# Simplifying trigonometric summation: $x[n] = \sum_{k = 1}^{\infty} (-1)^{k} \frac{\sin \left( 2 \pi k M \frac{a}{b} n \right)}{\pi k}$

I was reading an engineering publication and attempting to follow the math and got stuck at this "easy to show but somewhat lengthy" step. The author starts with $$x[n] = \sum_{k = 1}^{\infty} (-1)^{k} \frac{\sin \left( 2 \pi k M \frac{a}{b} n \right)}{\pi k}$$ where $a,b,M,k,n$ are integers. The claim is that when letting $$\hat{b} = \frac{b}{\gcd(M,b)}$$ and if $\hat{b}$ is even, then $$x[n] = \sum_{k=1}^{\hat{b}/2} \frac{(-1)^k}{\hat{b} \tan \left( \pi k / \hat{b} \right)} \sin \left( 2 \pi \frac{k \hat{M} a}{\hat{b}} n \right).$$ I tried splitting into partial sums but could come up with nothing absolutely convergent. I saw several questions that refer to complex analysis when dealing with these types of summation series, which I have no experience in. If that is the route required then I will grab a text. Wanted to know if there was some "easy" or "obvious" thing that I missed. I will note that the author said he used the "antisymmetry" of sine to accomplish this reduction.

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Also, I should have noted that $a$ and $b$ are coprime. – dcdo Jan 19 '13 at 17:32

Not sure what your author is trying to do. The sum is simple to evaluate in terms of the given parameters as follows.

Write the sum as

$$f(y) =\sum_{k=1}^{\infty} (-1)^k \frac{\sin{\pi y k}}{\pi k}$$

Then if we take the derivative of $f$, the sum is easy to evaluate:

$$f'(y) = \sum_{k=1}^{\infty} (-1)^k \cos{\pi y k}$$

Write in terms of a complex exponential:

$$f'(y) = \Re{\sum_{k=1}^{\infty} (-1)^k e^{i \pi y k} }$$

This is simply a geometric series with sum

$$f'(y) = -\Re{\frac{e^{i \pi y}}{1+ e^{i \pi y}}}$$

which can be rewritten as

$$f'(y) = -\frac{1}{2}$$

Integrating with respect to $y$:

$$f(y) = -\frac{y}{2}+ C$$

with $C$ being a constant of integration, which may be found to be zero upon noting that $f(0)=0$. Therefore, using the constants provided in lieu of $y$, we get that the sum is

$$x[n] = -\pi M \frac{a}{b} n$$

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The author is trying to derive the number of "tones" to be expected over one period of n. When $k$ goes from 1 to $\infty$ is looks like we should expect an infinite number of spectra, but he is arguing that the summation simplifies to $\hat{b}$ number of tones. Also, I am not sure I can take the derivative and compute in that manner since $x[n]$ is a discrete function and $f(y)$ is a continuous function. Perhaps I can better phrase the question? – dcdo Jan 19 '13 at 17:26
@dcdo: what I did was a trick that allowed me to manipulate the sum so I could evaluate it. $x[n]$ being discrete has no bearing; in fact, the final result still references $n$. As for the author, what is the publication? – Ron Gordon Jan 19 '13 at 17:54
"Noise Spectra of Digital Sine-Generators Using the Table-Lookup Method", by Soenke Mehrgardt, IEEE Transactions on Acoustics, Speech and Signal Processing, Volume 31, Issue 4, August 1983. I downloaded through IEEExplore through my university – dcdo Jan 19 '13 at 18:56
Thanks. I do not have access, but I'll try to see the context somehow. – Ron Gordon Jan 19 '13 at 18:58
I can take the equations and put them into MATLAB for a sanity check. I could not find it with a Google search, so apparently it is only downloadable from behind a paywall. I will report back the results. – dcdo Jan 19 '13 at 19:53