This is a follow-up on a previous question I have asked, but since I have made some improvements, I wanted to make a new post.
I was studying 'Introduction to Electromagnetism' by David Griffiths and during his derivation of the solution to Laplace's equation in ch. 3.3, he states that the following relation is true (See eq. 3.36-3.37).
$V(x,y)=\frac{4 V_0}{\pi} \sum_{n=1,3,5…} \left[\frac{1}{n}\text{e}^{-\frac{n \pi x}{a}} \text{sin}\left(\frac{n \pi y}{a}\right) \right]=\frac{2 V_0}{\pi} \text{tan}^{-1}\left(\frac{\text{sin}(\frac{\pi y}{a})}{\text{sinh}(\frac{\pi x}{a})}\right)\tag{1}$
I want to show that this relation is true. We see that;
$\sum_{n=1,3,5…} \left[\frac{1}{n}\text{e}^{-\frac{n \pi x}{a}} \text{sin}\left(\frac{n \pi y}{a}\right) \right]=\frac{1}{2} \text{tan}^{-1}\left(\frac{\text{sin}(\frac{\pi y}{a})}{\text{sinh}(\frac{\pi x}{a})}\right)\tag{2}$
Now, the left-hand side of (2) can be rewritten as
$\text{Im} \left(\sum_{\text{n odd}}^{\infty}\frac{z^n}{n}\right)$, where $z=\text{e}^{\frac{\pi(-x+iy)}{a}}$
Because
\begin{equation}
\begin{aligned}
& \operatorname{Im}\left(\sum_{\text{n odd}} \frac{e^{n \pi (-x+i y) / a}}{n}\right)=\operatorname{Im}\left(\sum_{\text{n odd}} \frac{e^{-n{\pi x / a}} e^{n \pi i y/a}}{n}\right) \\
& =\operatorname{Im}\left(\sum_{n \text { odd }} \frac{e^{-n \pi x/ a} \cos (n \pi y/a)+i e^{-n \pi x / a} \sin (n \pi y/a.}{n}\right) \\
& =\sum_{\text{n odd}} \frac{e^{-n \pi x/a} \sin (n \pi y / a)}{n} \\
&
\end{aligned}
\end{equation}
So essentially the problem can be expressed as determining the convergence of $ \left(\sum_{\text{n odd}}^{\infty}\frac{z^n}{n}\right)$ and then take the imaginary part.
I have used the ratio test to show that this relation actually converges, but how do I determine to what value the series converges and ultimately show that (1) is true?
Best Answer
$$S_1(z)=\sum_{n=1}^{\infty}\frac{z^n}{n}=-\ln(1-z),~~~~~~~~S_2(z)=\sum_{n\in\text{odd}}^{\infty}\frac{z^n}{n}$$
First note above two series are absolute convergent for $|z|<1$, hence rearrangement doesn't change the value that they are convergent to.
$$S_2(z)=S_1(z)-\sum_{n\in\text{even}}^{\infty}\frac{z^n}{n}=S_1(z)-\sum_{k=1}^{\infty}\frac{z^{2k}}{2k}=S_1(z)-\frac12S_1(z^2)$$
hence
$$S_2(z)=-\ln(1-z)+\frac12\ln(1-z^2)=\frac12\ln\left(\frac{1+z}{1-z}\right),~~~~-1<z<1$$