How I can find limit of a given problem: $\lim_{n\to\infty}\frac1n[(n+1)(n+2)\dots(n+n)]^{\frac 1n}$?

Asked Jan 21 '16 at 18:31

Active Aug 04 '17 at 15:00

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What is the limit ?

$$\displaystyle\lim_{n\to\infty}\frac{1}{n}[(n+1)(n+2)\dots(n+n)]^{\displaystyle\frac {1} {n}}$$

The given answer is $\displaystyle\frac{4}{e}$, but I don't know how to find it or which formula is applicable here.Please help.

Thanks for help.

edited Aug 04 '17 at 15:00

Martin Sleziak

asked Jan 21 '16 at 18:31

user306934

any restrictions like Taylor series? – Alex Jan 21 '16 at 18:33
2

Bring the $1/n$ inside, get $[(1+1/n)(1+2/n)\cdots(1+n/n)]^{1/n}$. Take the log and recognize a Riemann sum. – André Nicolas Jan 21 '16 at 18:37
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Also duplicate to https://math.stackexchange.com/q/1115562/115115 from 1 year ago and certainly many others – Lutz Lehmann Jan 21 '16 at 18:44
Here's a another way to do it: This limit equals $$\lim_{n\rightarrow \infty} \left(\binom{2n}{n} \frac{n!}{n^n}\right)^{1/n}$$and $\frac{1}{2n+1}4^{n}\leq \binom{2n}{n}\leq 4^{n},$ so we see that the limit equals $$4 \lim_{n\rightarrow \infty} \left(\frac{n!}{n^n}\right)^{1/n}=\frac{4}{e}$$ by Sterling's approximation. – Eric Naslund Jan 21 '16 at 18:52
See also: How to prove that $\lim \frac{1}{n} \sqrt[n]{(n+1)(n+2)... 2n} = \frac{4}{e}$ and other posts linked there. – Martin Sleziak Aug 04 '17 at 15:00

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