Solving the limit $\lim_{n\ \rightarrow\infty}((n+2)!^{\frac{1}{n+2}}-(n)!^{\frac{1}{n}})$ given that it exists finitely I stumbled across this problem and couldn't solve it.
$$\lim_{n\ \rightarrow\infty}((n+2)!^{\frac{1}{n+2}}-(n)!^{\frac{1}{n}})$$
I tried using different expansions derived by taylor's series and tried to find any function which may satisfy or give me some hint regarding this as of now I have tried playing with $ln(x)$ and $e^x$ expansions, but couldn't find any relation to this. This seems like I have to eliminate some terms as of the negative sign and $(n+2)$ and $n$ are pretty close.
The answer given is $\frac{2}{e}$.
Any kind of help would be appreciated
 A: Using the inequalities,
$$\left(\frac n e\right)^n < n! < \left(\frac n e\right)^{n+1}$$
and
$$\left(1+\frac 1n\right)^n < e < \left(1+\frac 1n\right)^{n+1}$$
it is easy to show that
$$\frac{(n!)^{\frac 1n}}{n} \to \frac 1e$$
or
$$(n!)^{\frac 1n} \to \frac ne$$
Now, replace $n$ by $(n+2)$ to get an approximation for the first term, and substract the two to get your answer.

As DonAntonio pointed out, the lasrt algebraic expression is not rigorous. To make it rigorous, we can rephrase it as
$$\left|(n!)^{\frac{1}{n}} - \frac{n}{e}\right| \to 0$$
and use triangle Inequality (as pointed out by Gareth Ma).
A: You can use reverse Stolz' theorem Converse of Stolz-Cesaro: If $\lim_{n\to\infty}\frac{a_n}{b_n}=l\neq1$ and $\lim_{n\to\infty}\frac{b_n}{b_{n+1}}=L\neq1$, then
$$ \lim_{n\to\infty}(a_{n+1}-a_n)=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n\to\infty}\frac{a_n}{b_n}=l. $$
In fact, let $a_n=(n!)^{\frac1n}, b_n=n$. Then
Using
$$\lim_{n\to\infty}\ln\left(\frac{a_n}{b_n}\right)=\lim_{n\to\infty}\frac1n\ln(\frac{n!}{n})=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\ln(\frac{k}{n})=\int_0^1\ln xdx=-1 $$
one has
$$ \lim_{n\to\infty}(a_{n+1}-a_n)=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n\to\infty}\frac{a_n}{b_n}=e^{-1}. $$
So
$$ \lim_{n\to\infty}(a_{n+2}-a_n)= 2e^{-1}. $$
