5
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ The interval of convergence will be centered on the origin, $x=0$. $\endgroup$ – hardmath Nov 16 '14 at 21:42
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I am not quite following. What is a limit supremum? $\endgroup$ – Joshua Nov 16 '14 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 $\endgroup$ – Milo Brandt Nov 16 '14 at 21:44
2
$\begingroup$

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 $1/1=1$.

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ sorry, but im not quite following what you're trying to say. $\endgroup$ – Joshua Nov 16 '14 at 21:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.