If $f(x)=2+6x+18x^2+54x^3+\ldots$ and $f(x) = O(g(x))$

Question

If $f(x)=2+6x+18x^2+54x^3+\ldots$ and $f(x) = O(g(x))$, what is the value of $g(x)$ in the above sequence ?

I tried calculating $f(x)$ using the sum of an infinite GP but can't understand how to find $g(x)$.

Answer 1

You can write the above sum as follows:

$ f(x)=2(1+3x+(3x)^2+\dots)$

Now, the thing inside the bracket is an infinite geometric progression with common difference $3x$. The sum can only be calculated if the absolute value of common difference is less than 1.

So, for $|3x|<1$, we have:

$f(x)=2(1+3x+(3x)^2+\dots)=\displaystyle\frac{2}{1-3x}$. Now, the expression $\displaystyle\frac{2}{1-3x}$ is always less than $0$ for $\displaystyle x>\frac{1}{3}$. So, $g(x)$ can be $1$ which would make $f(x)=O(1)$.

For the case when $|3x|\geq 1$, you can define $g(x)$ as any divergent sum such that each term of $g(x)$ is greater than or equal to the corresponding term in $f(x)$ after a particular value of $x$.

For example, $g(x)=2(1+e^{3x}+e^{2\times 3x}+e^{3\times 3x}+\dots)$

This is the best you can do for the divergent case as far as I know.