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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
3 years, 11 months ago

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Look at the binomial/Maclaurin expansion of $$(1-x)^{\frac12}$$ in the (valid) limit as $$x \to 1$$. · 3 years, 11 months ago

till n terms $$[\frac{1}{2} - \frac{(2(n + 1)!)}{2^{2n +1}((n + 1)!)^2}]$$ , till $$\infty$$ $$\frac{1}{2}$$ ? · 3 years, 11 months ago

Could you please give a solution? · 3 years, 11 months ago

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 $$n=\infty$$, $$S=\frac{1}{2}$$ · 3 years, 11 months ago