How do you prove the series $\sum_\limits{n=0}^{\infty}(2n+1)(-x)^n$ is convergent?

The Wikipedia article about Formal Power Series states that $$S(x)=\sum_{n=0}^{\infty}(-1)^n(2n+1)x^n,$$ if considered as a normal power series, has radius of convergence 1. How do we prove this?

I understand that for positive $x$ the series is alternating, and apart from a finite number of terms the sequence $(2n+1)x^n$ is monotone decreasing in $n$ for $0\le x<1$ and tends to zero for $n\to\infty$. Thus the alternating series test applies and the series is (conditionally) convergent for $0\le x<1$. But, since this is a power series, it has to be convergent in a disk in the complex plane, and thus we conclude the series is in fact convergent for complex $x$ when $|x|<1$. Is this argument correct?

Also, I checked with Maple and I get: $$S(x)=\frac{\sqrt{\pi}(-x)^{1/4}\text{LegendreP}\left(\frac12,\frac12,\frac{1-x}{1+x}\right)}{(1+x)^{3/2}},$$ where the function in the numerator is the Legendre function of the first kind. This closed form solution, if correct, suggests that there is a branch cut for $x>0$ and $x<-1$, hence I would naively think based on this representation that true converge occurs for $-1<x<0$. Does anybody know how to prove this identity and where on the complex plane it holds?

• Be careful about the edge of the disk. Just because something occurs on the inside of the disk doesn't mean it will converge everywhere on the edge! For example, take the Taylor Series for $\frac{1}{1-x}$ – Brevan Ellefsen Nov 10 '16 at 22:47
• Yes, this is true. For now I am concerned with the interior of the disk. Even that is tricky I believe. – Andras Vanyolos Nov 10 '16 at 22:49
• Nah, it's not too tricky, you just need a stronger method. I was going to post something about the ratio test, but others seem to have already done so :) The ratio test is definitely the way to go as far as the interior is concerned. It's the boundary that is the tough part (though it's not very hard in this case. Just plug in the condition for a radius of $1$ in the complex plane, i.e. $|x|=1$ :D – Brevan Ellefsen Nov 10 '16 at 22:51
• Are you sure you wrote this right? The series you wrote down is much simpler than you write.... $S(x) = \frac{1-x}{(x+1)^2}$ – Brevan Ellefsen Nov 10 '16 at 22:58
• As for your argument, it could've converged around $x=1/2$ with radius of convergence $1/2$. – Simply Beautiful Art Nov 10 '16 at 23:28

First of all, note that your closed form seems a little off. A much simpler closed form is
$$S(x) = \frac{1-x}{(x+1)^2}$$ However, it is trivial to find the convergence in the complex plane even without this. We apply the Ratio Test, which is stronger than the alternating series test in that it can show absolute convergence. Note that the ratio test says that, for a series $\sum_{k=0}^\infty a_n$, the series coverges absolutely if $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1$. Working this out, we find that
$$\lim_{n\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{n\to\infty}\left|\frac{(-1)^{n+1}(2n+3)x^{n+1}}{(-1)^n(2n+1)x^n}\right|$$ $$=\left|\lim_{n\to\infty}\frac{(2n+3)x^{n+1}}{(2n+1)x^n}\right|=|x|$$ Thus, we have that the series converges when $|x|<1$, as you stated in your argument. To answer your question, your argument is more or less correct, and this confirms is. The boundary $|x|=1$ is an entirely different ballgame, especially working on the complex plane, and must be checked separately, which I leave to the OP if desired.

By ratio test

$$\lim_{n\to+\infty}\frac{|(-1)^{n+1}(2n+3)x^{n+1}|}{|(-1)^n(2n+1)x^n|}=|x|$$

thus if $|x|<1$, the series converges.

and

if $|x|>1$, it diverges.

the convergence radius is $R=1$.

The Ratio Test is the simplest way to show that the radius of convergence is $1$. For $|x|=1$, the terms don't go to $0$, so there is no convergence on the boundary of the disk.

• Thanks, you are right! I completely forgot about that. So roughly speaking this series converges absolutely for $|x|<1$ and as fast or slightly faster as a geometric series. On the boundary the Ratio Test doesn't say anything but we can use the closed form solution - from Brevan Ellefsen's post - to do analytic continuation and assign a finite limit to the series on the boundary, except at $x=-1$, where the only divergence occurs (2nd order pole). In fact this is analytic everywhere except that pole; a meromorphic continuation. – Andras Vanyolos Nov 11 '16 at 6:51
• As I said, it does not converge anywhere on the boundary. The fact that it can be analytically continued does not change that. Abel summability is not the same thing as convergence. – Robert Israel Nov 11 '16 at 8:25
• @Andras123 the closed form solution I provide can be analytically continued as Robert says, but recall that the closed form is not necessarily equal to the series outside of the radius of convergence (or on the boundary). You have to check the original series to know this! I again advocate looking at the example $\frac{1}{1-x}=1+x+x^2 + x^3+...$ and try plugging in $x=1$. CAN we do analytic continuation at $x=1$? Sure, we get $\zeta(1)$. However, our series is clearly divergent there. – Brevan Ellefsen Nov 11 '16 at 13:57
• @Andras123 you're thinking isn't too flawed though...if we were analyzing the rational function alone you would be right, but we're actually studying the series that is equivalent to that function (within the disk of convergence, with possible convergence on the boundary) – Brevan Ellefsen Nov 11 '16 at 14:00

The root test for me, since I know that $\lim_{n \to \infty} a^{1/n} =\lim_{n \to \infty} n^{1/n} =1$ for any $a > 0$.

Then $\lim_{n \to \infty} (2n+1)^{1/n} \le \lim_{n \to \infty} (3n)^{1/n} \le \lim_{n \to \infty} 3^{1/n}\lim_{n \to \infty} n^{1/n} =1$ and $\lim_{n \to \infty} (2n+1)^{1/n} \ge 1$ so the radius of convergence is $1$.