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How to find the interval of convergence of the following series: $x + \frac{1}{2} x^2 + 3x^3 + \frac{1}{4}x^4 +...$

I have no idea what to proceed. Any help? Thanks!

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  • $\begingroup$ The interval of convergence will be centered on the origin, $x=0$. $\endgroup$
    – hardmath
    Commented Nov 16, 2014 at 21:42

3 Answers 3

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Hint: Apply the root test to the series. If we let $a_n$ be the sequence starting $x,\frac{1}2x^2,3x^3,\ldots$, then we wish to calculate $\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}$, knowing that if this is less than $1$, the series converges, and if it is greater than one, it diverges.

Notice that the coefficients of odd powers are always the greatest so the limit supremum equals $$\lim_{n\rightarrow\infty}\sqrt[2n+1]{|a_{2n+1|}}=\lim_{n\rightarrow\infty}\sqrt[2n+1]{(2n+1)|x|^{2n+1}}=\lim_{n\rightarrow\infty}\sqrt[2n+1]{2n+1}\cdot |x|=|x|.$$ From here, you can easily find the radius of convergence, leaving just two easy boundary cases to resolve.

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  • $\begingroup$ I am not quite following. What is a limit supremum? $\endgroup$
    – Joshua
    Commented Nov 16, 2014 at 21:37
  • $\begingroup$ It captures the notion of the "greatest" value occuring in a sequence, even when a limit does not converge. The statement $\limsup_{n\rightarrow\infty}b_n=M$ can roughly be taken to mean "For any $M'>M$ and all large enough $n$, the sequence $b_n$ cannot exceed $M'$" - it's basically the upper bound on a limit (even if the limit doesn't converge). So, here, we're saying that, if, for some $k<1$, the inequality $\sqrt[n]{|a_n|}<k$ holds for all but finitely many $n$, then the series converges, as it would be bounded above by $ck^n$ for some $c$. So, for our purposes, we ignore the smaller terms $\endgroup$ Commented Nov 16, 2014 at 21:44
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There's a formula (Cauchy) for computing the radius of convergence of arbitrary series:

$$\limsup_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{k\to\infty}\sqrt[2k]{2k}=\lim_{n\to\infty}\sqrt[n]n=1$$ It's rather clear we are only interested in odd terms because of $\limsup$ and $\sqrt[n]n\to1$ is well known, therefore the radius is $1/1=1$.

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I would recommend splitting the $x^k*k$ and the $\frac{x^k}{k}$ terms and seeing where the two intervals intersect.

$(∑(2k+1)x^{2k+1}) + (∑\frac{x^{2k}}{2k})$

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  • $\begingroup$ sorry, but im not quite following what you're trying to say. $\endgroup$
    – Joshua
    Commented Nov 16, 2014 at 21:25

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