Rudin definition of Eulers constant e

Question

In Rudin's Principles of Mathematical Analysis, he offers the following definition of e:

e = $\sum_{n=0}^\infty \frac{1}{n!} $

He then goes on to show the following:

$s_n$ = 1 + 1 + $\frac{1}{1*2}$ + ...+ $\frac{1}{1*2*...*n}$
< 1 + 1 + $\frac{1}{2}$ + $\frac{1}{2^2}$ + ... + $\frac{1}{2^{n-1}}$ < 3

My question is how he proves that the last inequality is less than 3?

Other references mention that the series converges by the ratio test, but Rudin has not introduced that yet. He seems to use the fact that all the partial sums of the series are bounded and the terms of the series are positive. But I'm not quite seeing how he determines the series is bounded, and bounded by 3 specifically.

Thanks in advance.

Answer 1

$$\sum_{i=0}^{\infty}2^{-i}=\frac1{1-\frac12}=2>\sum_{i=0}^{n-1}2^{-i}$$

Answer 2

$S_n:=1+1/2+1/2^2+ .....1/2^{n-1};$

$(1/2)S_n =1/2+1/2^2+....1/2^{n-1}+1/2^n;$

Subtract:

$(1/2)S_n= 1- 1/2^n \lt 1.$

Hence : $S_n \lt 2.$