Who's up to the challenge? 34

Calculus Level 5

lim n [ Γ ( 1 ) Γ ( 1 2 ) Γ ( 1 3 ) Γ ( 1 n + 1 ) n + 1 Γ ( 1 ) Γ ( 1 2 ) Γ ( 1 3 ) Γ ( 1 n ) n ] \lim _{ n\to \infty } \left [ { \sqrt [ n+1 ]{ \Gamma (1)\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 3 } \right) \cdots\Gamma \left( \frac { 1 }{ n+1 } \right) } \\ \quad \quad \quad \quad \quad \quad \quad -\sqrt [ n ]{ \Gamma (1)\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 3 } \right) \cdots\Gamma \left( \frac { 1 }{ n } \right) } } \right ]

If the limit above is equal to A e B \dfrac {A}{e^B} , where A A and B B are positive integers, find A + B A+B .

Notation : Γ ( ) \Gamma(\cdot) denotes the Gamma function .


The answer is 2.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Pi Han Goh
Apr 22, 2016

Let a n = Γ ( 1 ) Γ ( 1 2 ) Γ ( 1 3 ) Γ ( 1 n ) n a_n = \sqrt [ n ]{ \Gamma (1)\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 3 } \right) \cdots\Gamma \left( \frac { 1 }{ n } \right) } and b n = n b_n = n . Then the limit can be written lim n a n + 1 a n b n + 1 b n . \lim_{n\to\infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} \; .

Since b n b_n is a strictly monotone divergent sequence, we can apply Stolz–Cesàro theorem , that is, if the following limit exists, then it must be equal to the limit in question:

lim n a n b n = lim n Γ ( 1 ) Γ ( 1 2 ) Γ ( 1 3 ) Γ ( 1 n ) n n = lim n Γ ( 1 ) Γ ( 1 2 ) Γ ( 1 3 ) Γ ( 1 n ) n n n . \lim_{n\to\infty} \dfrac{ a_n}{b_n} =\lim_{n\to\infty} \dfrac{\sqrt [ n ]{ \Gamma (1)\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 3 } \right) \cdots\Gamma \left( \frac { 1 }{ n } \right) }}n =\lim_{n\to\infty} \sqrt[n]{ \dfrac{\Gamma (1)\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 3 } \right) \cdots\Gamma \left( \frac { 1 }{ n } \right)}{n^n} }\; .

And again, if we apply Stolz–Cesàro theorem again, that is, if the following limit exists, then it must be equal to the limit in question:

lim n [ Γ ( 1 ) Γ ( 1 2 ) Γ ( 1 3 ) Γ ( 1 n ) Γ ( 1 n + 1 ) ( n + 1 ) n + 1 ÷ Γ ( 1 ) Γ ( 1 2 ) Γ ( 1 3 ) Γ ( 1 n ) n n ] = lim n [ Γ ( 1 n + 1 ) n n ( n + 1 ) n + 1 ] = lim n [ ( n + 1 ) Γ ( 1 n + 1 ) ( n n + 1 ) n ] = lim n [ ( n + 1 ) Γ ( 1 n + 1 ) a n ( 1 1 n + 1 ) n + 1 b n ÷ ( 1 1 n + 1 ) c n ] \begin{aligned} &&\lim_{n\to\infty} \left [ { \dfrac{\Gamma (1)\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 3 } \right) \cdots\Gamma \left( \frac { 1 }{ n } \right)\Gamma \left( \frac { 1 }{ n+1 } \right)}{(n+1)^{n+1}} } \div { \dfrac{\Gamma (1)\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 3 } \right) \cdots\Gamma \left( \frac { 1 }{ n } \right) }{n^n }} \right ] \\ &&= \lim_{n\to\infty} \left [ \Gamma\left( \dfrac1{n+1} \right) \cdot \dfrac{n^n}{(n+1)^{n+1} } \right ] \\ &&= \lim_{n\to\infty} \left [ (n+1) \cdot \Gamma\left( \dfrac1{n+1} \right) \cdot \left( \dfrac n{n+1} \right)^n \right ] \\ &&= \lim_{n\to\infty} \left [ \underbrace{(n+1) \cdot \Gamma\left( \dfrac1{n+1} \right)}_{a_n} \cdot \underbrace{\left( 1 - \dfrac1{n+1} \right)^{n+1}}_{b_n} \div \underbrace{\left( 1 - \dfrac1{n+1} \right)}_{c_n} \right ] \\ \end{aligned}

