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Knowing that $1 - \frac 12 + \frac 13 - \cdots = \ln 2$ and $1 - \frac 13 + \frac 15 - \cdots = \frac{\pi}{4}$, compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}$. I programmed and found that $\frac{\ln 2}{4}+{\frac{\pi}{4}}/2$ is precise to the eighth digit.

Michael Hardy
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user94602
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    You've already done the hard part. So what happens if you take (a) your sequence for $ln 2$, divide it by two, (b) your sequence for $\pi/4$, divide it by four, then (c) add them? – Richard Rast Sep 12 '13 at 23:41
  • The Maple 17 command $$sum(1/(4n+1)-1/(4n+2), n = 0 .. infinity); $$ produces $$1/4,\ln \left( 2 \right) +1/8,\pi

    $$

     – user64494 Sep 13 '13 at 04:29

3 Answers3

16

$$ \begin{align} \log(2)&=\sum_{n=0}^\infty\frac1{4n+1}-\frac1{4n+2}+\frac1{4n+3}-\frac1{4n+4}\\ \frac12\log(2)&=\sum_{n=0}^\infty\hphantom{-\frac1{4n+2}}\frac1{4n+2}\hphantom{\;+\frac1{4n+3}}-\frac1{4n+4}\\ \frac\pi4&=\sum_{n=0}^\infty\frac1{4n+1}\hphantom{-\frac1{4n+2}}-\frac1{4n+3}\\ \frac12\left(\log(2)-\frac12\log(2)+\frac\pi4\right)&=\sum_{n=0}^\infty\frac1{4n+1}-\frac1{4n+2} \end{align} $$

robjohn
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9

$$\begin{aligned}\sum_{n\geq 0}\frac{1}{4n+1}-\frac{1}{4n+2} =\sum_{n\geq 0}\int_0^1 t^{4n}(1-t)\,dt=\int_0^1 \frac{t-1}{t^4-1}\,dt=\frac{\ln 2}{4}+\frac{\pi}{8}\end{aligned}$$

L. F.
  • 8,498
3

$$ {\cal F}\left(x\right) \equiv \sum_{n = 0}^{\infty}{x^{4n + 2} \over \left(4n +1\right)\left(4n + 2\right)} $$ \begin{align} {\cal F}''\left(x\right) &\equiv \sum_{n = 0}^{\infty}x^{4n} = {1 \over 1 - x^{4}} = {1/2 \over 1 - x^{2}} + {1/2 \over 1 + x^{2}} = {1 \over 4}\sum_{\sigma = \pm}{1 \over 1 + \sigma\,x} + {1/2 \over 1 + x^{2}} \\[3mm] {\cal F}'\left(x\right) &\equiv {1 \over 4}\sum_{\sigma = \pm}\sigma\,\ln\left(1 + \sigma\,x\right) + {1 \over 2}\arctan\left(x\right) \\[3mm] {\cal F}\left(x\right) &\equiv {1 \over 4}\sum_{\sigma = \pm}\sigma\left\lbrack% x\ln\left(1 + \sigma\,x\right) - \int_{0}^{x}x'\,{\sigma \over 1 + \sigma\,x'}\,{\rm d}x' \right\rbrack + {1 \over 2}\,x\,\arctan\left(x\right) - {1 \over 2}\int_{0}^{x}{x' \over x'^{2} + 1}\,{\rm d}x' \\[3mm]&= {1 \over 4}\sum_{\sigma = \pm}\sigma\left\lbrack% x\ln\left(1 + \sigma\,x\right) - x + \sigma\ln\left(1 + \sigma\,x\right) \right\rbrack + {1 \over 2}\,x\arctan\left(x\right) - {1 \over 4}\,\ln\left(1 + x^{2}\right) \\[5mm]& \end{align}

$$ {\cal F}\left(x\right) = {1 \over 4}\left\lbrack% \left(-x + 1\right)\ln\left(1 - x\right) + \left(x + 1\right)\ln\left(1 + x\right) \right\rbrack + {1 \over 2}\,x\arctan\left(x\right) - {1 \over 4}\,\ln\left(1 + x^{2}\right) $$

$$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% \lim_{x \to 1^{-}}{\cal F}\left(x\right) = \sum_{n = 0}^{\infty}{1 \over \left(4n +1\right)\left(4n + 2\right)} = {1 \over 8}\,\pi + {1 \over 4}\,\ln\left(2\right)\quad} \\ \\ \hline \end{array} $$

Felix Marin
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