Does the following series converge
$$ \sum_1^{\infty} \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{4^n 2^n n!} $$
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1A question posted here should not be phrased in language appropriate to assigning homework. And this thing is begging for a ratio test because of the way factorials cancel. – Michael Hardy May 16 '14 at 00:14
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See also: Ratio Test Series - How to solve: $\sum\limits_{n=1}^\infty \frac{1\cdot3 \dots (2n-1)}{4^n 2^n n!}$? – Martin Sleziak Dec 30 '19 at 11:03
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But maybe even easier than the ratio test: $$\frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{4^n 2^n n!} =\frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{4^n(2\cdot4\cdot6\cdots(2n))} <\frac{1}{4^n}\ ,$$ and then use the comparison test.
David
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Note that $1\times3\times\cdots\times(2n-1)=\dfrac{(2n)!}{2^nn!}$. Thus, the sum is $$\sum^{\infty}_{n=1}\dfrac{(2n)!}{2^{4n}(n!)^2}$$ Use the ratio test: $$\lim_{n\to\infty}\left|\dfrac{(2n+2)!}{2^{4(n+1)}((n+1)!)^2}\dfrac{2^{4n}(n!)^2}{(2n)!}\right|=\lim_{n\to\infty}\left|\dfrac{(2n+2)(2n+1)}{2^{4}(n+1)^2}\right|<1$$ The series converges, and sums to $\dfrac{2}{\sqrt{3}}$.
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Not only you vote to close the question, but also post an answer. No principles, eh? – May 16 '14 at 12:50