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!
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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. $\displaystyle \left(\left\lfloor \frac{2^{k+1}}{3^{n}}\right\rfloor -2\left\lfloor \frac{2^{k}}{3^{n}}\right\rfloor \right) = 0$ !
Yes, you are right. The correct condition is $\displaystyle k>8$.
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.
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