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I'm trying to show that

$$1 -\frac{1}{5} + \frac{1}{9} - \frac{1}{13} + \cdots = \frac{\pi + 2\ln(1+\sqrt2)}{4\sqrt2}.$$

I thought of using the power series for $\tanh^{-1}z$ which I found was $\sum_{n=0}^\infty \frac{z^{2n+1}}{2n+1}$ and playing with $z=i^{3/2}$ but I get some unwanted terms that I can't get rid of...

Can you help me please?

Spenser
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Dollydot
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    In general, $~\displaystyle\sum_{n=0}^\infty\frac{(-1)^n}{n+a}=\frac{H\Big(\frac{a-1}2\Big)-H\Big(\frac a2-1\Big)}2,~$ where $H(k)$ is the $k^{th}$ harmonic number. – Lucian Nov 06 '14 at 21:23

2 Answers2

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HINT:

$$\sum_{n=0}^\infty\frac{(-1)^n}{4n+1}=\sum_{n=0}^\infty\int_0^1(-1)^nx^{4n}dx=\int_0^1\frac{dx}{1+x^4}$$

To evaluate this integral, see, for example, this answer.

Community
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Spenser
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You may use your idea too and write the expansion of : \begin{align} \arctan(z)&=\sum_{n=0}^\infty (-1)^n\frac{z^{2n+1}}{2n+1}\\ \operatorname{arctanh}(z)&=\sum_{n=0}^\infty \frac{z^{2n+1}}{2n+1}\\ \end{align} so that $$\frac 12(\arctan(z)+\operatorname{arctanh}(z))=\sum_{n=0}^\infty \frac{z^{4n+1}}{4n+1}$$

Noting that $\;\arctan(z)=-i\,\operatorname{arctanh}(i\,z))\,$ and searching a root of $z^4=-1$ should help you to conclude... especially if you remember that $\;\operatorname{arctanh}(x)=\dfrac{\ln(1+x)-\ln(1-x)}2$.

Raymond Manzoni
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  • How do you get rid of the $\sum \frac{e^{i\frac{\pi}{4}}}{4n+1}$ term that evaluating $\sum \frac{z^{4n+1}}{4n+1}$ at $z=e^{i\frac{\pi}{4}}$ gives? – Dollydot Nov 06 '14 at 21:25
  • Does the edit answer my question? I did remember that – Dollydot Nov 06 '14 at 21:27
  • @Dollydot: not entirely :-)... well this is merely a multiplicative constant and you may multiply by $e^{-i\frac {\pi}4}$ at the end but I didn't compute the whole thing yet... (the numerical result is right) – Raymond Manzoni Nov 06 '14 at 21:31
  • I'll try to keep going a little, maybe I can lead it to its end! – Dollydot Nov 06 '14 at 22:07
  • @Dollydot: it simplifies at the end since $\dfrac{1+e^{i\frac{\pi}4}}{1-e^{i\frac{\pi}4}}=i,(1+\sqrt{2});$ and $\dfrac{1+i,e^{i\frac{\pi}4}}{1-i,e^{i\frac{\pi}4}}=i,(\sqrt{2}-1);$ if I my computations are right but I'll agree that these computations are a bit lengthy... I am fine with the other answer anyway! :-) Excellent continuation, – Raymond Manzoni Nov 06 '14 at 22:10