For every integer $n\geqslant2$ prove the inequality $$ \frac{1}{2!}+\frac{2}{3!}+\ldots+\frac{2^{n-2}}{n!}\leqslant\frac{3}{2}, $$where $k!=1\cdot2\cdot\ldots\cdot k$.
Problem
Source: 2018 Belarusian National Olympiad 9.2
Tags: number theory, inequalities, factorial
KrysTalk
29.03.2019 14:12
We prove that $$\frac{2^{n-2}}{n!} \leqslant \frac{1}{n-2} - \frac{1}{n-1} \ \ \left(\forall n \geqslant 4 \right). \ \ \ \ \ \ (*)$$Indeed, it's $$n \cdot (n-3)! \ge 2^{n-2}. \ \ \ \ \ (**)$$If $n=4$ it's obvious. Now assume that $(**)$ true for all $n \ge 4.$ then it's also true for $n+1$ because : $$(n+1) \cdot (n-2)! > 2n \cdot (n-3)! \ge 2^{n-1}.$$Use $(*)$ we get $$LHS \le \frac{1}{2!} + \frac{2}{3!} + \sum_{i=4} ^{n} \left(\frac{1}{i-2} - \frac{1}{i-1} \right) = \frac{4}{3} - \frac{1}{n-1} < \frac{3}{2}$$
NikoIsLife
29.03.2019 14:21
According to the following identity:
$$\sum_{n=0}^\infty\frac{x^n}{n!}=e^x$$we have:
$$\sum_{n=0}^\infty\frac{2^n}{n!}=e^2$$$$\frac34+\sum_{n=2}^\infty\frac{2^{n-2}}{n!}=\frac{e^2}4$$$$\sum_{n=2}^\infty\frac{2^{n-2}}{n!}=\frac{e^2-3}4$$And, since $\frac{e^2-3}4<\frac{3^2-3}4=\frac32$, the inequality is true.
Wiz-of-Oz
29.03.2019 16:24
KrysTalk wrote:
We prove that $$\frac{2^{n-2}}{n!} \leqslant \frac{1}{n-2} - \frac{1}{n-1} \ \ \left(\forall n \geqslant 4 \right). \ \ \ \ \ \ (*)$$Indeed, it's $$n \cdot (n-3)! \ge 2^{n-2}. \ \ \ \ \ (**)$$If $n=4$ it's obvious. Now assume that $(**)$ true for all $n \ge 4.$ then it's also true for $n+1$ because :
$$(n+1) \cdot (n-2)! > 2n \cdot (n-3)! \ge 2^{n-1}.$$Use $(*)$ we get $$LHS \le \frac{1}{2!} + \frac{2}{3!} + \sum_{i=4} ^{n} \left(\frac{1}{i-2} - \frac{1}{i-1} \right) = \frac{4}{3} - \frac{1}{n-1} < \frac{3}{2}$$
Nice
AopsUser101
17.03.2020 08:11
Here is how I did it, recalling the Taylor polynomial for $e^x$:
Attachments:
atdaotlohbh
17.04.2024 00:13
Let $a_i=\frac{2^{i-2}}{i!}$ Note that $\frac{a_{i+1}}{a_i}=\frac{2}{i+1} \leq \frac{2}{3}$, so $a_i \leq (\frac{2}{3})^{i-2}a_2$ Then $a_2+\ldots \leq a_2(\frac{1}{1-\frac{2}{3}})=\frac{3}{2}$