What is the limit of Gamma function as n approaches infinity?

[kq6up]

Homework Statement

Find the limit of: $\frac { \Gamma (n+\frac { 3 }{ 2 } ) }{ \sqrt { n } \Gamma (n+1) }$ as $n\rightarrow \infty$.

Homework Equations

$\Gamma (p+1)=p^{ p }e^{ -p }\sqrt { 2\pi p }$

The Attempt at a Solution

Mathematica and wolfram Alpha gave the limit as 1. My solution was $\frac{1}{\sqrt{e}}$. My work is here https://plus.google.com/u/0/1096789...6080238892798007794&oid=109678926107781868876 Sorry about the photo, but my school is about to set the alarm, and I need to get out of here. Hopefully it is visible.

Thanks,
Chris

 

[Ray Vickson]

Homework Statement

Find the limit of: $\frac { \Gamma (n+\frac { 3 }{ 2 } ) }{ \sqrt { n } \Gamma (n+1) }$ as $n\rightarrow \infty$.

Homework Equations

$\Gamma (p+1)=p^{ p }e^{ -p }\sqrt { 2\pi p }$

The Attempt at a Solution

Mathematica and wolfram Alpha gave the limit as 1. My solution was $\frac{1}{\sqrt{e}}$. My work is here https://plus.google.com/u/0/1096789...6080238892798007794&oid=109678926107781868876 Sorry about the photo, but my school is about to set the alarm, and I need to get out of here. Hopefully it is visible.

Thanks,
Chris

Homework Statement

Find the limit of: $\frac { \Gamma (n+\frac { 3 }{ 2 } ) }{ \sqrt { n } \Gamma (n+1) }$ as $n\rightarrow \infty$.

Homework Equations

$\Gamma (p+1)=p^{ p }e^{ -p }\sqrt { 2\pi p }$

The Attempt at a Solution

Mathematica and wolfram Alpha gave the limit as 1. My solution was $\frac{1}{\sqrt{e}}$. My work is here https://plus.google.com/u/0/1096789...6080238892798007794&oid=109678926107781868876 Sorry about the photo, but my school is about to set the alarm, and I need to get out of here. Hopefully it is visible.

Thanks,
Chris

You wrote
\Gamma (p+1)=p^{ p }e^{ -p }\sqrt { 2\pi p }\; \Longleftarrow \; \text{FALSE!}
Perhaps you mean $\Gamma(p+1) \sim p^p e^{-p} \sqrt{2 \pi p},$, where $\sim$ means "is asymptotic to". That is a very different type of statement.

Anyway, I (or, rather, Maple) get a different final answer (=1), using the asymptotic result above. Somewhere you must have made an algebraic error.

 

[kq6up]

Sorry, yes I meant asymptotic to. Were you able to see my solution via the google plus link? Mary Boas' manual also gives a limit of one. Our solutions start out the same, but I make some approximations that seem not to be justified in her steps. However, I am unable to follow her steps. I can post that too if you are interested in taking a look at her solution.

Thanks,
Chris

 

[Ray Vickson]

Sorry, yes I meant asymptotic to. Were you able to see my solution via the google plus link? Mary Boas' manual also gives a limit of one. Our solutions start out the same, but I make some approximations that seem not to be justified in her steps. However, I am unable to follow her steps. I can post that too if you are interested in taking a look at her solution.

Thanks,
Chris

I don't have her book, and cannot really follow your screenshot. Anyway, if $r(n)$ is your ratio, and using $\Gamma(p+1) \sim c p^{p+1/2} e^{-p}$, the numerator $N(n)$ has asymptotic form
N(n) = \Gamma(n+3/2) \sim c (n+1/2)^{n+1/2 + 1/2} e^{-(n+1/2)} = c (n+1/2)^{n+1} e^{-n} e^{-1/2}
The denominator $D(n)$ has the asymptotic form
D(n) = \sqrt{n} \: \Gamma(n+1) \sim n^{1/2} c n^{n+1/2} e^{-n} = c n^{n+1} e^{-n}
Therefore,
r(n) \sim \frac{(n+1/2)^{n+1} e^{-n} e^{-1/2}}{n^{n+1} e^{-n}} = \left( 1 + \frac{1}{2} \frac{1}{n} \right)^{n+1} e^{-1/2}
Since $\lim \,(1 + a/n)^{n+1} = \lim \,(1+a/n)^n = e^a$, we are done: $\lim r(n) = 1$.

 

[kq6up]

That is pretty much her solution. However, I can actually follow yours better. I made an approximation that was not justified.

Thanks,
Chris