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 February 8th, 2018, 09:02 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 Thanks: 2 Ratio test complex numbers question: calculate the radius of convergence for the series: g(z)=$\sum_{n=1}^{\infty}\frac{z^n}{n^2}$ i used the ratio test to show that An+1/an converges to absolute value of z as n tends to infinity, does this mean that our radius of convergence lies between R=[0,1)? I know that if the limit is equal to 1 for An+1/an then we still do not know anything about the convergence/divergence of the series so i am slightly confused.
February 8th, 2018, 11:09 AM   #2
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 Originally Posted by Jaket1 calculate the radius of convergence for the series: g(z)=$\sum_{n=1}^{\infty}\frac{z^n}{n^2}$ i used the ratio test to show that An+1/an converges to absolute value of z as n tends to infinity, does this mean that our radius of convergence lies between R=[0,1)? I know that if the limit is equal to 1 for An+1/an then we still do not know anything about the convergence/divergence of the series so i am slightly confused.
I wanted to just say the RoC is $|z| < 1$ but it seems that this sum, also known as $polylogarithm(2,z)$ or $L_2(z)$ is able to be extended to the region of $|z|\geq 1$ via analytic continuation.

Mathematica returns values for $polylog(2,z)$ for $|z| \geq 1$

$polylog(2,1) = \dfrac{\pi^2}{6}$ is pretty well known.

A discussion at math stack exchange seems to settle on $|z| < 1$ so I'm not really sure what the correct answer is. You've opened up a nice can of worms!

February 8th, 2018, 12:33 PM   #3
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 Originally Posted by romsek I wanted to just say the RoC is $|z| < 1$ but it seems that this sum, also known as $polylogarithm(2,z)$ or $L_2(z)$ is able to be extended to the region of $|z|\geq 1$ via analytic continuation. Mathematica returns values for $polylog(2,z)$ for $|z| \geq 1$ $polylog(2,1) = \dfrac{\pi^2}{6}$ is pretty well known. A discussion at math stack exchange seems to settle on $|z| < 1$ so I'm not really sure what the correct answer is. You've opened up a nice can of worms!
The answer is that the radius of convergence for a particular power series expansion is not related to analytic continuation. In fact, that is the entire point of analytic continuation: Some analytic functions are not entire.

For example, $f(z) = \frac{1}{z^2 + 1}$ is meromorphic i.e. analytic away from $\pm i$. However, any expansion will have a finite radius of convergence. Namely, if you expand about $z = z_0$, then the radius of convergence will be exactly given by
$R = \min \{|z_0 - i|, |z_0 + i|\}$
as this is the distance to the nearest pole. This doesn't mean recentering at another point (which is exactly what analytic continuation is here) doesn't define an extension of the function to other regions of the complex plane.

February 8th, 2018, 01:03 PM   #4
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 Originally Posted by SDK The answer is that the radius of convergence for a particular power series expansion is not related to analytic continuation. In fact, that is the entire point of analytic continuation: Some analytic functions are not entire. For example, $f(z) = \frac{1}{z^2 + 1}$ is meromorphic i.e. analytic away from $\pm i$. However, any expansion will have a finite radius of convergence. Namely, if you expand about $z = z_0$, then the radius of convergence will be exactly given by $R = \min \{|z_0 - i|, |z_0 + i|\}$ as this is the distance to the nearest pole. This doesn't mean recentering at another point (which is exactly what analytic continuation is here) doesn't define an extension of the function to other regions of the complex plane.
ok that's me told, so what is the answer to the original question?

February 8th, 2018, 02:53 PM   #5
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 Originally Posted by romsek ok that's me told, so what is the answer to the original question?
The radius for his series is 1 as he computed. Also noting that the series converges for $z = 1$ it follows that the series converges on $[-1,1]$.

February 9th, 2018, 01:39 AM   #6
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 Originally Posted by SDK The radius for his series is 1 as he computed. Also noting that the series converges for $z = 1$ it follows that the series converges on $[-1,1]$.
How do u know that the series also converges on z=1? Did u just sub 1 into the original series to find the series=1/n^2 which is convergent?

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