Which Expression of e Converges Fastest?

Question

Writing a program, would be helpful to know whether expressing e as the sum of the reciprocals of the factorials (as in the Taylor series expansion of $e^x$ with $x=1$) converges quicker than $(1+1/n)^n$. Tried Googling, no luck, so any help would be much appreciated. Many thanks!

Answer 1

Note that $x_n=\left(1+\frac1n\right)^n$ is $x_n=u\left(\frac1n\right)$ where $u(x)=(1+x)^{1/x}$ hence $$u(x)=\exp\left(\frac1x\log(1+x)\right)=\exp\left(\frac1x\left(x-\frac12x^2+o(x^2)\right)\right)=\exp\left(1-\frac12x+o(x)\right).$$ Thus, $x_n-\mathrm e\sim\frac1nu'(0)=-\frac12\mathrm e\frac1n$. On the other hand, $y_n=\sum\limits_{k=0}^n\frac1{k!}$ is such that $y_n-\mathrm e\sim-\frac1{n(n!)}$, hence $y_n$ approaches $\mathrm e$ much quicker that $x_n$ does (and both underestimate $\mathrm e$).

Answer 2

The taylor series

$$e = \sum_{k=0}^{\infty} \frac{1}{k!} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ...$$

converges much faster