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I am solving a question from book in which I have to use summation. It is as follows:
$$ \frac{1}{10}\sum_{n=0}^{9} e^{-jk\omega_0n} $$ The value of $\omega_0$ is $\frac{2\pi}{10}$. What I have done so far is:
$$ \frac{1}{10} \left(\frac{1-e^{-9jk\frac{2\pi}{10}}.e^{jk\frac{2\pi}{10}}} {1-e^{-jk\frac{2\pi}{10}}}\right)$$
After this step I am stuck and can not figure out the next step?
This sum appears quite often in DSP.
\begin{align} \sum_{n=0}^{N-1} \exp(-j\alpha n) &\stackrel{(a)}{=} \frac{1- \exp(-j\alpha N)}{1 - \exp(-j\alpha )}\\ &= \frac{e^{-j\alpha N/2}(e^{+j\alpha N/2}- e^{-j\alpha N/2})}{e^{-j\alpha /2}(e^{+j\alpha /2}- e^{-j\alpha /2})}\\ &\stackrel{(b)}{=} e^{-j\frac{\alpha}{2}(N-1)} \frac{\sin(N\alpha/2)}{\sin(\alpha/2)} \end{align} where the last step $(b)$ comes from identity $e^{+j\theta}-e^{-j\theta} = 2j\sin(\theta)$ and be careful at $(a)$ as Matt's comment.
Substitute $\alpha=k\omega_0=k\frac{2\pi}{N}$, $\sin(N\alpha/2)=\sin(k\pi)=0$ if $k\in\mathbb{Z}$ hence so is the sum.
It is trivial that $\exp(-j\alpha n) \stackrel{\alpha=0}{=} 1$ then the sum equals $N$.
It can be thought as the case of the final result after $(b)$ when $\alpha \to 0$ then $$e^{-j\frac{\alpha}{2}(N-1)} \frac{\sin(N\alpha/2)}{\sin(\alpha/2)} \stackrel{\alpha \to 0}{\to} N$$ where the limit is evaluated by L'Hôpital's rule.
Further interesting discussion can be found in Laurent Duval's answer.
As a good answer has been given by @AlexTP, I'll go from the detail to a larger picture.
Although the "De L'Hospital's rule" seems attractive, it should be handled with a lot of care (see my answer to Why does L'Hopital's rule fail in this case?), because it requires somehow strong conditions!
I do recommend to use weaker conditions instead, when one can. And with sines of positive arguments around zero $(0\le x\le\epsilon)$, luckily, one can the standard simple bounds:
$$ x(1-x^2/6) \le \sin(x) \le x$$
which suffice to bound the sine ratio $\frac{\sin(N\alpha/2)}{\sin(\alpha/2)}$ as $\alpha \to 0$ with $N$ given. More precisely, for both $0\le \alpha/2\le\epsilon$ and $0\le N\alpha/2\le\epsilon$, we have
$$ N\alpha/2(1-(N\alpha/2)^2/6) \le \sin(N\alpha/2) \le N\alpha/2$$ and $$ \frac{1}{\alpha/2} \le \frac{1}{\sin(\alpha/2)} \le \frac{1}{\alpha/2(1-(\alpha/2)^2/6)}$$
thus
$$ N(1-(N\alpha/2)^2/6) \le \frac{\sin(N\alpha/2)}{\sin(\alpha/2)} \le \frac{N}{(1-(\alpha/2)^2/6)}\,,$$ whose central term thus tends to $N$ as $\alpha \to 0_+$. And you can plug it into AlexTP solution.
What is important here is that the underlying function of $k$ (just suppose $k \in \mathbb{R}$) is obviously continuous, as it is a finite sum of ($N=10$) continuous functions $f_n(k)=\exp(-jk w_0 n)$ (this could not work with infinite sums, as an infinite sum of continuous functions might be discontinuous, especially if the convergence is not uniform).
So, you can conveniently use the geometric sum formula anywhere it holds (with a non-zero denominator), and just use the limit (as the function IS continuous) otherwise.
This sum of an exponential geometric series is important: it is the discrete equivalent to the Fourier transform of a continuous rectangular window. The former yields a sinc, a cardinal sine. Your formula, for a discrete rectangular window, yield a (phased) discrete sinc called aliased sinc (asinc) or periodized sinc (psync), with details on the JOS website signal processing site.
