#2015_23 Riemann is hiding?

Calculus Level 3

If

f ( n ) = 1 n { ( 2 n + 1 ) ( 2 n + 2 ) ( 2 n + n ) } 1 / n , f\left( n \right) =\frac { 1 }{ n } { \big\{ (2n+1)(2n+2)\ldots(2n+n) \big\} }^{ 1/n },

then what is lim n f ( n ) ? \displaystyle \lim _{ n\rightarrow \infty }{ f\left( n \right) }?


More questions related to KVPY:

4 e \frac 4e 27 4 e \frac{27}{4e} 4 e 4e 27 e 4 \frac{27e}4

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1 solution

Otto Bretscher
Sep 5, 2015

We have f ( n ) = k = 1 n ( 2 + k n ) 1 / n f(n)=\prod_{k=1}^{n}\left(2+\frac{k}{n}\right)^{1/n} , with ln ( f ( n ) ) \ln(f(n)) = k = 1 n ln ( 2 + k n ) 1 n =\sum_{k=1}^{n}\ln{\left(2+\frac{k}{n}\right)}\frac{1}{n} , a (right) Riemann sum of ln ( x ) \ln(x) between 2 and 3. Thus lim n ln ( f ( n ) ) \lim_{n\to\infty}\ln(f(n)) = 2 3 ln ( x ) d x = [ x ln ( x ) x ] 2 3 =\int_{2}^{3}\ln(x)dx=[x\ln(x)-x]_{2}^{3} = 3 ln ( 3 ) 2 ln ( 2 ) 1 =3\ln(3)-2\ln(2)-1 . Now lim n f ( n ) = e lim n ln ( f ( n ) ) = 3 3 2 2 e = 27 4 e , \lim_{n\to\infty}f(n)=e^{\lim_{n\to\infty}\ln(f(n))}=\frac{3^3}{2^2e}=\boxed{\frac{27}{4e}}, by the continuity of e x e^x .

Riemann sum is a useful tool for infinite summations! Even I did it in the same way. @Anandhu Raj can u change the name of the questions as they r not meant only for kvpy aspirants. The questions r for the entire community to solve. An attractive name would be better.

Aditya Kumar - 5 years, 9 months ago

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What about "Riemann is hiding?" ;) Does that give away too much?

Otto Bretscher - 5 years, 9 months ago

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Excellent sir, Both answer as well as name suggestion. :)

Anandhu Raj - 5 years, 9 months ago

Sorry, we can't edit the title once the question is posted :(

Anandhu Raj - 5 years, 9 months ago

Nice solution.

Hana Wehbi - 4 years, 2 months ago

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