Sum of Series (1 Viewer)

Paradoxica · Aug 8, 2016

Use the generating function for the binomial coefficients and integrate once from 0 to x. Manipulate all the individual terms of the series into Central Binomial coefficient form.

Then plug in x = 1/4 to obtain S = 1

juantheron · Aug 8, 2016

Thanks Paradoxica would you like to explain me in detail.

KingOfActing · Aug 10, 2016

Paradoxica · Aug 10, 2016

KingOfActing said:
$\text{We begin by inspecting the series expansion for } \sqrt{1-2x}:\\\sqrt{1-2x} = (1-2x)^\frac12 = \sum_{k = 0}^{\infty} {\frac12\choose k}(-2x)^k = \sum_{k=0}^{\infty} {2k \choose k}\frac{(-1)^{k+1}(-2x)^k}{4^k(2k-1)}\\\\-\sqrt{1-2x} = \sum_{k=0}^{\infty} \frac{x^k(2k)!}{2^kk!^2(2k-1)}\\1-\sqrt{1-2x} = \sum_{k=1}^{\infty}\frac{(2k)!}{2^kk!}\frac{1}{k!(2k-1)}x^k = \sum_{k=1}^{\infty}\frac{(2k-1)!!}{k!(2k-1)}x^k = \sum_{k=1}^{\infty}\frac{(2k-3)!!}{k!}x^k\\\\\text{Now, letting }x = \frac{1}{2} \text{ we obtain:}\\\frac{1}{2} +\frac{1}{2!}\left(\frac{1}{2}\right)^2+\frac{1 \times 3}{3!}\left(\frac{1}{2}\right)^3+\frac{1 \times 3 \times 5}{4!}\left(\frac{1}{2}\right)^2+\cdots = 1 - \sqrt{1-2\times\frac{1}{2}} = 1$

#GeneralisedBinomialTheorem

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KingOfActing · Aug 10, 2016

Paradoxica said:
#GeneralisedBinomialTheorem

#NonStandardHashTags

#TheHashTagsAreSelfAware

Paradoxica · Aug 10, 2016

KingOfActing said:
#TheHashTagsAreSelfAware

#RunWhileYouStillCan

juantheron · Aug 12, 2016

Thanks KingOfActing.

Sum of Series (1 Viewer)

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