Prove that $\lim\limits_{n\to\infty}1 + \frac{1}{1!} + \frac {1}{2!} + \cdots + \frac{1}{n!}\ge\lim\limits_{n\to\infty}(1+\frac{1}{n})^n$ So My professor assigned this question and I am really stuck on part B of the question.
For $n \in \mathbb{N}$ let {$T_n$} = {$1 + \frac{1}{1!} + \frac {1}{2!} + \cdots + \frac{1}{n!}$}.
(a) Prove that {$T_n$} is non-decreasing.
(b) Use {$S_n$} = {$(1+\frac{1}{n})^n$} to prove that {$T_n$} is bounded above and prove that $lim_n(T_n)\ge lim_n (S_n)$ 
Like I mentioned I can prove part (a) simply by supposing that $T_{n+1} \lt T_n$ then after some algebra concluding the statement is false and than with one other example proving that its not strictly increasing either.
Now for part (b) I have not a slightest clue about how to solve, I assume I will have to use the expanded form of $S_n$ but thats the best I have. So please give me some hints or nudges in the right direction.
 A: For part (b) observe that,
$$S_n = \left(1+\frac{1}{n}\right)^n = \sum_{r=0}^{n}\binom{n}{r}\frac{1}{n^r}=\sum_{r=0}^{n}\frac{1}{r!}\left(1\right)\left(1-\frac{1}{n}\right)\cdots \left(1-\frac{r-1}{n}\right)\leq \sum_{r=0}^{n}\frac{1}{r!}=T_n$$
Also to prove that it is bounded above you can check for yourself that , $$\frac{1}{r!}\leq\frac{1}{2^{r-1}},r>1$$
$$T_n=\sum\frac{1}{r!}\leq 2+\sum_{r=2}^{\infty}\frac{1}{2^{r-1}}=3$$
A: $(a)$ $T_{n+1}=T_n + \frac{1}{(n+1)!} \implies T_{n+1} - T_n > 0$.
$(b)$ You can easily check that $\frac{1}{k!} \geq \binom{n}{k}\frac{1}{n^k}$ expanding the binomial. Remember that $(1+\frac{1}{n})^n=\sum_{k=0}^{n}\binom{n}{k}\cdot(\frac{1}{n})^k\cdot1^{n-k}=\sum_{k=0}^{n}\binom{n}{k}\cdot(\frac{1}{n})^k$. We wrote both $T_n$ and $S_n$ as series from $0$ to $n$, and we proved that $T_n \geq S_n$ term by term. Thus $T_n \geq S_n$. Of course $\lim_n (T_n) \geq \lim_n (S_n)$. This can't prove of course that $T_n$ is bounded above! If you know that $e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!}$, you can say that $e=\sum_{k=0}^{\infty}\frac{1}{k!}$. As you can say $T_n$ and $S_n$ are both bounded above and have the same limit.
A: The $k^{th}$ term of $T_n$ is $\dfrac{1}{(k-1)!}$ while the $k^{th}$ term of $S_n$ is $\binom{n}{k-1}\cdot \dfrac{1}{n^{k-1}}= \dfrac{1}{(k-1)!}\cdot \dfrac{n!}{(n-k+1)!\times n^{k-1}}= \dfrac{1}{(k-1)!}\times \dfrac{n(n-1)\cdots (n-(k-2))}{n^{k-1}}\leq \dfrac{1}{(k-1)!} \Rightarrow T_n \geq S_n$. So suppose to the contrary that $T_n$ is unbounded, then since it is an non-decreasing function, $T_n \to +\infty$ as $n \to \infty$ but then $Q_n=\left(1+\dfrac{1}{n}\right)^{n+1} \geq T_n \Rightarrow \left(1+\dfrac{1}{n}\right)^{n+1} \to +\infty$ which can't happen since $Q_n \to e$.The fact that $Q_n \geq T_n$ is not that well-known but can be considered well-known if you learn about $S_n$, then you should also learn $Q_n$ at the same time.