By logarithmic differentation, it's easy to show that b n 1 e b_n \to \dfrac1e and c n 1 c_n \to 1 . We are left to evaluate lim n [ ( n + 1 ) Γ ( 1 n + 1 ) ] \displaystyle \lim_{n\to\infty} \left [ (n+1) \cdot \Gamma\left( \dfrac1{n+1} \right)\right] . Let m = 1 n + 1 m = \dfrac1{n+1} , then as n n\to\infty , m 0 m\to 0 , the limit is equivalent to

lim m 0 m Γ ( m ) . \lim_{m\to 0} m \Gamma(m)\; .

If we represent the Gamma function in terms of its Weierstrass Representation , we have

s Γ ( s ) = e γ s k = 1 e s / k ( 1 + s k ) 1 s \Gamma(s) = e^{-\gamma s } \prod_{k=1}^\infty e^{s/k} \left( 1 + \dfrac sk \right)^{-1}

As s 0 s\to 0 , RHS tends to 1, this can be shown by logarithmic differentation. Thus lim m 0 m Γ ( m ) = 1 \displaystyle \lim_{m\to 0} m \Gamma(m) = 1 , so a n 1 a_n \to 1 .

Putting it all together, the limit in question is equal to lim n ( a n × b n ÷ c n ) = 1 × 1 e ÷ 1 = 1 e \displaystyle \lim_{n\to\infty} (a_n \times b_n \div c_n) = 1 \times \dfrac1e \div 1 = \dfrac 1e .

Hence, A = 1 , B = 1 A + B = 2 A = 1, B = 1 \Rightarrow A+B = \boxed2 .

lim m 0 m Γ ( m ) = lim m 0 Γ ( m + 1 ) = 1 \displaystyle \lim_{m \to 0} m\Gamma(m) = \lim_{m \to 0} \Gamma(m+1) = 1 (By gamma functional equation)

Ishan Singh - 5 years, 1 month ago

Log in to reply

OHHHH LOLOLOL! hahahahahah I'm so dumb!!!

Pi Han Goh - 5 years, 1 month ago

I get how you used Stolz-Cesaro the first time, but I don't get the reverse application

can you explain why this is true?

thanks

D S - 3 years ago

Log in to reply

Let L 1 L_1 be the limit in question, L 2 L_2 be the limit lim n a n b n \displaystyle \lim_{n\to\infty} \dfrac{a_n}{b_n} . And let L 3 L_3 be the limit lim n [ Γ ( 1 ) Γ ( 1 2 ) Γ ( 1 3 ) Γ ( 1 n ) Γ ( 1 n + 1 ) ( n + 1 ) n + 1 ÷ Γ ( 1 ) Γ ( 1 2 ) Γ ( 1 3 ) Γ ( 1 n ) n n ] \displaystyle \lim_{n\to\infty} \left [ { \dfrac{\Gamma (1)\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 3 } \right) \cdots\Gamma \left( \frac { 1 }{ n } \right)\Gamma \left( \frac { 1 }{ n+1 } \right)}{(n+1)^{n+1}} } \div { \dfrac{\Gamma (1)\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 3 } \right) \cdots\Gamma \left( \frac { 1 }{ n } \right) }{n^n }} \right ] .

By Stolz–Cesàro theorem, if L 2 L_2 exists, then so does L 1 L_1 . Also, if L 3 L_3 exists, then so does L 2 L_2 .

Pi Han Goh - 3 years ago

Log in to reply

I see but what happened to the nth root part

D S - 3 years ago

As b n = n b_n=n and lim n b n b n + 1 = 1 \lim_{n\to\infty} \frac{b_n}{b_{n+1} }=1 . Though the ideas (choice of b n = n b_n=n ) works but as per the reverse Stolz cesaro theorem is not correct, is it?

Naren Bhandari - 11 months ago

Log in to reply

Yeah, I'm using Theorem 3 here .

Pi Han Goh - 10 months, 3 weeks ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...