Limit

[Someone2841]

Can someone explain to me how to solve for this limit? (Or at least point me in the right direction)

\(\displaystyle \underset{n \to \infty}{\lim}\left ((n+1)!^\frac{1}{n+1} - n!^\frac{1}{n}\right )\)

According to mathematica, the solution is \(\displaystyle \frac{1}{e} \), but I'm not sure as to how to arrive at that solution.

 

[galactus]

Are you familiar with Stirling's formula?.

It is one way to approach this. That, and a little algebra.

Stirling says that \(\displaystyle n! \sim n^{n}e^{-n}\sqrt{2\pi n}\)

So, using this in the limit and rewriting:

\(\displaystyle \left((n+1)^{\frac{1}{n+1}}e^{-(n+1)}\sqrt{2\pi (n+1)}\right)^\frac{1}{n+1}-\left(n^{n}e^{-n}\sqrt{2\pi n}\right)^{\frac{1}{n}}\)

\(\displaystyle =(n+1)e^{-1}(2\pi(n+1))^\frac{1}{2(n+1)}-ne^{-1}(2\pi n)^{\frac{1}{2n}}\)

\(\displaystyle \displaystyle \frac{(n+1)^{\frac{1}{2(n+1)}+1}\cdot (2\pi)^{\frac{1}{2(n+1)}}-(2\pi)^{\frac{1}{2n}}\cdot n^{\frac{1}{2n}+1}}{e}\)

Now, take note that as \(\displaystyle n\to \infty\), the exponents of 2Pi \(\displaystyle \to 0\)

and the exponents of the n and n+1 tend to 1.

So, this leaves \(\displaystyle (n+1)-n=1\)

Along with the e in the denominator, we get \(\displaystyle \frac{1}{e}\)

As I said, this is one way to approach it.