# Series Expansion of $\frac{1}{1-e^{int}}$

Is there a better way to deal with the series expansion, in particular? I'm interested in attempting to use the series expansion to estimate the value, which I will eventually combine with more complicated expressions.

So, to begin, I calculated the following expansion for this series (note that $n$ is an integer, $i$ is the imaginary number, and this is a series of complex $t$):

$\frac{1}{1-e^{int}}$ = $-\frac{1}{int}\sum_{k=0}^\infty{\left(-\sum_{j=0}^\infty\frac{(int)^{j+1}}{(j+2)!}\right)^k}$

The problem occurs when I attempt to expand the series by summing $k$ and $j$ through values less than infinity. For instance, if I run through the values 0 to 2 for both variables, I get a maximum power of $t$ when both are 2. This gives a term for $t^6$. Unfortunately, I'm missing the rest of the coefficient values for this term, and so it's inaccurate. In fact, I've found that the expansions will always be inaccurate because the series takes a term to the $k$th power. It's like I have to find a way to run through each term of the series expansion, instead of simply going through and pluuging in a value for infinity. In other words, I need a way to "traverse" through the series term by term, instead of by a simple expansion.

I'm really just looking for an accurate way to partially expand this series, but I want to actually work with the series. To make this clear - I want to attempt to use a series expansion. The one I have is ackward to work with. So I'd like to find another way to deal with series expansions, if possible. Any help would be greatly appreciated.

What I'm asking is for a better way to deal with the series expansion. Perhaps a new series? ...or perhaps a new way to get terms of the existing series.

Using that $$\frac{t}{e^t-1}=\sum_{k\geq0}B_k\frac{t^k}{k!},$$ where $B_k$ is a Bernoulli number, you can get a Taylor series for your function by doing a little massaging to this series. Now, it will contain Bernoulli numbers...
To actually compute, it may be simpler and better to just compute $e^{int}$ and then use the geometric series $(1-e^{int})^{-1}=\sum_{k\geq0}e^{inkt}$.
• But this last series is divergent for every real $nt$... – Did Feb 17 '11 at 22:27
• @Didier: well, I did not say to compute on which values of $t$ :=) – Mariano Suárez-Álvarez Feb 17 '11 at 22:39