2}}{2+\frac{2}{n}+\frac{1}{n^2}+\frac{1}{n^3}}=1 $ The series converges for $\left|x\right|<1$. For $x=1$, $\sum_{n=1}^{\infty}\frac{n}{2n^2+1}$ diverges by the limit comparison test with the harmonic series: $\lim_{n\to +\infty}\frac{\frac{n}{2n^2+1}}{\frac1n}=\lim_{n\to +\infty}\frac{1}{2+\frac{1}{n^2}}=\frac{1}{2}>0$ The harmonic series diverges and so does our series. For $x=-1$, $\sum_{n=1}^{\infty}\frac{(-1)^nn}{2n^2+1}$ converges by the Alternating series test. Your series converges for $-1\le x<1$
Ratio test:
$a_n:=\frac{n}{2n^2+1}|x|\Longrightarrow \frac{a_{n+1}}{a_n}=\frac{(n+1)|x|^{n+1}}{2(n+1)^2+1}\frac{2n^2+1}{n|x|^n}=$
$\frac{n+1}{n}\frac{2n^2+1}{2(n+1)^2+1}|x|\xrightarrow [n\to\infty]{}|x|<1\,\,\Longleftrightarrow\,\, -1
Now just check extreme points. For example,
$x=-1\Longrightarrow \sum_{n=1}^\infty\frac{(-1)^nn}{2n^2+1}\,\,\text{converges by Leibnitz test}$
If you use limit ratio test,
$a_n+1\over a_n$ = $(n+1)(x^(n+1))\over 2(n+1)^2+1$ $\over$ $n x^n \over 2n^2+1$
= $x2n^3+2n^2+n+1\over 2n^3+4n^2+3n$
We know that $2n^3+2n^2+n+1\over 2n^3+4n^2+3n$ converges to 1. Therefore the series \sum_{n=1}^\infty {n x^n\over 2n^2+1}$$ converges for all $x$ such that $-1\le x<1$ , $x \in \mathbb{R}$