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divergent series

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
g'(\theta) = 2\sum_{n = 1}^{\infty}(-1)^{n + 1}\cos n\theta.
$$
How can I show that this series diverges for all values other than $\pm\frac{\pi}{2}$?
 

chisigma

Well-known member
Feb 13, 2012
1,704
$$
g'(\theta) = 2\sum_{n = 1}^{\infty}(-1)^{n + 1}\cos n\theta.
$$
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

$\chi$ $\sigma$
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
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?