I wanted to share this sum with you all, though I have no clue how to prove it ---
∑n=1∞[(−1)n(n!)2n2(2n)!]=2log(ϕ)log(1ϕ)\sum_{n=1}^{\infty}\left[(-1)^n \dfrac{(n!)^2}{n^2(2n)!}\right]=2\log(\phi)\log\left(\dfrac{1}{\phi}\right)n=1∑∞[(−1)nn2(2n)!(n!)2]=2log(ϕ)log(ϕ1) where ϕ\phiϕ is the golden ratio.
It has an amazing closed form, doesn't it?
Note by Pratik Shastri 6 years, 7 months ago
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Here's a proof for a lot of these( including this one ): click me
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Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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\sin \theta
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http://mathworld.wolfram.com/CentralBinomialCoefficient.HTML gives a bit of info
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Here's a proof for a lot of these( including this one ): click me