Estimating the error for $e-(1+frac 1n)^n$ Code Answer

I would like to estimate the error $e-(1+frac 1n)^n$, for arbitrary $n ge 0$. I think I have found a way to do it, but it does not really seem optimal to me. Here is what I did:

I considered the function $f : [0,frac 1n]$, $f(0) = (1+frac 1n)^n$ and $f(x) = e^{nx}+(1+frac 1n)^n-(1+x)^{frac 1 x} $

This function is continuous and I have $exists theta in left(0,frac 1 nright ): $

$$ e-(1+frac 1n)^n=f left (frac 1nright ) – f(0) = frac 1 n f'(theta) $$

by the mean value theorem. However, $f'(theta)$ seems a bit too complicated. I tried simplifying it and bounding it from above, involving $n$, but it does not simplify a lot.

Any hints on maybe choosing another function that allows a smoother estimation? Help would be greatly appreciated!

Answer

Let $x=frac{1}{n}$. Then $(1+frac{1}{n})^n=(1+x)^{frac{1}{x}}$. Take the logarithm to get

$$lnleft[(1+x)^{frac{1}{x}}right]=frac{ln(1+x)}{x}=1-frac{x}{2}+frac{x^2}{3}-cdots+frac{(-x)^n}{n+1}+cdots.$$

Exponentiate back to get

begin{align}
left(1+frac{1}{n}right)^n&=eexpleft(-frac{x}{2}+frac{x^2}{3}-cdots+frac{(-x)^n}{n+1}+cdotsright)\
&=eleft(1-frac{1}{2n}+frac{11}{24n^2}-frac{7}{16n^3}+cdotsright),quad
ngg 1.
end{align}