series

S=124+13246+1352468n=? S= \frac {1}{2\cdot4} + \frac {1\cdot3}{2\cdot4\cdot6} +\frac {1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} \ldots n = ?

till n terms or till infinite

#Algebra #MathProblem #Math

Note by Megh Parikh
7 years, 9 months ago

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13 votes

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Comments

Look at the binomial/Maclaurin expansion of (1x)12(1-x)^{\frac12} in the (valid) limit as x1x \to 1.

Mark Hennings - 7 years, 9 months ago

till n terms [12(2(n+1)!)22n+1((n+1)!)2] [\frac{1}{2} - \frac{(2(n + 1)!)}{2^{2n +1}((n + 1)!)^2}] , till \infty 12\frac{1}{2} ?

jatin yadav - 7 years, 9 months ago

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Could you please give a solution?

Krishna Jha - 7 years, 9 months ago

Observe that: 12.4+1.32.4.6+1.3.52.4.6.8+...n\frac{1}{2.4} + \frac{1.3}{2.4.6} + \frac{1.3.5}{2.4.6.8} + ... n terms

=12.122!+12.12.323!+...n= \frac{\frac{1}{2}.\frac{1}{2}}{2!} + \frac{\frac{1}{2}.\frac{1}{2}.\frac{3}{2}}{3!} + ... n terms

=[1(12)(1)(12)(121)2!(1)2(12)(121)(121)3!(1)3...n= [-1 - (\frac{1}{2})(-1) - \frac{(\frac{1}{2})(\frac{1}{2} - 1)}{2!}(-1)^2 - \frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-1)}{3!}(-1)^3 - ... n terms ]+12 ] + \frac{1}{2}

Hence, this is (12(1x)12)(\frac{1}{2} - (1-x)^{\frac{1}{2}}) as x1x \rightarrow 1 with nn \rightarrow \infty. At n=n=\infty, S=12S=\frac{1}{2}

A Brilliant Member - 7 years, 9 months ago
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