Inspired by Calvin Lin!

Calculus Level 4

lim n i = 0 n ( 2 n ) ! ( i ! ) 2 ( ( n i ) ! ) 2 2 n = ? \large{ \lim_{n\to\infty} \sqrt[2n]{ \sum_{i=0}^n \frac{ (2n)!}{ (i!)^2 ((n-i)!)^2} } = \ ? }


The answer is 4.

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1 solution

Satyajit Mohanty
Jul 11, 2015

And then it follows to the problem lim n ( 2 n n ) 1 / n = ? \large \lim_{n \rightarrow \infty} { 2n \choose n } ^ { 1 / n } = \ ?

Check this problem by @Calvin Lin : Inspired by Ravi Dwivedi , for its solution.

And our answer then turns out to be 4 \fbox 4 .

I was waiting for the solution!

Aditya Kumar - 5 years, 9 months ago

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I had added the solution since July 11.

Satyajit Mohanty - 5 years, 9 months ago

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