# Radius of convergence and sum of alternating series $1 - z + z^2 - z^3 + \ldots$

I have a (complex) function represented by the power series

\begin{equation*} L(z) = z -\frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} \ldots \end{equation*}

which I have tried to represent (perhaps incorrectly) in summation notation as

\begin{equation*} L(z) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n \;z^{(n+1)}}{n+1} .\end{equation*}

To find the radius of convergence I considered the ratio of terms:

\begin{equation*} \left| \dfrac{a_{n+1}}{a_n} \right| = \left| \frac{(-1)^{n+1}\;z^{(n+2)}}{(n+2)} \right| \left| \frac{(n+1)}{(-1)^n\; z^{(n+1)}} \right|\end{equation*}

which simplifies to

\begin{equation*} \left| \frac{z\;(n+1)}{(n+2)}\right| = \left| z \right| \left| \frac{(\frac{n}{n}+\frac{1}{n})}{(\frac{n}{n}+\frac{2}{n})}\right| \rightarrow \left| z \right| \text{ as } n \rightarrow \infty. \end{equation*}

By the ratio test, $L(z)$ will converge for $\left| z \right| < 1$.

I believe that the derivative $L'(z)$ of my original series will have the same radius of convergence. Differentating term-by-term I have

\begin{equation*} L'(z) = 1 - z + z^2 - z^3 + \ldots \end{equation*}

I do not believe that this is a geometric series due to the alternating sign. My attempt to find a formula for the sum is as follows. Multipling by $z$ gives me:

\begin{equation*} z\;L'(z) = z - z^2 + z^3 - z^4 + \ldots \end{equation*}

I can take the $r$th partial sums of the previous two series to get

\begin{align*} &1 - z + z^2 - z^3 + \ldots + (-1)^r\;z^r \\ &+ \\ &z - z^2 + z^3 - z^4 + \ldots + (-1)^r\;z^{(r+1)} \\ &= 1 + (-1)^r\;z^{(r+1)}. \end{align*}

Thus I can say that

\begin{equation*} L'(z) + zL'(z) = \lim_{n \rightarrow \infty}1 + (-1)^n\;z^{(n+1)} \end{equation*}

where the last term tends to zero within the R.O.C. ($|z| < 1$). So finally, by factoring on the left and dividing through I get:

\begin{equation*} L'(z) = \frac{1}{1+z} \end{equation*}.

Can anyone tell me if I've done the above correctly, and if there was a quicker way of jumping to the the final sum?

Edit: Also, I've been told that the ratio test is only to be used on series with positive terms - why is it okay in this alternating series?

• Yes, $1-z+z^2-z^3+\cdots$ is a geometric series.
– anon
Commented Apr 17, 2015 at 23:02
• Oh no. I thought I had to worry more because of the alternating sign. Was there a quick answer I should know by heart? Commented Apr 17, 2015 at 23:05

Yes, you have done this correctly, and you are correct that $\dfrac{1}{1+z}=1-z+z^2-z^3+\cdots$ for $\vert z\vert<1$. If you are willing to take the formula for the sum of a geometric series for granted, then that can lead to a slightly quicker answer (although you have essentially derived the formula in your argument). The series $1-z+z^2-z^3+\cdots$ is, in fact, a geometric series with common ratio $-z$, so the sum of the series (for $\vert z\vert<1$) is $\dfrac{1}{1-(-z)}=\dfrac{1}{1+z}$. That is, geometric series are allowed to have negative common ratios.

• In fact, the series does not necessarily have a negative common ratio. The common ratio is $-z$, so if you took $z$ to be, say, $-\frac 12$, then you would get the series $1+\frac 12+\frac 14+\frac 18+\cdots$. Commented Apr 17, 2015 at 23:12
• Aha, the negative common ratio is what threw me. I knew the general answer $\dfrac{a}{1-z}$ for non-alternating, so I guess in this case I should have seen $a=1$ and ratio $-z$. I was also a little confused about the ratio test, as my solution sheet says that the ratio of coefficients is $\dfrac{n+1}{n}$ where I have $\dfrac{n+1}{n+2}$. I could have stated the series differently, but the answer still looks upside down. Also, I've been told that the ratio test only works for series with positive terms.. If anyone can tell me why it's okay here, that would be great. Thanks for your answer. Commented Apr 17, 2015 at 23:42
• The Ratio Test works for series with complex terms in general; not just positive real terms. It states that a series $\displaystyle{\sum_{n=0}^\infty}a_n$ will converge if $\displaystyle{\lim_{n\rightarrow\infty}\vert\frac{a_{n+1}}{a_n}\vert<1}$. Because you take the absolute value of the ratio, it doesn't matter if the terms of the series are positive, negative, or even nonreal. Commented Apr 18, 2015 at 4:09