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

how

also r should start in 1 and not 0

hmm

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

but that has even terms in it

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

im not sure how to do that

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

now do you see it

Signs will now alternate

is there another way to do this?

|dw:1355650436133:dw|

ln 2 ?

let me check for generating function

|dw:1355652797759:dw|
putting that value you would get
1/2 + 2/pi * pi/4 = 1/2 + 1/2 = 1

do you think that is the answer they want for question 3.?

lol .. it seems there are quite a lot of questions!! I would put the values in the function ..

i guess not, it's quite long

yes you can!! it's pretty well known

how do i get the \(\pi/4\) bit?

Do you know complex logarithm?
\[ \log (1+i) = \log \sqrt 2 + i{\pi \over 4}\]

i have not studied complex logarithms , and they are not in the book the question has come from