Prove S>4ar if a>0 for a geometric series

[Colin67]

Dear All





But I really cannot work out out to show S>4ar if a>0.

Can somebody give us some with help please, I have not attempted final part of question yet

Thanks

 

[Cubist]

It can sometimes be easier to show that an expression >0 rather than an expression > another expression, especially if you can factor the LHS. So you could try to show that S - 4ar > 0 if a>0

LHS: [math]\frac{a}{1-r}-4ar[/math]
[math]=a\left( \frac{1}{1-r}-4r \right)[/math]
continue...

 

[Colin67]

Thank you so much, I was trying to prove using a>0 but not the given result. The above can be demonstrated to be >0 if a>0 after simplifying, and leads to r>1/2.

 

[Cubist]

and leads to r>1/2.

...not quite, there's only one exceptional value that "r" can't be. Can you post what you did?

 

[pka]

I was trying to prove using a>0 but not the given result. The above can be demonstrated to be >0 if a>0 after simplifying, and leads to r>1/2.

It works for \(\large a=1~\&~r=\frac{1}{6}\)

 

[Colin67]

Is it r=1/2?

Thank you again for all your assistance

 

[pka]

View attachment 20702

Is it r=1/2?

Thank you again for all your assistance

The only thing \((2r-1)^2>0\) says is that for every \(r \ne\frac{1}{2}\) that is a statement.

 

[pka]

If \(r\ne\frac{1}{2}\) and we know that \(|r|<1\) so
\((2r-1)^2>0\\4r^2-4r+1>0\\1>4r-4r^2\\1>4r(1-r)\\ \dfrac{1}{1-r}>4r\)
Can you finish?

 

[Colin67]

I'm sorry I can't.

 

[pka]

I'm sorry I can't.

Can you multiply both sides by \(\large a~?\)

 

[Colin67]

Yes, and it leads to S>4ar which was what had to be shown

 

[Colin67]

So r cannot equal 1/2 is the exact value asked for?

 

[pka]

So r cannot equal 1/2 is the exact value asked for?

Use \( r=\dfrac{1}{2}\) in the expression \(\dfrac{a}{1-r}>4ar\).
Do you see a contradiction?
If so, what is it?

 

[Colin67]

You end up with 2a>2a, so r cannot equal 1/2.

Thank you for your guidance.