# Closed form expression of an exponential sum from -n to n

I'm looking for a way to write the following sum as a closed form expression:

$$\sum_{n=-N}^{N} e^{cn*i}$$

where $$c$$ is a constant and $$i$$ is the imaginary unit.

I found a closed form solution for $$n= [0, N-1]$$ at http://mathworld.wolfram.com/ExponentialSumFormulas.html , but I'm not sure how I can manipulate this to change the range from -N to N. Any help would be appreciated.

This is just a geometric sequence. The first term is $$e^{-cNi}$$ and you keep multiplying by $$e^{ci}$$. You have $$2N+1$$ terms. the sum is then $$e^{-cNi}\frac{1-e^{ci(2N+1)}}{1-e^{ci}}$$
• Sorry, can you explain why I would be multiplying by $e^{-ci}$ with each iteration? – Skipher Apr 9 at 15:04
• Sorry, its $e^{ci}$. I'll fix the answer – Andrei Apr 9 at 15:07
• My fundamental knowledge is lacking in understanding why I would multiply for each iteration of a summation. Could you explain why I multiply and where $e^{ci}$ comes from? Thank you so much your answer. – Skipher Apr 9 at 15:10
• The sum looks like $a_0+a_0r+a_0r^2+...$ Look at the ratio of two consecutive terms: $\frac{e^{c(n+1)i}}{e^{cni}}=e^{ci}$ – Andrei Apr 9 at 15:13
$$\sum_{n=-N}^N e^{cni} = \sum_{n=-N}^0 e^{cni} + \sum_{n=1}^N e^{cni} = \sum_{n=0}^N e^{-cni} + \sum_{n=1}^N e^{cni} = \frac{1-e^{-ci(N+1)}}{1-e^{-ci}}+ e^{ci}\frac{1-e^{cNi}}{1-e^{ci}}$$
$$=\frac{1-e^{-ci(N+1)}}{1-e^{-ci}}+ \frac{e^{cNi}-1}{1-e^{-ci}} = \frac{e^{cNi}-e^{c(N+1)i}}{1-e^{-ci}} = -e^{c(N+1)i}$$