How to solve inequalities with infinite terms

Question

Consider the following inequality:

$x + 2 < 1 + \dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x^3} ... $ with $x>0$.

Is there a general way to solve such an inequality with infinite terms?

The best I can do is some conjectures:

For $x = 2$ the right hand side equals 2, so I know that $x < 2$. Logically $ 0 < x \leq 1$, so what remains open is the case $1 < x < 2 $.

But I was thinking if there exists for example an algebraic way of solving this stuff, instead of what I'm doing.

score 4 · Accepted Answer · answered Sep 29 '13 at 13:26

If $0<x\leq 1$ the infinite sum diverges to $+\infty$ and the inequality is valid.

Let $x>1$ then $$\sum_{n=0}^\infty \frac{1}{x^n}=\frac{1}{1-\frac{1}{x}}=\frac{x}{x-1}$$ and the inequality becomes $$x+2<\frac{x}{x-1}\iff x^2<2\iff x\in(1,\sqrt 2)$$

score 2 · Answer 2 · answered Sep 29 '13 at 13:23

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If $x>1$ let $y = 1/x$ and write it as geometric series that converges. if $0<x<1$, the right side diverges to $\infty$.

answered Sep 29 '13 at 13:23

S L

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i got confused on my left and right hand. – S L Sep 29 '13 at 13:26
As you can see it helps to break down the problem OP. – Don Larynx Sep 29 '13 at 13:26

score 2 · Answer 3 · answered Sep 29 '13 at 13:23

2

There isn't a completely general method, but the right-hand side is a geometric series. When $|x| > 1$ it converges to a well-known and simple value. For $|x|\le 1$ it diverges, so the inequality is necessarily true.

answered Sep 29 '13 at 13:23

MJD

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So basically, your ability of doing this problem depends on your knowledge of certain sequences? – Phaptitude Sep 29 '13 at 13:29
There are probably a lot of ways to solve the problem. But in general, solving problems depends on knowing things, yes. Since the right-hand side involves an infinite series, you will need to know, or find out, at least a little bit about infinite series in order to solve it. – MJD Sep 29 '13 at 13:31

score 2 · Answer 4 · answered Sep 29 '13 at 13:27

If $|\frac1x|<1 \iff |x|>1$

$$1 + \dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x^3} ... =\frac1{1-\frac1x}=\frac x{x-1}$$

So, we need $\displaystyle x+2<\frac x{x-1}$

Multiplying either sides by $(x-1)^2,$

$\displaystyle\iff (x+2)(x-1)^2< x(x-1)$

If $x>1,$ we have $(x+2)(x-1)<x\iff x^2-2<0\iff -\sqrt2<x<\sqrt2$

$\implies 1<x<\sqrt2$

If $x<1,$ we have $(x+2)(x-1)>x\iff x^2-2>0\iff x<-\sqrt2$ or $x>\sqrt2$

$\implies -\sqrt2<x<1$ as $x<1$

But given that $x>0\implies 0<x<1$

How to solve inequalities with infinite terms

4 Answers4