given that
$$f(x)=\ln(2\cos(\frac{x}{2}))$$
has a fourier series
$$f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1} \cos(nx)}{n}$$find $$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$$

Question

given that
$$f(x)=\ln(2\cos(\frac{x}{2}))$$
has a fourier series
$$f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1} \cos(nx)}{n}$$find $$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$$

experimentx · Answer

http://www.wolframalpha.com/input/?i=1+-+1%2F2+%2B+1%2F3+-+1%2F4+%2B+1%2F5+-+... Put x=0 and get your result $$ \sum_{n=1}^{\infty}\frac{(-1)^{n+1} \cos(n \times 0 )}{n} = \ln(2\cos(\frac{0}{2}))$$

anonymous · Answer

OH Thanks... I dont know why my mind is blocked i thought too far of it

experimentx · Answer

the job of mathematics is to make simple ... 
most of the problems (very difficult one) i've faced so far had very simple solutions
the other way to solve this is to expand 
ln(1+x) and put x=1

anonymous · Answer

Yes i know, but this was a question on my test, and it said 10 marks so i couldnt think simple

experimentx · Answer

lol ... it was more like intimidation ...
i certainly find Fourier series quite difficult

anonymous · Answer

i prefer fourier series in Signals and systems lol not in mathematics