Laurent series/Cauchy product

Question

$$f(z) = \frac{z^n e^{1/z}}{1+z}, n \in \mathbb{N}$$

I have written the power series: $\frac{z^n}{1-(-z)} = \sum_{k=0}^{\infty} (-1)^{k} z^{k+n}$ valid for $|z| < 1$ (right ?)

$e^{1/z} = \sum_{k=0} \frac{1}{z^k k!}$ The formula for the Cauchy product is the following:

$$\left(\sum_{k\ge 0} a_k \right)\left(\sum_{k\ge 0}b_k \right) = \sum_{k\ge 0} \sum_{0 \le l \le k} a_l b_{k-l}.$$

so would I be having:

$$\sum_{k\ge 0} \sum_{0 \le l \le k} (-1)^{l} z^{l+n} \frac{1}{z^{k-l} (k-l)!}$$

would this be right up until now? how could I proceed to get $a_{-1}$?

Answer 1

First, you can commute the two sums to disentangle their indices in the following way : $$ \sum_{k\ge0}\sum_{0\le l\le k} \frac{(-1)^l}{(k-l)!} z^{n-k+2l} = \sum_{l\ge0}\sum_{k\ge l} \frac{(-1)^l}{(k-l)!} z^{n-k+2l} = \sum_{l\ge0}\sum_{k\ge0} \frac{(-1)^l}{k!} z^{n-k+l} $$ Now, the coefficient $a_{-1}$ will be associated to the case $n-k+l = -1$, hence $k = n+l+1$ and $$ \begin{array}{rcl} a_{-1} &=& \displaystyle \sum_{l\ge0} \frac{(-1)^l}{(n+l+1)!} \\ &=& \displaystyle \sum_{0\le l\le n} \frac{(-1)^l}{(n+l+1)!} + \sum_{l\ge n+1} \frac{(-1)^l}{(n+l+1)!} \\ &=& \displaystyle \sum_{0\le l\le n} \frac{(-1)^l}{(n+l+1)!} + \sum_{l\ge0} \frac{(-1)^{n+l+1}}{l!} \\ &=& \displaystyle \frac{(-1)^{n+1}}{e} + \sum_{0\le l\le n} \frac{(-1)^l}{(n+l+1)!} \end{array} $$ Unfortunately, the partial sum doesn't seem to possess a closed form.

Answer 2

Your caluculation is right but to get $a_{-1}$ you have to proceed defferently. $g(z):=e^{1/z}=\sum_{k=0}^\infty 1/z^kk!$ converges locally uniformly and $f_n(z):=\sum_{k=0}^n(-1)^kz^{k+n}$ converges uniformly in $|z|\leq 1-\varepsilon$ as $n$ goes to infinity, so that $f_ng$ converges to $f$ locally uniformly. In particular,the coefficient of $1/z$ in $f_ng, 1/(k+1)!-1/(k+2)!+\dots+(-1)^{-1}/(k+n+1)!$, converges to $a_{-1}$. On the other hand, its limit is $(-1)^{k+1}e^{-1}+(-1)^k\sum_0^k(-1)^i/i!$