Simultaneous Inequalities: x + 2 < 1 + 1/x + 1/x^2 + 1/x^3 + ...

[Geoffry A]

I have a question on this problem.

. . . . .\(\displaystyle x\, +\, 2\, <\, 1\, +\, \dfrac{1}{x}\, +\, \dfrac{1}{x^2}\, +\, \dfrac{1}{x^3}\, +\, ...\)

I know the right ride of the equation sums to

x/x-1

Therefore,

x+2<x/x-1

At this point, I know x can't be equal to one. My question is do you know to multiply both sides by (x-1)^2 instead of just x-1? If you do the former (x-1)^2 and rearrange you get:

x^3-x^2-2x+2<0

giving you three solutions with x<1 being one of them.

But if you only multiply both sides by x-1, you get.

x^2-2<0

giving you two solutions with x<1 not being one of them.

Help would be appreciated.

Thank you.

 

[sinx]

I have a question on this problem.

. . . . .\(\displaystyle x\, +\, 2\, <\, 1\, +\, \dfrac{1}{x}\, +\, \dfrac{1}{x^2}\, +\, \dfrac{1}{x^3}\, +\, ...\)

I know the right ride of the equation sums to

x/x-1

Therefore,

x+2<x/x-1

At this point, I know x can't be equal to one. My question is do you know to multiply both sides by (x-1)^2 instead of just x-1? If you do the former (x-1)^2 and rearrange you get:

x^3-x^2-2x+2<0

giving you three solutions with x<1 being one of them.

But if you only multiply both sides by x-1, you get.

x^2-2<0

giving you two solutions with x<1 not being one of them.

Help would be appreciated.

Thank you.

check your original eqn to see what makes sense.
1 doesn't work
+-sqrt2 works, so this is the soln.

you could multiply both sides by another combination of (x-a) and come up another meaningless soln;
take (x+4)
(x²-2)(x-1)(x+4)<0
now you have 4 solns, 2 of which make no sense.

 

[pka]

I have a question on this problem.

. . . . .\(\displaystyle x\, +\, 2\, <\, 1\, +\, \dfrac{1}{x}\, +\, \dfrac{1}{x^2}\, +\, \dfrac{1}{x^3}\, +\, ...\)

I know the right ride of the equation sums to

x/x-1

Therefore,

x+2<x/x-1

At this point, I know x can't be equal to one. My question is do you know to multiply both sides by (x-1)^2 instead of just x-1? If you do the former (x-1)^2 and rearrange you get:

x^3-x^2-2x+2<0

giving you three solutions with x<1 being one of them.
But if you only multiply both sides by x-1, you get.

x^2-2<0

giving you two solutions with x<1 not being one of them.

It seems that you are examine for what values of \(\displaystyle x\) is it true that \(\displaystyle x + 2 < \sum\limits_{k = 0}^\infty {{x^{ - k}}} \)
You say that \(\displaystyle \sum\limits_{k = 0}^\infty {{x^{ - k}}}=\frac{x}{x-1} \) BUT that is only true if \(\displaystyle |x|>1\).
Of course if \(\displaystyle 0<x<1\) we know that \(\displaystyle \sum\limits_{k = 0}^\infty {{x^{ - k}}} = \infty \)

So do you need to rethink your work?