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This summation is related to the Leibniz pi formula and the Dirichlet L-series: $$\dfrac{\pi}{2\sqrt{2}}=\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{11}+\cdots=\dfrac{\pi}{4\sin(\pi/4)}$$

How do I evaluate: $$\dfrac{1}{1}+\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}+\cdots$$

Does the answer tend to $\dfrac{\pi}{2}=\dfrac{\pi}{4\sin(\pi/6)}$ ?

Similarly, does $$\dfrac{\pi}{\sin(\pi/8)}=\dfrac{1}{1}+\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}+\cdots?$$

Borzoi
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    Give the general forms please – Claude Leibovici Oct 29 '23 at 09:52
  • $\dfrac{\pi}{4\cdot \sin(\pi /2m)}$ Where m represents the fluctuations in the signs. By that I mean, in the Leibniz pi formula, m=1(+-+-+-). In the Dirichlet L-series, m=2(++--++--), etc. – Borzoi Oct 29 '23 at 10:10
  • The question has according to my definition, m=3 (+++---+++---) – Borzoi Oct 29 '23 at 10:17
  • Thank you ! I prefer to avoid stupid mistakes. – Claude Leibovici Oct 29 '23 at 10:23
  • I alternatively derived the L-series by extracting the real part of $\sqrt{i}\arctan(\sqrt{i})$. It's very long but essentially the generalization I gave is a guess from using this method. – Borzoi Oct 29 '23 at 10:28
  • A general method. Write the sum in terms of digamma values, then use Gauss's formulas to evaluate the digammas. – GEdgar Oct 29 '23 at 12:58
  • @GEdgar I am not familiar with the digamma functions sorry – Borzoi Oct 29 '23 at 13:06
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    Your answer in this case is $5\pi/12$. For use of digamma here, see https://en.wikipedia.org/wiki/Digamma_function#Evaluation_of_sums_of_rational_functions – GEdgar Oct 29 '23 at 13:11
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    Very much overkill: The sum is $\sum\limits_{n=1}^\infty\frac{a_n}{2n-1}$, where$$a_n=-\frac{\cos(n\pi)}3-\frac{\cos\frac{n\pi}3+\cos\frac{5n\pi}3}3+\frac{\sin\frac{n\pi}3-\sin\frac{5n\pi}3}{\sqrt3}$$so one could exploit Fourier series, e.g. this one for the $\sin$ terms. – user170231 Nov 01 '23 at 22:59

2 Answers2

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Let \begin{align*} S &= \frac{1}{1}+\frac{1}{3}+\frac{1}{5}-\frac{1}{7}-\frac{1}{9}-\frac{1}{11}+\cdots \tag{1} \\ &= \sum_{n=0}^\infty (-1)^{n-1} \left( \frac{1}{6n+1}+\frac{1}{6n+3}+\frac{1}{6n+5} \right) \\ &= \sum_{n=0}^\infty (-1)^{n-1} \int_0^1 \left( x^{6n} + x^{6n+2} + x^{6n+4} \right) \, \mathrm{d}x \\ &= \int_0^1 \sum_{n=0}^\infty (-1)^{n-1} \left( x^{6n} + x^{6n+2} + x^{6n+4} \right) \, \mathrm{d}x \\ &= \underbrace{\int_0^1 \frac{1+x^2+x^4}{1+x^6} \, \mathrm{d}x}_{I} \tag{2} \end{align*}

Now consider $$I = \int_0^1 \frac{1+x^2+x^4}{1+x^6} \, \mathrm{d}x \tag{3}$$ replace $x \to 1/x$

$$I = \int_1^\infty \frac{1+x^2+x^4}{1+x^6} \, \mathrm{d}x \tag{4}$$

After adding $(3)$ and $(4)$

\begin{align*} 2I &= \int_0^\infty \frac{1+x^2+x^4}{1+x^6} \, \mathrm{d}x \\ I &= \frac{1}{12} \int_0^\infty \frac{y^{-5/6}+y^{-1/2}+y^{-1/6}}{1+y} \, \mathrm{d}y \\ &= \frac{1}{12} \left( \beta(1/6,5/6)+\beta(1/2,1/2)+\beta(5/6,1/6) \right) \tag{5} \\ &= \frac{1}{12} \left( 2 \Gamma(1/6) \Gamma(5/6) + \Gamma(1/2) \Gamma(1/2) \right) \tag{6} \\ &= \frac{5 \pi}{12} \tag{7} \end{align*}

