[jacks]

Proving the result \[ \displaystyle \sum^{n}_{r=0}\frac{1}{\binom{n}{r}} \leq 2-2^{-n} \]

[Pie]

But it would have to be:

\[ \displaystyle \sum^{n}_{r=0}\frac{1}{\binom{n}{r}} \geq{} 2-2^{-n} \]

¿right?

I think it can be done like this:

As

\( 2-2^{-n} = \displaystyle\sum_{r=0}^n{\displaystyle\frac{1}{2^r}} \)

Then

\( \displaystyle\sum_{r=0}^n{\displaystyle\frac{1}{\binom{n}{r}}} \geq{}\displaystyle\sum_{r=0}^n{\displaystyle\frac{1}{2^r}} \Leftrightarrow{}
\displaystyle\sum_{r=0}^n{\binom{n}{r}}\leq{}\displaystyle\sum_{r=0}^n{2^r} \)

But

\( \displaystyle\sum_{r=0}^n{\binom{n}{r}}=2^n \)

So

\( \displaystyle\sum_{r=0}^n{2^r} \geq{}2^n \)

Cheers.

[Luis Fuentes]

Hi

\( \displaystyle\sum_{r=0}^n{\displaystyle\frac{1}{\binom{n}{r}}} \geq{}\displaystyle\sum_{r=0}^n{\displaystyle\frac{1}{2^r}} \Leftrightarrow{}
\displaystyle\sum_{r=0}^n{\binom{n}{r}}\leq{}\displaystyle\sum_{r=0}^n{2^r} \)

But this equivalence is not true.

For example:

\( \dfrac{1}{5}+\dfrac{1}{1}\geq \dfrac{1}{2}+\dfrac{1}{2} \) but \( 5+1\leq 2+2 \)

On the other hand: jacks, please, check the statment.

\( \displaystyle \sum^{n}_{r=0}\frac{1}{\binom{n}{r}}\geq \frac{1}{\binom{n}{0}}+\frac{1}{\binom{n}{1}}=2 \)

So the initial version is not true and Pie's version is trivial.

Best regards.

[Pie]

Hi

\( \displaystyle\sum_{r=0}^n{\displaystyle\frac{1}{\binom{n}{r}}} \geq{}\displaystyle\sum_{r=0}^n{\displaystyle\frac{1}{2^r}} \Leftrightarrow{}
\displaystyle\sum_{r=0}^n{\binom{n}{r}}\leq{}\displaystyle\sum_{r=0}^n{2^r} \)

But this equivalence is not true.

For example:

\( \dfrac{1}{5}+\dfrac{1}{1}\geq \dfrac{1}{2}+\dfrac{1}{2} \) but \( 5+1\leq 2+2 \)

On the other hand: jacks, please, check the statment.

\( \displaystyle \sum^{n}_{r=0}\frac{1}{\binom{n}{r}}\geq \frac{1}{\binom{n}{0}}+\frac{1}{\binom{n}{1}}=2 \)

So the initial version is not true and Pie's version is trivial.

Best regards.

Ups, sorry.

PD. No me había dado cuenta que era tan fácil como evaluar la primera sumatoria para \( n \geq{1} \), supongo que lo que puse se podría arreglar pero en este caso no merece la pena XD

Saludos.

[jacks]

Thanks Admin got it.