Question

Evaluate the sum at \(x=\tfrac\pi2\)

\[S(x)
=\frac{1}2+\frac2\pi\sum\limits_{n=1,3,5...}^\infty\frac{\sin({n x})}n\\=\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{\sin({(2r-1) x})}{2r-1}\\
\]

Answer 1

\[S\left(\tfrac\pi2\right)
=\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{\sin(\pi r-\tfrac\pi2)}{2r-1}\\
=\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{(-1)^{r-1}}{2r-1}\]

Answer 2

looks to me a series of log(1+x) and log(1-x)

Answer 3

how

Answer 4

also r should start in 1 and not 0

Answer 5

hmm

Answer 6

see this
http://www.wolframalpha.com/input/?i=series+log%281%2Bx%29

Answer 7

but that has even terms in it

Answer 8

eliminate using series for log (1-x)
try it
if it not works use ix instead of x

Answer 9

im not sure how to do that

Answer 10

see that
\[\frac{1}{2}\frac{\log(1+x)}{\log(1-x)}=x+\frac{x^3}{3}+\frac{x^5}{5}...\]
Replace x by ix

Answer 11

now do you see it

Answer 12

Signs will now alternate

Answer 13

is there another way to do this?

Answer 14

@mukushla