When Rudin proves the theorem $\lim_{n\rightarrow \infty} (1+\frac{1}{n})^n=e$, he uses the following claim:
Let $t_n= (1+\frac{1}{n})^n=1+1+\frac{1}{2!}(1-\frac{1}{n})+\frac{1}{3!}(1-\frac{1}{n})(1-\frac{2}{n})+\dots+\frac{1}{n!}(1-\frac{1}{n})(1-\frac{2}{n})\dots (1-\frac{n-1}{n}).$
If $n\geq m$, then $t_n\geq 1+1+\frac{1}{2!}(1-\frac{1}{n})+\dots +\frac{1}{m!}(1-\frac{1}{n})\dots (1-\frac{m-1}{n})$.
I am a little confused. Since $n\geq m$, we have $\frac{1}{n!}\leq \frac{1}{m!}$ and $1-\frac{m-1}{n}\geq 1-\frac{n-1}{n}$. Therefore, we should get $t_n\leq 1+1+\frac{1}{2!}(1-\frac{1}{n})+\dots +\frac{1}{m!}(1-\frac{1}{n})\dots (1-\frac{m-1}{n})$. Why does Rudin use $\geq$ here?
Thanks in advance!
So observe that
$t_n=\sum_{k=0}^{n}\frac{1}{k!}\prod_{i=0}^{k-1}(1-\frac{i}{n})$
If $n\geq m$, since each terms are positive,
$t_n=\sum_{k=0}^{n}\frac{1}{k!}\prod_{i=0}^{k-1}(1-\frac{i}{n})\geq \sum_{k=0}^{m}\frac{1}{k!}\prod_{i=0}^{k-1}(1-\frac{i}{n})$
Basically, in the inequality, rightside has less positive terms than $t_n$ if $n\geq m$