I have an inequality which is giving me the correct answer, except that the sign is opposite to what it should be, any help?

Question

I'm looking at a geometric series that looks like

$A_n = \sum_{k=0}^{n}A_0x^k$ where $A_0$ is a constant.

I'm looking to find an $n$ where my error between two terms is less than a certain value $E$ such that

$A_n - A_{n-1} \le E$.

Now,

$A_n - A_{n-1} = A_0x^n$

Therefore I need

$A_0x^n \le E$,

which I can rewrite as

$\ln{A_0} +n\ln{x} \le \ln{E}$

$n \le \frac{\ln{E/A_0}}{\ln{x}}$

Which gives me what I need (i.e. the correct $n$ for the error I expect), except that it should be that $n$ is greater than this and not less than. Can anyone see what I've missed? Perhaps more caffeine is needed?

Many thanks.

Answer 1

The sign is correct, with $x > 1$ and $A_0 > 1$, the error term $A_0x^n$ grows with $n$. Therefore if you want the error term to be lower than a constant, you need $n$ to be small

Now if you take $x<1$, I guess your interest, than the error term decreases with $n$, and $ln(x)<0$, therefore when dividing by $ln(x)$ you have to switch sign.

$$ \ln A_0 + n\ln x \leq \ln E \Rightarrow \left\{ \begin{array}{ll} n \leq \frac{\ln E/A_0}{\ln x} & \text{ if } x>1\\ n \geq \frac{\ln E/A_0}{\ln x} & \text{ if } x<1\\ \end{array} \right. $$

Answer 2

$\ln(x)$ is negative as $x < 1$. So when you divide by $\ln(x)$ you have to flip the sign on the inequality.