How to solve inequalities with infinite terms 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. 
 A: 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
$$x+2<\frac{x}{x-1}\iff x^2<2\iff x\in(1,\sqrt 2)$$
A: If $x>1$ let $y = 1/x$ and write it as geometric series that converges. if $0<x<1$, the right side diverges to $\infty$. 
A: 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.
A: 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 $x>0\implies 0<x<1$
