1

Gregory and Leibniz formula

(1)

$$-\sum_{m=1}^{\infty}\frac{(-1)^m}{2m-1}=\frac{\pi}{4}$$

We found another series equivalent to (1)

This is expressed in term of square numbers

$$-\sum_{m}^{\infty}\frac{\sum_{n=1}^{3m-2}(-1)^nn^2 }{\sum_{n=1}^{3m-2}(+1)^nn^2}=\frac{\pi}{4}$$

$$\frac{1^2}{1^2}+\frac{1-2^2+3^2-4^2}{1+2^2+3^2+4^2}+\frac{1-2^2+3^2-4^2+5^2-6^2+7^2}{1^2+2^2+3^2+4^2+5^2+6^2+7^2}+\cdots=\frac{\pi}{4}$$

Is there another series equivalent to (1) but expressable in term of cube numbers?

  • Yes, the case $m=1$ of http://math.stackexchange.com/questions/850442/an-interesting-identity-involving-powers-of-pi-and-alternating-zeta-series – Jack D'Aurizio May 18 '16 at 04:31
  • I was hoping the LHS be of cube and the RHS still be $\frac{\pi}{4}$, but anyway accept the answer (+1) –  May 18 '16 at 07:51

2 Answers2

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A proof of mahdi's result: $$ \sum_{k=1}^{n}k^3 = \frac{n^2(n+1)^2}{4}\tag{1}$$ $$ \sum_{k=1}^{2n}(-1)^{k+1} k^3 = -n^2(4n+3),\qquad \sum_{k=1}^{2n-1}(-1)^{k+1} k^3 = n^2(4n-3)\tag{2}$$ lead to:

$$\begin{eqnarray*}\sum_{n\geq 1}\frac{1^3-2^3+\ldots}{1^3+2^3+\ldots+n^3} &=& \sum_{m\geq 1}\frac{-(4m+3)}{(2m+1)^2}+\sum_{m\geq 1}\frac{(4m-3)}{(2m-1)^2}\\&=&3+\sum_{m\geq 1}\frac{(4m-3)-(4m-1)}{(2n-1)^2}\\&=&3-2\sum_{m\geq 1}\frac{1}{(2n-1)^2}=\color{red}{3-\frac{\pi^2}{4}}.\tag{3}\end{eqnarray*}$$

Jack D'Aurizio
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    Thank you @Jack D'Aurizio, I didn't expect that today, nice proof. –  May 18 '16 at 12:58
  • @mahdi: it is also intersting to notice that by replacing cubes with fifth powers we get $$15-\frac{1}{2} \pi \left(3 \pi +3 \text{Sec}\left[\frac{\sqrt{3} \pi }{2}\right]-4 \sqrt{3} \text{Tan}\left[\frac{\sqrt{3} \pi }{2}\right]\right)$$ – Jack D'Aurizio May 18 '16 at 14:31
  • Thank you @jack D'Azurio, beautiful, I don't know how you did it but I am going to post another page of your proposing formula with the closed form, please can write the prove also. –  May 18 '16 at 15:51
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In term of cube numbers:

$$\frac{1^3}{1^3}+\frac{1^3-2^3}{1^3+2^3}+\frac{1^3-2^3+3^3}{1^3+2^3+3^3}+\frac{1^3-2^3+3^3-4^3}{1^3+2^3+3^3+4^3}+\cdots=3-\frac{\pi^2}{4}$$