Show that $\underset{n\to \mathrm{\infty}}{lim}{\textstyle [}\frac{{n}^{n+1}+(n+1{)}^{n}}{{n}^{n+1}}{{\textstyle ]}}^{n}={e}^{e}.$ All we have done was elementary manipulations, but we got stuck.

Leah Pope
2022-06-29
Answered

Show that $\underset{n\to \mathrm{\infty}}{lim}{\textstyle [}\frac{{n}^{n+1}+(n+1{)}^{n}}{{n}^{n+1}}{{\textstyle ]}}^{n}={e}^{e}.$ All we have done was elementary manipulations, but we got stuck.

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lilao8x

Answered 2022-06-30
Author has **22** answers

Define ${u}_{n}=(1+\frac{1}{n}{)}^{n}$

${(1+\frac{1}{n}{{(1+\frac{1}{n})}^{n}})}^{n}$

$={(1+\frac{1}{n}{{u}_{n}})}^{n}$

$=\mathrm{exp}(n\mathrm{log}(1+\frac{{u}_{n}}{n}))$

$=\mathrm{exp}({u}_{n}+\mathcal{O}\left(\frac{1}{n}\right))$The $\mathcal{O}({n}^{-1})$ term drops to 0 and ${u}_{n}\to e$ in the limit, so the value sought is $\mathrm{exp}(e)={e}^{e}$

${(1+\frac{1}{n}{{(1+\frac{1}{n})}^{n}})}^{n}$

$={(1+\frac{1}{n}{{u}_{n}})}^{n}$

$=\mathrm{exp}(n\mathrm{log}(1+\frac{{u}_{n}}{n}))$

$=\mathrm{exp}({u}_{n}+\mathcal{O}\left(\frac{1}{n}\right))$The $\mathcal{O}({n}^{-1})$ term drops to 0 and ${u}_{n}\to e$ in the limit, so the value sought is $\mathrm{exp}(e)={e}^{e}$

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