3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

see:

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 $$k!=k(k-1)\cdots 2>2^{k-1},k\ge 2$$

$\endgroup$
  • $\begingroup$ How did you get $\sum_{k=0}^{n}\left(\dfrac{1}{k!}\cdot\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)\cdots\left(1-\dfrac{k-1}{n}\right)\right)<\sum_{k=0}^{n}\dfrac{1}{k!}$ $\endgroup$ – user19405892 Dec 11 '15 at 4:17
  • $\begingroup$ $(1-\dfrac{i}{n})<1,i=1,2,\cdots,k-1$ $\endgroup$ – math110 Dec 11 '15 at 4:20
  • $\begingroup$ How does that explain it? $\endgroup$ – user19405892 Dec 11 '15 at 4:22
  • $\begingroup$ .. It is clear. $\endgroup$ – math110 Dec 11 '15 at 4:23
  • $\begingroup$ $-4 < 1$ and $-5 < 1$ but $-4*-5 > 1$ is my problem. $\endgroup$ – user19405892 Dec 11 '15 at 4:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.