Existence of $\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$ I have a problem with understanding this proof. It's taken from Maor's $e$: The Story of a Number. I don't get the very last part. I see that the sum tends to $S_m$, but how does it imply that $S_m \leq T$, let alone that $S \leq T$? I don't see the connection.


 A: As the sequence $S_n$ increases monotonically, there is a standard result that states that if $S_n \leq T$, then $S_n$ converges to some $S$ and $S\leq T$.
A: For $m<n$, we have
$$
1 + 1 + \left( {1 - \frac{1}{n}} \right)\frac{1}{{2!}} + \left( {1 - \frac{1}{n}} \right)\left( {1 - \frac{2}{n}} \right) \cdots \left( {1 - \frac{{m - 1}}{n}} \right)\frac{1}{{m!}} < T_n .
$$
Fix $m$ and let $n$ tending to infinity. The left-hand side tends to $S_m$, the right-hand side tends to $T$. Thus, $S_m \leq T$, by the know relation between limits and inequalities (i.e., if $a_n<b_n$ for all $n$ then $\lim a_n \leq \lim b_n$).
A: Let the sum of the $m+1$ first terms of $T_n$ be denoted $T_{n,m}$, where $m<n$. From this definition we have
$$T_{n,m}<T_n$$ (because there are less terms) and $$\lim_{n\to\infty}T_{n,m}=S_m$$ (because the denominators $n$ make the individual factors tend to $1$).
Combining the two we draw
$$\lim_{n\to\infty}T_{n,m}=S_m\le\lim_{n\to\infty}T_n=T.$$
Finally,
$$\lim_{m\to\infty}S_m=S\le T.$$

Illustration:
$$T_{n,4}=1+\frac nn+\frac{n(n-1)}{2!n^2}+\frac{n(n-1)(n-2)}{3!n^3}+\frac{n(n-1)(n-2)(n-3)}{4!n^4}<T_n<T$$ and taking the limit on $n$,$$\lim_{n\to\infty}T_{n,4}=1+1+\frac1{2!}+\frac1{3!}+\frac1{4!}=S_4<T.$$
More generally,
$$S_m<T\implies S\le T,$$ taking the limit on $m$.
