# Thread: Solving infinite series inequality

1. ## Solving infinite series inequality

How do you solve

x+2<1+1/x^2 +1/x^3+..., where x>0?

Thanks

2. Are you sure it's not $\displaystyle x + 2 < 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots$?

3. Yes, sorry it should be that--I tried to copy and paste from another document where it was laid out correctly, but the formatting got mangled in the process...

4. $\displaystyle x + 2 < 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots$

$\displaystyle x + 2 < \sum_{n = 0}^{\infty}\frac{1}{x^n}$

$\displaystyle x + 2 < \sum_{n = 0}^{\infty}\left(\frac{1}{x}\right)^n$.

The right hand side is an infinite geometric series with common ratio $\displaystyle \frac{1}{x}$. Recall that the sum of an infinite geometric series is $\displaystyle \frac{a}{1 - r}$ if $|r| < 1$, i.e. $\displaystyle \left|\frac{1}{x}\right| < 1$, i.e. $\displaystyle |x| > 1$.

So assuming that $\displaystyle x > 1$, the right hand side is $\displaystyle \frac{1}{1 - \frac{1}{x}} = \frac{1}{\frac{x - 1}{x}} = \frac{x}{x - 1}$.

Thus

$\displaystyle x + 2 < \frac{x}{x - 1}$

$\displaystyle (x + 2)(x - 1) < x$ (we can do this since we know $\displaystyle x > 1$)

$\displaystyle x^2 + x - 2 < x$

$\displaystyle x^2 - 2 < 0$

$\displaystyle x^2 < 2$

$\displaystyle |x| < \sqrt{2}$

$\displaystyle -\sqrt{2} < x < \sqrt{2}$.

5. Thank you!

6. Originally Posted by Prove It
$\displaystyle x + 2 < 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots$

$\displaystyle x + 2 < \sum_{n = 0}^{\infty}\frac{1}{x^n}$

$\displaystyle x + 2 < \sum_{n = 0}^{\infty}\left(\frac{1}{x}\right)^n$.

The right hand side is an infinite geometric series with common ratio $\displaystyle \frac{1}{x}$. Recall that the sum of an infinite geometric series is $\displaystyle \frac{a}{1 - r}$ if $|r| < 1$, i.e. $\displaystyle \left|\frac{1}{x}\right| < 1$, i.e. $\displaystyle |x| > 1$.

So assuming that $\displaystyle x > 1$, the right hand side is $\displaystyle \frac{1}{1 - \frac{1}{x}} = \frac{1}{\frac{x - 1}{x}} = \frac{x}{x - 1}$.

Thus

$\displaystyle x + 2 < \frac{x}{x - 1}$

$\displaystyle (x + 2)(x - 1) < x$ (we can do this since we know $\displaystyle x > 1$)

$\displaystyle x^2 + x - 2 < x$

$\displaystyle x^2 - 2 < 0$

$\displaystyle x^2 < 2$

$\displaystyle |x| < \sqrt{2}$

$\displaystyle -\sqrt{2} < x < \sqrt{2}$.
The 'identity' $\displaystyle \sum_{n=0}^{\infty}\frac{1}{x^{n}} = \frac{x}{x-1}$ is valid only for $|x|>1$ , so that the inequality $\displaystyle x+2< \sum_{n=0}^{\infty}\frac{1}{x^{n}}$ is valid, because is also $x>0$, 'only' for $1...

Kind regards

$\chi$ $\sigma$