$5$:Using $\beta$ function
$6$:As $\beta(a,b)=\beta(b,a)$
$7$:Using Euler's reflection formula

So from $(2)$ and $(7)$ we get, $$S = \frac{5 \pi}{12}.$$

Lucky Chouhan
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$$ \begin{align} &\sum_{k=0}^\infty\textstyle\left(\frac1{12k+1}+\frac1{12k+3}+\frac1{12k+5}-\frac1{12k+7}-\frac1{12k+9}-\frac1{12k+11}\right)\tag1\\ &=\sum_{k=0}^\infty\left(\frac1{12k+1}+\frac1{12(-k-1)+1}\right)\tag{2a}\\ &+\sum_{k=0}^\infty\left(\frac1{12k+3}+\frac1{12(-k-1)+3}\right)\tag{2b}\\ &+\sum_{k=0}^\infty\left(\frac1{12k+5}+\frac1{12(-k-1)+5}\right)\tag{2c}\\ &=\frac1{12}\sum_{k=-\infty}^\infty\frac1{k+\frac1{12}}+\frac1{12}\sum_{k=-\infty}^\infty\frac1{k+\frac14}+\frac1{12}\sum_{k=-\infty}^\infty\frac1{k+\frac5{12}}\tag3\\ &=\frac\pi{12}\cot\left(\frac\pi{12}\right)+\frac\pi{12}\cot\left(\frac\pi4\right)+\frac\pi{12}\cot\left(\frac{5\pi}{12}\right)\tag4\\[6pt] &=\frac\pi{12}\left(2+\sqrt3\right)+\frac\pi{12}\cdot1+\frac\pi{12}\left(2-\sqrt3\right)\tag5\\[6pt] &=\frac{5\pi}{12}\tag6 \end{align} $$ Explanation:
$\text{(1):}$ write the sum with better notation
$\text{(2):}$ break the sum into absolutely converging pieces
$\text{(3):}$ rewrite the pieces as principal value sums
$\text{(4):}$ apply $(7)$ from this answer or $(7)$ from this answer
$\text{(5):}$ evaluate
$\text{(6):}$ simplify

Addendum $$ \begin{align} &\textstyle\overbrace{1+\frac13+\cdots+\frac1{2n-1}}^{\text{$n$ $+$'s}}\overbrace{-\frac1{2n+1}-\frac1{2n+3}-\cdots-\frac1{4n-1}}^{\text{$n$ $-$'s}}+\frac1{4n+1}+\frac1{4n+3}+\cdots\tag7\\ &=\sum_{k=0}^\infty\sum_{j=0}^{n-1}\left(\frac1{4kn+2j+1}-\frac1{(4k+2)n+2(n-j-1)+1}\right)\tag8\\ &=\sum_{j=0}^{n-1}\sum_{k=0}^\infty\left(\frac1{4kn+2j+1}+\frac1{-(4k+2)n-2(n-j-1)-1}\right)\tag9\\ &=\sum_{j=0}^{n-1}\sum_{k=0}^\infty\left(\frac1{4kn+2j+1}+\frac1{-(4k+4)n+2j+1}\right)\tag{10}\\ &=\sum_{j=0}^{n-1}\sum_{k=-\infty}^\infty\frac1{4kn+2j+1}\tag{11}\\ &=\sum_{j=0}^{n-1}\frac1{4n}\sum_{k=-\infty}^\infty\frac1{k+\frac{2j+1}{4n}}\tag{12}\\ &=\sum_{j=0}^{n-1}\frac\pi{4n}\cot\left(\frac\pi{4n}(2j+1)\right)\tag{13} \end{align} $$ Explanation:
$\text{(7):}$ alternate $n$ positive terms and $n$ negative terms
$\text{(8):}$ pair $\frac1{4kn+2j+1}$ with $-\frac1{(4k+4)n-(2j+1)}$
$\text{(9):}$ move the $-$ to the denominator
$\text{(10):}$ simplify the denominator of the right-side term
$\text{(11):}$ change the double sum to a double ended sum
$\text{(12):}$ factor out the $\frac1{4n}$
$\text{(13):}$ apply $(7)$ from this answer or $(7)$ from this answer

robjohn
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