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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

Melo
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2 Answers2

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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).

Sayan Dutta
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    Your very last algebraic expression makes no sense: if you're taking the limit when $;n\to\infty;$ , how is it possible that the expression $;(n!)^{1/n};$ converges (?) to an expression depending on $;n;$ , namely $;\frac ne;$ ... – DonAntonio Mar 24 '22 at 18:09
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    @DonAntonio you can rephrase it as $\left|(n!)^{\frac{1}{n}} - \frac{n}{e}\right| \to 0$ (and use triangle ineq) – Gareth Ma Mar 24 '22 at 18:09
  • True, I'll do it. – Sayan Dutta Mar 24 '22 at 18:10
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    @GarethMa Oh, I know that...and it mabe justified (sometimes...!) by using the squeeze theorem or something from arithmetic of limits. The point is you can't do that. It is the same story year after year with students' exams in a first course in Calculus (engineering students, mind you) – DonAntonio Mar 24 '22 at 18:13
  • @GarethMa: the answer uses the fact that $(n!) ^{1/n}/n\to 1/e$ and this does not necessarily imply that $(n!) ^{1/n}-(n/e)\to 0$ (at least not without further analysis). – Paramanand Singh Mar 28 '22 at 16:07
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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 $$ \lim_{n\to\infty}(a_{n+2}-a_n)= 2e^{-1}. $$

xpaul
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  • Cesaro-Stolz works in reverse. So your approach does not work. Please fix it. – Paramanand Singh Mar 27 '22 at 17:34
  • I think you misinterpreted my comment. Cesaro-Stolz uses existence of limit of $(a_{n+1}-a_n)/(b_{n+1}-b_n)$ to infer existence of limit of $a_n/b_n$ and not the other way around. – Paramanand Singh Mar 28 '22 at 16:04
  • Final I got the reference. – xpaul Mar 28 '22 at 16:42
  • The reference you link says that the converse works only when $b_n/b_{n+1}$ tends to something other than $1$ but here $b_n/b_{n+1}$ tends to $1$. A solution to current problem is available in my comment to the question. – Paramanand Singh Mar 29 '22 at 01:14