Solving the limit $\lim_{n\ \rightarrow\infty}((n+2)!^{\frac{1}{n+2}}-(n)!^{\frac{1}{n}})$ given that it exists finitely

Question

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

Answer 1

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 $$ \lim_{n\to\infty}(a_{n+2}-a_n)= 2e^{-1}. $$

Answer 2

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.

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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).