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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
    
@HenningMakholm I see what you mean and I am not sure you are entirely right. Imperative is not rude in a specific context (statement of homework exercises) and it is definitely rude in what ought to be the context of a question asked here. Hence mentioning the use of the imperative does point exactly, albeit in an indirect way, to the very problems you mention (and I think this was exactly Arturo's approach when he wrote this comment). To sum up, although I do not intend to stay away from mathematics, I find rude the use of the imperative here. –  Did Dec 16 '12 at 12:04
    
@did: That's like saying that you find the use of the letter "e" objectionable in the question because the "e"s in the question happen to be used to state a bad question. I agree that the question is bad, but it is not bad because there are "e"s in it, nor because it contains imperatives. The letter "e" and the imperative form are both completely innocent devices; just because they are used for bad purposes here don't make them bad in themselves. –  Henning Makholm Dec 16 '12 at 12:18
    
Is not going to help you explain to the OP why his question offends you if you keep pointing to a completely non-offensive grammatical feature during your explanation. It will merely confuse the reader. Point to what is actually bad about the question, without confusing the matter with irrelevant references to the incidental use of entirely unobjectionable grammatical forms. –  Henning Makholm Dec 16 '12 at 12:18
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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
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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
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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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