e And His Twin

Letting the given limit be $L$ , we then have that

$\ln L = \displaystyle \lim_{n \to \infty} n \left(\ln\left(1 + \dfrac{1}{n}\right)^{n} - \ln(e)\right) = \lim_{n \to \infty} n\left(n\ln\left(1 + \dfrac{1}{n}\right) - 1\right)$ .

Now for $|x| \lt 1$ we have that $\ln(1 + x) = x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3} - ... = x - \dfrac{x^{2}}{2} + O(x^{3})$ .

Thus $n\ln\left(1 + \dfrac{1}{n}\right) - 1 = n\left(\dfrac{1}{n} - \dfrac{1}{2n^{2}} + O\left(\dfrac{1}{n^{3}}\right)\right) - 1 = -\dfrac{1}{2n} + O\left(\dfrac{1}{n^{2}}\right)$ .

So finally $\ln L = \displaystyle \lim_{n \to \infty} n\left(-\dfrac{1}{2n} + O\left(\dfrac{1}{n^{2}}\right)\right) = -\dfrac{1}{2} \Longrightarrow L = e^{-1/2} = \dfrac{1}{\sqrt{e}} \approx \boxed{0.60653}$ .

$\displaystyle \frac{((1+\frac{1}{n})^{n})^{n}}{e^{n}}=e^{\frac{nln{\frac{n+1}{n}}-1}{\frac{1}{n}}}$
Let $A=\displaystyle \lim_{n\to\infty} \left ({\frac{nln{\frac{n+1}{n}}-1}{\frac{1}{n}}} \right)= \lim_{n\to\infty} \left ({\frac{ln{\frac{n+1}{n}}-\frac{1}{n+1}}{-\frac{1}{n^2}}} \right)$ (Using L' Hopital's Rule)
$2A=\displaystyle \lim_{n\to\infty} \left ({\frac{nln{\frac{n+1}{n}}-1}{\frac{1}{n}}}+{\frac{ln{\frac{n+1}{n}}-\frac{1}{n+1}}{-\frac{1}{n^2}}} \right)$ $=\displaystyle \lim_{n\to\infty} \left ({-\frac{n}{n+1}} \right) =-1$
So, $A=-\frac{1}{2}$ , $\displaystyle \lim_{n \to \infty} \frac{((1+\frac{1}{n})^{n})^{n}}{e^{n}}=e^A=\frac{1}{\sqrt{e}} \approx \boxed{0.60653065}$ .

This is neat! Since most people may see the definition of the transcendental number $e$ defined as the limit of $(1+\frac{1}{n})^n$ as $n \rightarrow \infty$ in the numerator and plug in the limit value and then evaluate to a limit of 1. This is a good lesson on how to evaluate limits so as to not plug in the limiting value to evaluate the limiting sequence.