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We know the following inequality,

$$\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\ldots\left(1-\frac{1}{2^{n}}\right)\geq\frac{1}{4}+\frac{1}{2^{n+1}}$$

We can find in these links the demonstration for induction of this result: [1], [2] and [3].

Obviously the result is still true if we change the right side by,

$$\frac{1}{4}\color{red}{-}\frac{1}{2^{n+1}}$$

Of course, I could use the previous inequality to complete this one, but I would like to know if it is possible to prove it directly by induction. It all comes down to proving this inequality,

$$\left(1-\frac{1}{2^{n+1}}\right)\left(\frac{1}{4}-\frac{1}{2^{n+1}}\right)\ge\frac{1}{4}-\frac{1}{2^{n+2}}$$

That it is not true, then, does that mean that it is not possible to show by induction? In this case, how would this result be demonstrated?

Mrcrg
  • 2,767
  • The last inequality is not true eg when $n=1$ – Albus Dumbledore Apr 17 '21 at 13:19
  • @AlbusDumbledore Now it's right – Mrcrg Apr 17 '21 at 13:23
  • @M no try $n=1$ again!! – Albus Dumbledore Apr 17 '21 at 13:26
  • Truth! I was using a wrong value. So it's probably not possible to do it by induction .... – Mrcrg Apr 17 '21 at 13:28
  • @Mrcrg The LHS of the bottom inequality should be $$\left(1 - \frac{1}{2^{n + 1}}\right)\left(\frac{1}{4} \color{red}{+} \frac{1}{2^{n + 1}}\right)$$ See https://math.stackexchange.com/questions/1929391/manipulating-expression-frac14-frac12n11-frac12n1-i – 光復香港 時代革命 Free Hong Kong Apr 17 '21 at 17:19
  • @光復香港時代革命FreeHongKong I'm not talking about the original inequality, it is still true if the signal changes. – Mrcrg Apr 17 '21 at 17:33
  • @Mrcrg If you switch signs on both LHS and RHS, the inequality switches direction: $$\begin{align}\left(1 - \frac{1}{2^{n+1}}\right) \left(\frac{1}{4} - \frac{1}{2^{n+1}}\right) &= \frac{1}{4} - \frac{1}{2^{n+1}} - \frac{1}{2^{n+3}} + \frac{1}{2^{2n+2}} \ &= \left(\frac{1}{4} - \frac{1}{2^{n+2}}\right) - \frac{1}{2^{n+2}} - \frac{1}{2^{n+3}} + \frac{1}{2^{2n+2}} \ &\le \left(\frac{1}{4} - \frac{1}{2^{n+2}}\right) - \frac{1}{2^{n+2}} + \frac{1}{2^{2n+2}} \ &\le \frac{1}{4} - \frac{1}{2^{n+2}}\end{align}$$ since $\frac{1}{2^{2n+2}} - \frac{1}{2^{n+2}} \le 0$ by induction. – 光復香港 時代革命 Free Hong Kong Apr 17 '21 at 17:57
  • You likely cannot prove it directly by induction since the LHS is multiplied by a term that is < 1 but the RHS increases (is multiplied by a term that is > 1). If given this version, I would have done a "stronger inequality" which was the original version. – Calvin Lin Apr 19 '21 at 02:12

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