1. ## Inequality proof

How to proof that

$\displaystyle \sum_{n=1}^{\infty}\left(\left\lfloor \frac{2^{k+1}}{3^{n}}\right\rfloor -2\left\lfloor \frac{2^{k}}{3^{n}}\right\rfloor \right)\geq2$

inequality holds for $\displaystyle k\geq8$?

Thanks!

2. Originally Posted by Ben92
How to proof that

$\displaystyle \sum_{n=1}^{\infty}\left(\left\lfloor \frac{2^{k+1}}{3^{n}}\right\rfloor -2\left\lfloor \frac{2^{k}}{3^{n}}\right\rfloor \right)\geq2$

inequality holds for $\displaystyle k\geq8$?

Thanks!

There is something wrong here.

$\displaystyle \left(\left\lfloor \frac{2^{k+1}}{3^{n}}\right\rfloor -2\left\lfloor \frac{2^{k}}{3^{n}}\right\rfloor \right) = 0$ !

3. Yes, you are right. The correct condition is $\displaystyle k>8$.

4. Originally Posted by Ben92
How to proof that

$\displaystyle \sum_{n=1}^{\infty}\left(\left\lfloor \frac{2^{k+1}}{3^{n}}\right\rfloor -2\left\lfloor \frac{2^{k}}{3^{n}}\right\rfloor \right)\geq2$

inequality holds for $\displaystyle k\geq8$?

Thanks!
Induction would be my first bet.