Problem with inequality

[Engelhelm]

Help me please with this inequality . I need to prove this inequality, but i don't have any idea how to do this .

I tried to link this inequality with binomial theorem ,but i failed with it. Also I proved that inequality is valid for n=1;2;3;4;5;6. But i don't know how to prove that the inequality is valid for all meanings of n.

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[stapel]

Help me please with this inequality . I need to prove this inequality, but i don't have any idea how to do this .

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View attachment 4713

The image in your post has a character whose meaning I don't know. In the below:

. . . . .\(\displaystyle \displaystyle{ \frac{\left(1\, +\, \frac{x}{2}\right)^n\, -\, 1}{\left(1\, +\, x\right)^n\, -\, 1}\, \leq\, \frac{1}{2} \color{red}{|} }\)

What is the meaning of the terminal character displayed above in red?

When you reply, please include a clear statement of your thoughts and efforts so far. Thank you!

 

[Engelhelm]

The image in your post has a character whose meaning I don't know. In the below:

. . . . .\(\displaystyle \displaystyle{ \frac{\left(1\, +\, \frac{x}{2}\right)^n\, -\, 1}{\left(1\, +\, x\right)^n\, -\, 1}\, \leq\, \frac{1}{2} \color{red}{|} }\)

What is the meaning of the terminal character displayed above in red?

When you reply, please include a clear statement of your thoughts and efforts so far. Thank you!

This line is from computer's mouse. It means nothing.

 

[Ishuda]

Assuming n is an integer, the binomial theorem says
\(\displaystyle \displaystyle{ \left(1\, +\, y\right)^n\, =\, \sum_{m\,=\,0}^{m\,=\,n}\space \left(_{n}C_{m}\space y^m \right) }\)

or

\(\displaystyle \displaystyle{ \left(1\, +\, y\right)^n \, -\, 1\, =\, y\space \sum_{m\,=\,0}^{m\,=\,n\,-\,1}\space _{n}C_{m\,+\,1}\space y^m }\)

where \(\displaystyle _{n}C_{m}\) is the binomial coefficient.

So we have

\(\displaystyle \displaystyle{ \frac{\left(1\,+\,\dfrac{x}{2}\right)^n \,-\,1}{\left(1\,+\,x\right)^n\, -\,1}}\, = \,\dfrac{\left(\dfrac{x}{2}\right) \space \displaystyle{\sum_{m\,=\,0}^{m\,=\,n\,-\,1}}\space _{n}C_{m\,+\,1} \space \left(\dfrac{x}{2}\right)^m}{ \left(x\right) \space \displaystyle{ \sum_{m\,=\,0}^{m\,=\,n\,-\,1}}\space _{n}C_{m\,+\,1}\space x^m} \,= \, \left(\dfrac{1}{2}\right)\, \left(\dfrac{ \displaystyle{ \sum_{m\,=\,0}^{m\,=\,n\,-\,1} } \space _{n}C_{m\,+\,1}\space \left(\dfrac{x}{2}\right)^m}{ \displaystyle{ \sum_{m\,=\,0}^{m\,=\,n\,-\,1} } \space _{n}C_{m\,+\,1}\space x^m}\right) \)

EDIT: Be careful (x/2) is not always less than x. Oh, and BTW, the statement isn't true for all x.

 

[Engelhelm]

Assuming n is an integer, the binomial theorem says
(1+y)ⁿ = \(\displaystyle \Sigma_{m=0}^{m=n}\space _{n}C_{m}\space y^m\)
or
(1+y)ⁿ -1 = \(\displaystyle y\space \Sigma_{m=0}^{m=n-1}\space _{n}C_{m+1}\space y^m\)
where \(\displaystyle _{n}C_{m}\) is the binomial coefficient.

So we have
\(\displaystyle \frac{(1+\frac{x}{2})^n -1}{(1+x)^n -1} = \frac{\frac{x}{2}\space\Sigma_{m=0}^{m=n-1}\space _{n}C_{m+1}\space (\frac{x}{2})^m}{x\space \Sigma_{m=0}^{m=n-1}\space _{n}C_{m+1}\space x^m} = \frac{1}{2}\frac{\Sigma_{m=0}^{m=n-1}\space _{n}C_{m+1}\space (\frac{x}{2})^m}{\Sigma_{m=0}^{m=n-1}\space _{n}C_{m+1}\space x^m}\)

EDIT: Be careful (x/2) is not always less than x. Oh, and BTW, the statement isn't true for all x.

Thank you , but sorry for stupid question. How we got x/2 before Σ ?

 

[Deleted member 4993]

Thank you , but sorry for stupid question. How we got x/2 before Σ ?

Assuming n is an integer, the binomial theorem says
(1+y)ⁿ = \(\displaystyle \Sigma_{m=0}^{m=n}\space _{n}C_{m}\space y^m\)
or
(1+y)ⁿ -1 = \(\displaystyle y\space \Sigma_{m=0}^{m=n-1}\space _{n}C_{m+1}\space y^m\) ...... did you notice y in front of Σ
where \(\displaystyle _{n}C_{m}\) is the binomial coefficient.

So we have
\(\displaystyle \frac{(1+\frac{x}{2})^n -1}{(1+x)^n -1} = \frac{\frac{x}{2}\space\Sigma_{m=0}^{m=n-1}\space _{n}C_{m+1}\space (\frac{x}{2})^m}{x\space \Sigma_{m=0}^{m=n-1}\space _{n}C_{m+1}\space x^m} = \frac{1}{2}\frac{\Sigma_{m=0}^{m=n-1}\space _{n}C_{m+1}\space (\frac{x}{2})^m}{\Sigma_{m=0}^{m=n-1}\space _{n}C_{m+1}\space x^m}\)

.

 

[Ishuda]

Thank you , but sorry for stupid question. How we got x/2 before Σ ?

In one case (for the numerator) the y in the first equation is x/2 and in the other case (for the denominator) it is x.

 

[Engelhelm]

.

I don't understand why in \(\displaystyle \displaystyle{ \left(1\, +\, y\right)^n\, =\, \sum_{m\,=\,0}^{m\,=\,n}\space \left(_{n}C_{m}\space y^m \right) }\) we don't have y before

, but in \(\displaystyle \displaystyle{ \left(1\, +\, y\right)^n \, -\, 1\, =\, y\space \sum_{m\,=\,0}^{m\,=\,n\,-\,1}\space _{n}C_{m\,+\,1}\space y^m }\) we have y before

.

 

[Deleted member 4993]

I don't understand why in \(\displaystyle \displaystyle{ \left(1\, +\, y\right)^n\, =\, \sum_{m\,=\,0}^{m\,=\,n}\space \left(_{n}C_{m}\space y^m \right) }\) we don't have y before

, but in \(\displaystyle \displaystyle{ \left(1\, +\, y\right)^n \, -\, 1\, =\, y\space \sum_{m\,=\,0}^{m\,=\,n\,-\,1}\space _{n}C_{m\,+\,1}\space y^m }\) we have y before

.

With paper and pencil, expand first three terms of each of the summations.

What do you get?

Please share your work with us.

 

[Engelhelm]

Oh , i understood, it's a factorization. Thank you !