2
$\begingroup$

The question sets a condition where $r\in\mathbb{R}$ and $0<r<1$ with a sequence of rational numbers $a_1,a_2,a_3,\dotsc$ given by $$a_n=\Bigl(1+\frac{r}{2^n}\Bigr)^{2^n}$$ The question then asks to:

Prove that $a_n<\dfrac{1}{1-r}$ for all $n\in\mathbb{N}$.

How I approached this question is to first put the information I have in place

$$\Bigl(1+\frac{r}{2^n}\Bigr)^{2^n}<\frac{1}{1-r}$$

where we can then take a base case of $n=1$.$$\Bigl(1+\frac{r}{2}\Bigr)^2<\frac{1}{1-r}$$

Here I expanded the equation to look like $$1+r+\frac{r^2}{4}<\frac{1}{1-r}$$

But at this point, I really didn't know how I would continue with proving the result. Any help would be appreciated.

$\endgroup$
3
$\begingroup$

$$ \left(1+\frac{r}{2^n}\right)^{2^n}=\sum_{i=0}^{2^n}\frac{1}{2^{ni}}\binom{2^n}{i}r^i \le \sum_{i=0}^{2^n} r^i < \sum_{i\ge 0} r^i = \frac{1}{1-r}. $$

$\endgroup$
2
$\begingroup$

Multiplying by the positive number $(1-r)$ the inequality to be proven becomes

$$(1-r)\left(1+r+\frac{r^2}{4} \right) = 1-r^2 + \frac{r^2}{4}-\frac{r^3}{4}<1\,.$$

$\endgroup$

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.