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I'm trying to get an explicit formula for the series $\sum_{n=1}^\infty\frac{(-1)^n}{4n+3}$

What I tried to do was use the taylor series for the natural logarithm: $-\ln(1-z)=\sum_{n=1}^\infty \frac{z^n}{n}$

So I could put in roots of unity and look at sums like this:$$-\ln(1-e^{2\pi i/4})=\sum_{n=1}^\infty\frac{e^{2\pi in /4}}{n}=\sum_{n=1}^\infty \frac{e^{2\pi i/4}}{4n-3}+\frac{e^{2\pi i2/4}}{4n-2}+\frac{e^{2\pi i 3/4}}{4n-1}+\frac{1}{4n}$$

I'm wondering if maybe there is a way I can add some of these sums together so that everything cancels but the terms I want. I saw someone do something like this before, but I can't remember how. If it is not possible would someone show me any other way to find the first sum? I would be grateful for some help.

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  • $\begingroup$ See the addendum to this answer: math.stackexchange.com/questions/962658/… $\endgroup$ – Ron Gordon Oct 14 '14 at 21:02
  • $\begingroup$ @RonGordon Thanks, I don't understand how to do complex integrals yet but it looks cool. $\endgroup$ – user184407 Oct 14 '14 at 21:15
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Hint: use the identity $$\int_0^1 x^{n-1}dx = \frac 1n$$

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  • $\begingroup$ Oh I get it and then like put $$\int_{0}^1\sum_{n=1}^\infty(-1)^{n}x^{4n+2}dx=\sum_{n=1}^\infty\frac{(-1)^n}{4n+3}$$ So the other side can be evaluated as geometric series and then do integral? $\endgroup$ – user184407 Oct 14 '14 at 21:02
  • $\begingroup$ exactly. do not forget the proper justifications of the inversions of series and integral :) $\endgroup$ – mookid Oct 14 '14 at 21:06
  • $\begingroup$ That is a very clever, integral trick never would have thought of that. Thanks for the answer. I would click accept, but its telling me I have to wait 6 minutes. $\endgroup$ – user184407 Oct 14 '14 at 21:08

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