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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. If this is homework, please add the homework tag; people will still help, so don't worry. 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
@did: People who find the use of the imperative rude should stay away from mathematics; written mathematics is full of imperatives. It is plenty rude in itself to write questions that are simply quotes from homework exercises without adding thoughts of one's own; there's no need to invent an entirely fictional form of rudeness to blame to OP for additionally. – Henning Makholm Dec 16 '12 at 11:56
@did: If you find then imperative itself offensive, then how dare you use it in your own comment!? "Please consider rewriting your pose" -- "consider" is an imperative verb. On the other hand, if you think the imperative is only sometimes offensive, depending on what it's used for, then what that actually mean is that you don't actually find the imperative offensive. It is what the imperative is being used for you find offensive, and you should criticize the offensive use rather than the NON-offensive grammatical tools by which that use is made. – Henning Makholm Dec 16 '12 at 13:04
@did: "... If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post." There are three imperatives in your comment and one in the question. If the imperative is offensive in itself, then who of you and OP is the ruder one? – Henning Makholm Dec 16 '12 at 13:29
@did: I have tried to make my point as clear as I can: It is unproductive and confusing to chide new users from using a particular form of grammar when it is not in fact that form of grammar that is the problem (because that form of grammar is used all the time by everyone in written math), but the fact that they quote an exercise text without adding anything of their own. That makes it hard for the recipient of the comment to understand what he is doing wrong, and therefore makes it less likely that he will stop doing it wrong. I don't know if that qualifies as an "axe" for you. – Henning Makholm Dec 16 '12 at 13:52

4 Answers 4

up vote 2 down vote accepted

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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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 $-1\le x<1$

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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 $-1\le x<1$ , $x \in \mathbb{R}$

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This is wrong, as taking $\,x=1\,$ gives us a series that diverges by direct comparison with the harmonic series. Check this. – DonAntonio Dec 16 '12 at 11:56
Are you sure that the series converges for all $x\in \mathbb{R}$? – juniven Dec 16 '12 at 11:58
@DonAntonio Ops! you are both right. I edited my answer. – Amadeus Bachmann Dec 16 '12 at 12:00
So you made a mistake, @AmadeusBachmann?? Welcome to the club: I'm honorary president and almost co-founder of it. What's important is to mend our mistakes. I know I've learned from mine a huge lot. – DonAntonio Dec 16 '12 at 12:27
@DonAntonio Thank you so much. It is a pleasure for me to be a member of the club. = ) – Amadeus Bachmann Dec 16 '12 at 15:07

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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