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Evaluate the sum at \(x=\tfrac\pi2\) \[S(x) =\frac{1}2+\frac2\pi\sum\limits_{n=1,3,5...}^\infty\frac{\sin({n x})}n\\=\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{\sin({(2r-1) x})}{2r-1}\\ \]

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\[S\left(\tfrac\pi2\right) =\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{\sin(\pi r-\tfrac\pi2)}{2r-1}\\ =\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{(-1)^{r-1}}{2r-1}\]
looks to me a series of log(1+x) and log(1-x)

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Other answers:

also r should start in 1 and not 0
see this
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?
ln 2 ?
let me check for generating function
probably the one by @NotSoBright works just put x=i and get the sum of series ... it should work out.
\[ {1 \over 1 - x } = 1 + x + x^2 + ....\\ {1 \over 1 +x} = 1 - x + x^2 - ... \\ ------\text{ Adding }------\\ {1 \over 1 - x} + {1 \over 1 +x} = {2 \over 1 -x^2}= 2( 1 +x^2 + x^4 + ...) \\ \text{Integrating both sides} \\ {1 \over 2} \log \left( 1 + x \over 1- x\right) = x+\frac{x^3}{3}+\frac{x^5}{5}... \\ \text{Replace x by i you get} \\ \log 2 - \frac{\log \sqrt 2} 2 + i\frac \pi 2 = i \left( 1 - {1 \over 3} + \frac 15 - ... \right) \] comparing imaginary terms, you get that sum is pi/2
woops sorry ... \[ \log 2 - \frac{\log \sqrt 2} 2 + i\frac \pi 4 = i \left( 1 - {1 \over 3} + \frac 15 - ... \right) \] should have been pi/4
|dw:1355652797759:dw| putting that value you would get 1/2 + 2/pi * pi/4 = 1/2 + 1/2 = 1
this is definitely expansion of what i wrote earlier. If you have Mathematica you can try this code Plot[1/2 + 2/Pi *Sum[ Sin[(2 n - 1) x]/(2 n - 1), {n, 1, 30}], {x, 0, 2 Pi}]
do you think that is the answer they want for question 3.?
well ... i think there are lots of series that uses the convergence of Fourier series to find the value of series. like ... for example: \( \sum {1 \over n^2} , \sum {1 \over n^4} \) .. i think it is other way ... probably to show that the summation of series and the value from Fourier are equal.
lol .. it seems there are quite a lot of questions!! I would put the values in the function ..
i think your both getting the right result, i just dont follow the method, and id be surprised if this is the method they are looking for
i guess not, it's quite long
can i just assume that we know the sum of the alternating harmonic series is ln2, (in the context of Fourier series)
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
this shouldn't be difficult \[ \large z = |z|e^{i \arg(z)} \\ \log(z) = \log |z| + i\arg(z)\] as you see z = 1+i, |z| = \sqrt(2) and arg(z) = pi/4
i think i had the completely idea wrong idea of what this question was asking, because in question 1 i know S(1/2) so..
\[\begin{align*} % S(x) S(x) &=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{\sin\big((2r-1)x\big)}{2r-1}\\ \\ \\% S(π/2) S\left(\tfrac\pi2\right) &=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{\sin(\pi r-\tfrac\pi2)}{2r-1}\\ 1&=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{(-1)^{r-1}}{2r-1}\\ \frac\pi4&=\sum\limits_{r=1}^\infty\frac{(-1)^{r-1}}{2r-1}\\ \\ \end{align*}\]
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