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Prove that $$1-\dfrac 1 7+\dfrac 1 9 - \dfrac{1}{15} + \dfrac 1 {17}\mp ...=\dfrac{1+\sqrt{2}}{8}\pi$$

My attempt: I tried to break it into two series $$(1+1/9+1/17+...)-(1/7+1/15+1/23+...)$$ But I don't know how to proceed. Any hints would be appreciated.

YuiTo Cheng
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Pranav Aggarwal
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    Breaking into two series in that way can lead to $\infty - \infty$ as your original series is not absolutely convergent – Henry May 14 '19 at 08:08
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    Write it as $$1- \sum_{n=1}^{\infty} \frac 1{8n-1}-\frac 1{8n+1}$$ – For the love of maths May 14 '19 at 08:09
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    Hint: change every "1" to an appropriate power of x and then differentiate or do some other operations. – Feng May 14 '19 at 08:16
  • Related: https://math.stackexchange.com/questions/2553910/help-proving-l1-chi-frac-pi3-sqrt3 – Greg Martin May 14 '19 at 08:30
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    To be a bit less cryptic, @FengShao is suggesting that you consider the series $x-x^7/7+x^9/9-x^{15}/15+\cdots$ and relate it to a simpler series that you can work with by hand. – Greg Martin May 14 '19 at 08:32

3 Answers3

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Using the hints by Mohammad Zuhair Khan and Feng Shao, let

$$f(x):=1-\sum_{n=0}^\infty\left(\frac{x^{8n-1}}{8n-1}-\frac{x^{8n+1}}{8n+1}\right).$$

Then if we differentiate term-wise,

$$f'(x)=-\sum_{n=0}^\infty(x^{8n-2}-x^{8n}).$$

Using the geometric sum formula,

$$f'(x)=-\frac{x^6}{1-x^8}+\frac{x^8}{1-x^8}=-\frac{x^6(1-x^2)}{1-x^8}.$$

Finally,

$$f(1)=1-\int_0^1\frac{x^6(1-x^2)}{1-x^8}dx.$$

https://www.wolframalpha.com/input/?i=integrate+x%5E6(1-x%5E2)%2F(1-x%5E8)+from+0+to+1

I see no easy way to solve the integral, other than by decomposition in simple fractions, which is tedious.

  • Slightly tangential but this is a very similar integral to $\int_0^1\frac{x^4(1-x)^4}{1+x^2},\mathrm{d}x=\frac{22}7-\pi$ that appears in this proof. However this doesn't reveal much insight as that integral is just methodically evaluated with polynomial long division. – Jam May 20 '19 at 16:40
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Just for your curiosity.

Since you received good hints and a good answer, let me show how we could compute the partial sum $$S_p=1- \sum_{n=1}^{p} \frac 1{8n-1}-\sum_{n=1}^{p}\frac 1{8n+1}$$ It write $$S_p=1+\frac{1}{8} \left(\psi \left(p+\frac{9}{8}\right)-\psi \left(p+\frac{7}{8}\right)-\psi \left(\frac{9}{8}\right)+\psi \left(\frac{7}{8}\right)\right)$$ where appears the digamma function.

Using the asymptotics and continuing with Taylor series for large values of $p$ $$S_p=\frac{ \pi}{8} \cot \left(\frac{\pi }{8}\right)+\frac{1}{32 p}-\frac{1}{64 p^2}+O\left(\frac{1}{p^3}\right)$$

Computing $$S_5=\frac{106748767459}{111928041225}\approx 0.953726$$ while the above truncated series would give $$\frac{\pi}{8} \cot \left(\frac{\pi }{8}\right)+\frac{9}{1600}\approx 0.953684$$

Just remember that, using the half angle, $\tan \left(\frac{\pi }{8}\right)=\sqrt 2 -1$ makes $\cot\left(\frac{\pi }{8}\right)=\sqrt 2 +1$

Claude Leibovici
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$$1-\dfrac 1 7+\dfrac 1 9 - \dfrac{1}{15} + \dfrac 1 {17}+...=1- \sum_{n=1}^{\infty} \frac 1{8n-1}-\frac 1{8n+1}$$ $$=1-\sum_{n=1}^{\infty} \frac 2{64n^2-1}=1-\frac{1}{32}\sum_{n=1}^{\infty} \frac 1{n^2-\frac{1}{8^2}}$$ and we have $$\frac{1-\pi x \cot(\pi x)}{2x^2}=\sum_{n=1}^{\infty }\frac{1}{n^2-x^2}$$ so $$1-\frac{1}{32}\sum_{n=1}^{\infty} \frac 1{n^2-\frac{1}{8^2}}=1-\frac{1}{32}\frac{1-\frac{\pi}{8}\cot(\pi/8)}{2(\frac{1}{64})}=\frac{1+\sqrt{2}}{8}\pi$$

E.H.E
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