# Prove inequality using binomial theorem

I have this math question that I'm kind of stuck on.

Use the binomial theorem to prove that for all integers $n\ge 2$:$$\left (1+\frac{1}{n}\right )^n < \sum_{j=0}^{n}{\frac{1}{j!}} < 2+\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2^{n-1}}$$

I know applying the binomial theorem I get:$$\sum_{j=0}^{n}{\binom{n}{j}\left ( \frac{1}{n}\right )^{n-j}} < \sum_{j=0}^{n}{\frac{1}{j!}} < 2+\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2^{n-1}}$$ $$\implies \sum_{j=0}^{n}{\left ( \frac{\binom{n}{j}}{n^{n-j}}\right )} < \sum_{j=0}^{n}{\frac{1}{j!}} < 2+\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2^{n-1}}$$ $$\implies \sum_{j=0}^{n}{\left ( \frac{n!}{n^{n-j}\cdot j!\cdot (n-j)!} \right )} < \sum_{j=0}^{n}{\frac{1}{j!}} < 2+\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2^{n-1}}$$

I'm not really sure what to do from here... Thanks.

Show that $\left(1+\frac{1}{n}\right)^{n}\geq \sum_{k=0}^n\left(\frac{1}{k!}\prod_{i=0}^{k-1}\left(1-\frac{i}{n}\right)\right)$ \begin{align*}\left(1+\dfrac{1}{n}\right)^n&=\sum_{k=0}^{n}\left(\binom{n}{k}\dfrac{1}{n^k}\right)\\ &=\sum_{k=0}^{n}\dfrac{n!}{(n-k)!\cdot k!}\cdot\dfrac{1}{n^k}\\ &=\sum_{k=0}^{n}\left(\dfrac{1}{k!}\cdot\dfrac{n(n-1)\cdots(n-(k-1))}{n^k} \right)\\ &=\sum_{k=0}^{n}\left(\dfrac{1}{k!}\cdot\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)\cdots\left(1-\dfrac{k-1}{n}\right)\right)\\ &<\sum_{k=0}^{n}\dfrac{1}{k!} \end{align*}
and $$\dfrac{1}{k!}<\dfrac{1}{2^{k-1}},k\ge 2$$because $$k!=k(k-1)\cdots 2>2^{k-1},k\ge 2$$
• How did you get $\sum_{k=0}^{n}\left(\dfrac{1}{k!}\cdot\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)\cdots\left(1-\dfrac{k-1}{n}\right)\right)<\sum_{k=0}^{n}\dfrac{1}{k!}$ – user19405892 Dec 11 '15 at 4:17
• $(1-\dfrac{i}{n})<1,i=1,2,\cdots,k-1$ – math110 Dec 11 '15 at 4:20
• $-4 < 1$ and $-5 < 1$ but $-4*-5 > 1$ is my problem. – user19405892 Dec 11 '15 at 4:23