Determine sum function and area of convergence the Laurent series $\sum_{n=1}^\infty \frac{z^{-4n}}{4n} + \sum_{n=0}^\infty \frac{z^{4n}}{(4n)!}$.

Question

I am trying to compute the sum function and area of convergence of $\sum_{n=1}^\infty \frac{z^{-4n}}{4n} + \sum_{n=0}^\infty \frac{z^{4n}}{(4n)!}$, and to determine the singularities of the sum function on the boundary of the area of convergence.

I have made some progress for the second term I think, notice that $$ \cos(z) = \sum_{n=0}^\infty \frac{(-1)^n z^{2n}}{(2n)!} = 1-\frac{z^2}{2!} + \frac{z^4}{4!} - \ldots, $$ with $R = \infty$, and $$ \cosh(z) = \sum_{n=0}^\infty \frac{z^{2n}}{(2n)!} = 1 + \frac{z^2}{2!} + \frac{z^4}{4!} + \ldots, $$ with $R = \infty$. From this it can be seen that the second sum is given by $$ \sum_{n=0}^\infty \frac{z^{4n}}{(4n)!} = \frac{\cos(z) + \cosh(z)}{2}. $$ This has area of convergence all of $\mathbb{C}$.

I am stuck for the first term. My idea is to use the geometric series, something like $$ \frac{1}{1-z^{-1}} = \sum_{n=0}^n z^{-n}, $$ and differentiating this would give something of the form $\sum_{n=0}^n -n z^{-n}$, but this seems not to work out.

score 1 · Accepted Answer · answered May 09 '20 at 10:51

1

$$\sum_{n=1}^\infty\dfrac{(z^{-4})^n}n=-\ln(1-z^{-4})$$

which is convergent if $|z^{-4}|<1$ and $z^{-4}=-1$

See What is the correct radius of convergence for $\ln(1+x)$?

answered May 09 '20 at 10:51

lab bhattacharjee

279,016

But how to get $4n$ in the denominator instead of $n$? – Sigurd May 09 '20 at 12:06
@Sigurd, $$\sum_{n=1}^\infty\dfrac{z^{-4n}}{4n}=\dfrac14\cdot\sum_{n=1}^\infty\dfrac{(z^{-4})^n}n$$ – lab bhattacharjee May 09 '20 at 12:13

Determine sum function and area of convergence the Laurent series $\sum_{n=1}^\infty \frac{z^{-4n}}{4n} + \sum_{n=0}^\infty \frac{z^{4n}}{(4n)!}$.

1 Answers1