# Summation formula for trig functions

Does anyone know if there is a summation formula to find the sum of an expression with n as an argument in a trig function? I'm asking this because I'm learning about Fourier series/analysis but it seems that once we have the Fourier series we only sum for n=1,n=2,n=3... We never sum there series all the way to infinity. Is there a way to do that, simular to the way an integral is defiened as the Riemann sum of an infinate number of points?

If anyone can point me in a good dirrection I would sincerely appreciate it.

Thanks a lot

Ben Niehoff
Gold Member
There are some tricks that can be used, but not every series can be summed explicitly. As an example, take

$$\sum \frac1n \sin nx$$

Using complex variables, we can write this as

$$\Im \sum \frac1n e^{inx}$$

Now let $z = e^{ix}$, so we can write

$$\Im \sum \frac1n z^n$$

And now use the series for $\ln (1+z)$:

$$\ln (1+z) = z - \frac{z^2}2 + \frac{z^3}3 - \frac{z^4}4 + \ldots$$

to write the sum as

$$\Im (- \ln (1-z)) = - \arg (1-z) = - \arctan \frac{\Im (1-z)}{\Re (1-z)}$$

Finally, put in the definition of z:

$$\sum \frac1n \sin nx = \arctan \frac{\sin x}{1 - \cos x}$$

If you plot this, it reproduces the periodic ramp function represented by the original series.

To learn how it works, why don't you try working it out for a square wave? Basically all you do is use the ln(1+z) series in creative ways.