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Let $$\sum_{n=0}^{\infty} (2n+1)z^n$$ be a power series with $z \in \mathbb{C}$

Find the radius of convergence and show that the series is divergent for all $z \in \mathbb{C}$ on the boundary


So by using the ratio test I have found the radius of convergence to be equal to 1. For $z=-1$ and $z=1$ it's fairly straightforward to see the series diverges. Also I can see that for z with modulus 1:

$$\sum_{n=0}^{\infty} \mid (2n+1)z^n \mid = \sum_{n=0}^{\infty} (2n+1) \mid z \mid^n =\sum_{n=0}^{\infty} (2n+1) = \infty$$

So we have "absolute divergence". However as far as I know this does not imply regular divergence... So my question is how do I check for all other complex numbers on the boundary apart from 1 and -1?

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3 Answers 3

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For $|z|=1$ we have that the sequence $((2n+1)z^n)$ does not converge to $0$, since

$|(2n+1)z^n|=2n+1 \to \infty$ as $n \to \infty.$ Hence $ \sum_{n=0}^{\infty} (2n+1)z^n$ is divergent for $|z|=1.$

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  • $\begingroup$ I can see $\sum_{n=0}^{\infty} (2n+1)z^n$ does not absolutely converge to 0 but how do you know that it doesn't 'regularly' converge to 0? $\endgroup$
    – CruZ
    May 24, 2019 at 11:40
  • $\begingroup$ A series can only converge if the general term tends to zero, this is a necessary condition. @Fred is just saying that the general term does not tend to zero (a sequence converges to zero if and only if its absolute value converges to zero) and so the series cannot be convergent. $\endgroup$ May 24, 2019 at 11:46
  • $\begingroup$ I simply don't see how you see the terms not tending to zero without adding an absolute value and thus turning it into not converging absolutely $\endgroup$
    – CruZ
    May 24, 2019 at 11:52
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When $|z|=1$ you can write $z = e^{i \theta} = \cos \theta + i \sin \theta$ and the series become $$ \sum_{i=0}^{\infty}(2n+1)(\cos (n \theta) + i \sin (n \theta))= \sum_{i=0}^{\infty}(2n+1)\cos (n \theta) + i \sum_{i=0}^{\infty}(2n+1)\sin (n \theta), $$

which is divergent (for any given $\theta$, at least one of real or imaginary parts are divergent series).

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In this case, I would use the fact that $ \limsup_n |(2n+1)z^n| = \infty \neq 0$ which is necessary for the series to converge (why?).

Try to adapt this argument to the convergence on the boundary of the power series $\sum_{n \ge 0 } z^n$ as an exercise if you wish.

Let us now look at a slightly less trivial example, say power series with a general term $\frac{z^n}{2n+1}$. The convergence radius is the same, the convergence on the boundary is not. Moreover, the series diverges absolutely for $|z|=1$.

One can pass to polar coordinates, if $|z|=1$ then $z= \exp(i \theta)$ for some real parameter $\theta \in [0, 2 \pi [$ which yields $$\sum_n \frac{\exp(ni \theta)}{2n+1}.$$ Now for $z \neq 1$ (i.e. $\theta \neq 0 $) we can apply Abel's test (explain why we can do so, noting that in this case, $\sum_{N\ge n \ge 0}\exp(ni\theta)$ is bounded independently of $N$. See here for example.).


P.S. One famous but more advanced theorem in the direction of your question: https://en.wikipedia.org/wiki/Fatou%27s_theorem

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