I've been trying to prove a statement that I found in a book.
Given that(I've already proven these statements):
$$ \forall\; n\in Z^+:(1 + \frac 1n)^n = 1 + \sum_{k=1}^n \frac 1{k!}\prod_{r=0}^{k-1}(1 - \frac rn)$$ $$\forall\; n\in Z,n \gt 3:2^n < n!$$
I have to prove that if n > 1:
$$2 \lt (1+\frac1n)^n \lt 1 + \sum_{k=1}^n\frac 1{k!} \lt 3$$
I've no clue how to get started.
By the Bernoulli inequality, the sequence given by $a_n=(1+1/n)^n$ is increasing, hence the first inequality is trivial. The second one follows from the binomial theorem: $$\left(1+\frac{1}{n}\right)^n = \sum_{j=0}^{n}\binom{n}{j}\frac{1}{n^j}=1+\sum_{k=1}^{n}\frac{\frac{n}{n}\cdot\frac{n-1}{n}\cdot\ldots\cdot\frac{n-k+1}{n}}{k!}\leq 1+\sum_{k=1}^{n}\frac{1}{k!}$$ and the last one follows from the lower bound $k!>2^k$, holding for any $k>3$: $$1+\sum_{k=1}^{n}\frac{1}{k!}\leq 1+1+\frac{1}{2}+\frac{1}{6}+\sum_{k=4}^{+\infty}\frac{1}{2^k}=2+\frac{1}{2}+\frac{1}{6}+\frac{1}{8}=\frac{67}{24}<3.$$