What is the limit of sequence $a(n)$= $\left(\frac{1.3.5....(2n-1)}{2.4.6....(2n)}\right)$.
I calculate $\lim \left(\frac{a(n+1)}{a(n)}\right)$ which comes 1.So I am confirm that limit of $a(n)$$\ne$$0$.But what is actual limit?
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1It is explained here very briefly. – Nov 24 '17 at 21:29
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All answers have used concept of stirling approximation which I haven't studied yet...any easy way.. – ogirkar Nov 24 '17 at 21:39
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Really all? Did you have a look at this answer https://math.stackexchange.com/a/1025669/42969 ? – Martin R Nov 24 '17 at 21:41
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Not all..I also look at this but sorry didn't get this one too.. – ogirkar Nov 24 '17 at 21:45
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Easy with Bernoulli's inequality: we'll prove the inverse tends to $+\infty$. Indeed \begin{align} \left(\frac{2\cdot4\cdot6\cdots(2n)}{1\cdot3\cdot5\cdots(2n-1)}\right)&= \Bigl(1+\frac{1}{1}\Bigr)\Bigl(1+\frac{1}{3}\Bigr)\Bigl(1+\frac{1}{5}\Bigr)\dotsm \Bigl(1+\frac{1}{2n-1}\Bigr) \\\ &\ge 1+\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\dotsm +\frac{1}{2n-1}, \end{align} and the latter tends to $+\infty$.
Bernard
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