# Elementary proof of claim in Baby Rudin 3.31

On p65 Rudin states: $$e-s_n = \frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}+... < \frac{1}{(n+1)!}(1 + \frac{1}{n+1}+\frac{1}{(n+1)^2}+...)=\frac{1}{n!n}$$ Where $s_n = \sum_{k=0}^{n}{\frac{1}{k!}}$. What I understand:

• Each partial sum for $n\ge 2$ on the left is less than the one on the right. So the total sum on the left is less than or equal to the one on the right.
• If the sum on the left is equal to the sum on the right, then $e = s_n + \frac{1}{n!n}$ would be rational.

I was wondering:

Is there a more constructive way to show that the sum on the left is not equal to that on the right?

• The sum on the left is strictly less than that on the right, because starting from the $2$nd term, the term on the left is less than the corresponding one on the right. Nov 5, 2014 at 10:09
• Could you please explain why this is true? I thought about the general case, but consider $S_n = 1-\frac{1}{n^2}, T_n = 1-\frac{1}{n^3}$ Then starting from the second partial sum, $S_n < T_n$ but $S_n \rightarrow 1$ and $S_n \rightarrow 1$. Nov 5, 2014 at 10:14

Series are different from sequences. Suppose that $a_n\le b_n$ for all $n$ and $a_N<b_N$ for some fixed $N$. Then for any $m>N$ $$\sum^m_{n=1}b_n-\sum^m_{n=1}a_n=\sum^m_{n=1}(b_n-a_n)\ge b_N-a_N.$$ Pushing $m\to \infty$, we get $$\sum b_n-\sum a_n\ge b_N-a_N.$$

Note: The given argument actually shows that $e$ is irrational. One of the most elementary arguments I know of.

• Thanks! This was exactly what I was looking for. Nov 6, 2014 at 0:31

The inmequality is true since

$$\frac1{(n+1)!}+\frac1{(n+2)!}+\frac1{(n+3)!}+\ldots=\frac1{(n+1)!}\left(1+\frac1{n+2}+\frac1{(n+2)(n+3)}+\ldots\right)<$$

$$<\frac1{(n+1)!}\left(1+\frac1{n+1}+\frac1{(n+1)^2}+\ldots\right)$$

since

$$(n+1)<(n+i)\;,\;\;\forall\;2\le i\le k\implies(n+1)^k<(n+2)(n+3)\cdot\ldots\cdot(n+k)$$