Proposition

Find a function representing each of the following series.

$\sum_{k \geq 0} \frac{z^{2k}}{k!}$
$\sum_{k \geq 1} k(z - 1)^{k - 1}$
$\sum_{k \geq 2} k(k - 1)z^k$

Solution

1

Based on the Taylor series for $e^z$, it is easy to see that this series represents $e^{z^2}$.

2

$\frac{1}{1 - (z - 1)} = \sum_{k \geq 0} (z - 1)^k$. By taking the derivative on both sides, we obtain $\frac{1}{(2 - z)^2}$.

3

$\frac{1}{1 - z} = \sum_{k \geq 0} z^k$. By taking the derivative twice and multiplying $z^2$, we obtain $\frac{2z^2}{(1 - z)^3} = \sum_{k \geq 2} k(k - 1)z^k$.