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series

\( S= \frac {1}{2\cdot4} + \frac {1\cdot3}{2\cdot4\cdot6} +\frac {1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} \ldots n = ?\)

till n terms or till infinite

Note by Megh Parikh
4 years, 2 months ago

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13 votes

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

Mark Hennings - 4 years, 2 months ago

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

Jatin Yadav - 4 years, 2 months ago

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Could you please give a solution?

Krishna Jha - 4 years, 2 months ago

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Observe that: \(\frac{1}{2.4} + \frac{1.3}{2.4.6} + \frac{1.3.5}{2.4.6.8} + ... n\) terms

\(= \frac{\frac{1}{2}.\frac{1}{2}}{2!} + \frac{\frac{1}{2}.\frac{1}{2}.\frac{3}{2}}{3!} + ... n\) terms

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

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

Paramjit Singh - 4 years, 2 months ago

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