# For which x values, this series is convergent?

Find all real number $x$ such that the series:

$$\sum_{n=1}^\infty {n x^n\over 2n^2+1}$$

is absolutely convergent?

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After 9 questions on the site, it might be the time to begin to say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Did Dec 16 '12 at 11:53

Ratio test:

$$a_n:=\frac{n}{2n^2+1}|x|\Longrightarrow \frac{a_{n+1}}{a_n}=\frac{(n+1)|x|^{n+1}}{2(n+1)^2+1}\frac{2n^2+1}{n|x|^n}=$$

$$\frac{n+1}{n}\frac{2n^2+1}{2(n+1)^2+1}|x|\xrightarrow [n\to\infty]{}|x|<1\,\,\Longleftrightarrow\,\, -1<x<1$$

Now just check extreme points. For example,

$$x=-1\Longrightarrow \sum_{n=1}^\infty\frac{(-1)^nn}{2n^2+1}\,\,\text{converges by Leibnitz test}$$

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If you use limit ratio test,

$a_n+1\over a_n$ = $$(n+1)(x^(n+1))\over 2(n+1)^2+1$$ $\over$ $$n x^n \over 2n^2+1$$

= $x$$2n^3+2n^2+n+1\over 2n^3+4n^2+3n We know that 2n^3+2n^2+n+1\over 2n^3+4n^2+3n converges to 1. Therefore the series$$\sum_{n=1}^\infty {n x^n\over 2n^2+1}$$converges for all x such that -1\le x<1 , x \in \mathbb{R} - The radius of convergence of the power series is :$$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}$$converges by the Alternating series test. Your series converges for -1\le x<1 - Use the limit comparison test with$$\sum_{n=1}^{\infty}\frac{x^n}{n}$$Which you already know its radius of convergence(This is$\log(1-x)\$).

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