Using Stirling’s Formula, prove that \((\frac { 2n }{ n } )\sim \frac { { 2 }^{ { 2 }^{ n } } }{ \sqrt { \pi n } }.\)
Solution:
Stirling’s Formula is \(n!\sim \sqrt { 2\pi n } { (\frac { n }{ e } ) }^{ n }\)
Taking the fornula twice would give us \(n!n!\sim 2\pi n{ (\frac { n }{ e } ) }^{ 2n }\)
We can also take (2n)! which would give us \((2n!)\sim 2\sqrt { \pi n } ({ \frac { 2n }{ e } ) }^{ 2n }\)
Now we combine n!n! and (2n)! in terms of \((\frac { 2n }{ n } )\)
\(\frac { 2n! }{ n!n! } \sim (\frac { 1 }{ 2\pi n } { \frac { e }{ n } }^{ 2n })(2\sqrt { \pi n } ({ \frac { 2n }{ e } ) }^{ 2n })\)
You can simply this to get \(\frac { { 2 }^{ 2n } }{ \sqrt { \pi n } }\)
So by using Stirling’s Formula, we can prove that \((\frac { 2n }{ n } )\sim \frac { { 2 }^{ 2n } }{ \sqrt { \pi n } }\)