real analysis - 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?

Answer

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.