Solving Sum (-e^(2pi i x))^k Equation: Conditions & Implications

[robertoCamera]

Homework Statement

I want to know if there are any conditions we can apply that will give us the following equality:

sum from k=1 to infinity of (-e^(2pi i x))^k = -e^(2pi i x)/ (1+e^(2pi i x))

Homework Equations

I know in Real Numbers if |r|<1 then we have sum k=1 to infinity of r^k = r/(1-r)

In the Real Numbers things fail when the magnitude is 1. Is the same true in Complex numbers?

The Attempt at a Solution

We get the equivalence for finite N via induction:

sum k=1 to 1 (-exp(2pi i x))^k=-exp(2pi i x)=(-exp(2pi i x)-exp(2pi i 2x ) ) / ( 1 +exp(2pi i x) )

Then using induction

Assuming
sum k=1 to N (-exp(2pi i x))^k=(-exp(2pi i x)-exp(2pi i (N+1)x ) ) / ( 1 +exp(2pi i x) )
we get
sum k=1 to N+1 (-exp(2pi i x))^k=(-exp(2pi i x)-exp(2pi i (N+1)x ) ) / ( 1 +exp(2pi i x) ) + (-exp(2pi i x))^(N+1)
=(-exp(2pi i x)-exp(2pi i (N+2)x ) ) / ( 1 +exp(2pi i x) )

but since |exp(2pi i x)|=1, we don't get the second term in the numerator going to 0 as N goes to infinity, so what can we do?
Relating to another post:

https://www.physicsforums.com/showthread.php?t=435992

I know that if I could use this equivalence about the geometric sum of complex exponentials then I would be able to say that the derivative of the Fourier series of f(x)=x on [-p/2,p/2) is 1 and thus the Fourier series converges to the function.

I suppose I could just look at this numerically to check, but I would still need some proof that this true in general.Thanks for any help.

 

[mighty2000]

Hello! It seems like you are interested in exploring the convergence of infinite series involving complex numbers. This is a fascinating topic and has been extensively studied in mathematics.

To answer your question, the same conditions for convergence that apply to real numbers also apply to complex numbers. In particular, the series will converge if and only if the absolute value of the ratio of consecutive terms is less than 1. This means that for your series, it will converge if |e^(2pi i x)| < 1, which is equivalent to x not being an integer.

As for your attempt at a solution, it is correct up to a certain point. However, as you mentioned, the second term in the numerator does not go to 0 as N goes to infinity, which means that the series does not converge for all values of x. This is because the condition for convergence is not satisfied for x = 1.

In general, the convergence of complex series can be quite tricky and often requires more advanced techniques such as complex analysis. If you are interested in exploring this topic further, I would recommend looking into complex analysis and studying the convergence of series in that context.

I hope this helps and good luck with your studies!