MHB Show How to Prove Divergence of g'($\theta$)

[Dustinsfl]

$$
g'(\theta) = 2\sum_{n = 1}^{\infty}(-1)^{n + 1}\cos n\theta.
$$
How can I show that this series diverges for all values other than $\pm\frac{\pi}{2}$?

 

[chisigma]

$$
g'(\theta) = 2\sum_{n = 1}^{\infty}(-1)^{n + 1}\cos n\theta.
$$
How can I show that this series diverges for all values other than $\pm\frac{\pi}{2}$?

Using the identity $\displaystyle \cos n \theta= \frac{e^{i n \theta}+ e^{-i n \theta}}{2} \ $ You obtain that the series is the sum of two geometric series the sum of the first N terms being...

$\displaystyle S_{N}= \frac{1+ (-1)^{N}\ e^{i N \theta}}{1+e^{i \theta}} + \frac{1+ (-1)^{N}\ e^{- i N \theta}}{1+e^{- i \theta}}$ (1)

Now what is $\displaystyle \lim_{N \rightarrow \infty} S_{N}$?...

Kind regards

$\chi$ $\sigma$

 

[Dustinsfl]

Using the identity $\displaystyle \cos n \theta= \frac{e^{i n \theta}+ e^{-i n \theta}}{2} \ $ You obtain that the series is the sum of two geometric series the sum of the first N terms being...

$\displaystyle S_{N}= \frac{1+ e^{i N \theta}}{1+e^{i \theta}} + \frac{1+ e^{- i N \theta}}{1+e^{- i \theta}}$ (1)

Now what is $\displaystyle \lim_{N \rightarrow \infty} S_{N}$?...

Kind regards

$\chi$ $\sigma$

When I obtain the same common denominator for those 2 fractions, I don't get the original problem. So how are they equivalent?