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find the limit of $\large\frac{n}{\sqrt[n]{n!}}$ using the ratio test

$$\Large \frac{\left(\frac{(n+1)^{n+1}}{(n+1)!}\right)^\frac{1}{n+1}}{\left(\frac{n^n}{n!}\right)^\frac{1}{n}}$$

I have added $\frac{n}{n}$ and $\frac{n+1}{n+1}$ to both components and got:

$$\Large\frac{\left(\frac{(n+1)^{n+1}}{(n+1)!}\right)^\frac{n}{n(n+1)}}{\left(\frac{n^n}{n!}\right)^\frac{n+1}{n(n+1)}}$$

and then

$$\Large\frac{\left(\frac{(n+1)^{n+1}}{(n+1)!}\right)^\frac{n}{n(n+1)}}{\left(\frac{n^n}{n!}\right)^\frac{n}{n(n+1)}}+\frac{\left(\frac{(n+1)^{n+1}}{(n+1)!}\right)^\frac{n}{n(n+1)}}{\left(\frac{n^n}{n!}\right)^\frac{1}{n(n+1)}}$$ $$\Large=\left(\frac{\frac{(n+1)^{n+1}}{(n+1)!}}{(\frac{n^n}{n!})}\right)^\frac{n}{n(n+1)}+\frac{\left(\frac{(n+1)^{n+1}}{(n+1)!}\right)^\frac{n}{n(n+1)}}{\left(\frac{n^n}{n!}\right)^\frac{1}{n(n+1)}}$$ $$\Large=\left(\frac{\frac{(n+1)^{n+1}}{(n+1)!}}{(\frac{n^n}{n!})}\right)^\frac{n}{n(n+1)}+\frac{\left(\frac{(n+1)^{n+1}}{(n+1)!}\right)^\frac{n}{n(n+1)}}{\left(\frac{n^n}{n!}\right)^\frac{1}{n(n+1)}}$$ $$=\left(\frac{n+1}{n}\right)^n+\Large\frac{\left(\frac{(n+1)^{n+1}}{(n+1)!}\right)^\frac{n}{n(n+1)}}{\left(\frac{n^n}{n!}\right)^\frac{1}{n(n+1)}}$$

How to I proceed from here? (the limit is $e$)

Alistair
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gbox
  • 12,867

1 Answers1

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Let $\displaystyle a_n=\frac{n^n}{n!}$.

Then $\displaystyle\frac{a_{n+1}}{a_n}=\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}=\left(\frac{n+1}{n}\right)^{n}=\left(1+\frac{1}{n}\right)^n\to e$,

so $\displaystyle\frac{n}{(n!)^{\frac{1}{n}}}=(a_n)^{\frac{1}{n}}\to e$

user84413
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