Examine convergence of the series $$1-\frac{1}{2}+\frac{1\cdot3}{2\cdot4}-\frac{1\cdot3\cdot5}{2\cdot4\cdot6}+\cdots$$
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1Use Stirling's approximation. – Kavi Rama Murthy Jul 13 '20 at 05:08
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3@KaviRamaMurthy: I think things are much easier here ;) – metamorphy Jul 13 '20 at 05:13
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What have you tried so far? Where are you stuck? – saulspatz Jul 13 '20 at 05:20
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Let's denote $a_n=\frac{1 \cdot 3 \cdot \cdots \cdot (2n-1)}{2 \cdot 4 \cdot \cdots \cdot (2n)}$ and consider: $$\frac{a_n}{a_{n+1}}=1+\frac{1}{2n+1}$$ So by Gauss test we have absolute divergence and conditional convergence by Leibniz test and for limit 0
zkutch
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For the Leibniz test we need to show $a_n\to0$. I don't see how this follows from the above. – saulspatz Jul 13 '20 at 05:34
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@saulspatz. Added to answer (it automatically appeared to me on the right column). – zkutch Jul 13 '20 at 05:43
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You have
$$a_n=(-1)^n\frac{\prod_{k=1}^n (2k-1) } {\prod_{k=1}^n (2k) }=\binom{-\frac{1}{2}}{n}$$ and then $$S=\sum_{n=0}^\infty a_n x^n=\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n} x^n=\frac{1}{\sqrt{1+x}}$$
Claude Leibovici
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