Find :
$$\lim _{n\rightarrow +\infty }\sqrt [n] {\dfrac {\left( 2n\right) !} {n!n^{n}}}$$
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Hint: take a value of $n$, and write out couple of terms... – imranfat Apr 28 '17 at 21:22
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Working on it ! – Am ine Apr 28 '17 at 21:24
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2$$\lim _{n\rightarrow +\infty }\sqrt [n] {\dfrac {\left( 2n\right) !} {n!n^{n}}}$$ turns into $$\lim _{n\rightarrow +\infty }\sqrt[n] {\dfrac {\left( 2n\right) !} {n!}}*\frac {1}{n}$$ – Jacob Claassen Apr 28 '17 at 21:25
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@JacobClaassen How does this help? – Jaideep Khare Apr 28 '17 at 21:45
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1Denote the quantity under root sign as $b_{n} $ and find the limit of $b_{n+1}/b_{n}$ to get the answer as $4/e$. – Paramanand Singh Apr 28 '17 at 21:47
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@ParamanandSingh if you hadn't close the question,you'd have been able to add your nice comment as an answer... – lhf Apr 28 '17 at 23:13
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@lhf: no worry. I have given many answers based on that technique. You can see that such typical questions with no context should not be encouraged here. But thanks for the fact that you liked my approach. – Paramanand Singh Apr 29 '17 at 02:50
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@lhf : the same question is posted again and based on your request I have given the answer. See https://math.stackexchange.com/a/2257079/72031 – Paramanand Singh Apr 29 '17 at 03:03
1 Answers
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Use $$(2n)!=2^n \cdot n! \cdot 1 \cdot 3 \cdot 5 \ldots (2n-1)$$
Now pull $2^n$ out of the root, cancel out $n!$ and then use logarithms.
You'll get :
$$A=\sqrt[n]{\frac{(2n)!}{n!n^n}} \implies \ln A=\ln2 +\sum_{r=1}^{n}\ln\left(\frac{2r-1}{n} \right)$$
Can you proceed now (Using Riemann Integrals)?
The answer is $\dfrac{4}{e}$
Jaideep Khare
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