##\lim_{n\to\infty}\left(n\left(1+\frac1n\right)^n-ne\right)=?##

[littlemathquark]

Homework Statement

$\lim_{n\to\infty}\left(n\left(1+\frac1n\right)^n-ne\right)=?$

Relevant Equations

none

$\left(1+\dfrac1n\right)^n=e^{n\ln\left(1+\dfrac1n\right)}$ and using Taylor expansion $\ln\left(1+\dfrac1n\right)=\dfrac 1n-\dfrac1{2n^2}+\dfrac1{3n^3}-\cdots$

$n\ln\left(1+\dfrac1n\right)=1-\dfrac 1{2n}+\dfrac1{3n^2}-\cdots$

$e^{n\ln(1+\frac1n)}=e.e^{-\frac1{2n}+\frac 1{3n^2}-...}$ For small x real numbers $e^x=1+x$, $x=-\frac1{2n}+\frac 1{3n^2}-..$

$e^{n\ln(1+\frac1n)}=e.(1-\frac1{2n}+\frac 1{3n^2}-...)$ and $n.e^{n\ln(1+\frac1n)}=n.e(1-\frac1{2n}+\frac 1{3n^2}-...)$

$\lim_{n \to \infty}(n.e^{n\ln(1+\frac1n)}-n.e)=n.e(1-\frac1{2n}+\frac 1{3n^2}-...)-n.e=-\frac e2+\frac e{3n}+...$

$\lim_{n \to \infty}(n.e^{n\ln(1+\frac1n)}-n.e)=-\frac e2$

 

[littlemathquark]

What about my solution?

 

[Mark44]

What about my solution?

I didn't check your work, but instead used Excel to approximate the limit. Your result seems about right to me.

 

[PeroK]

You can also use L'Hopital, with $x$ in place of $n$.

 

[littlemathquark]

You can also use L'Hopital, with $x$ in place of $n$.

How? Because it's not $\infty/\infty$, it's $\infty-\infty$

 

[PeroK]

How? Because it's not $\infty/\infty$, it's $\infty-\infty$

Make the $n$ in the numerator $\frac 1 n$ on the denominator.