if $g(n) = \frac{f(n+1) - f(n)}{f(n)}$ is unbounded at $[0, \infty]$, Then $\sum_{n \in N}\frac{2^n}{f(n)}$ converges. Let $f$ be an unbounded and non decreasing, positive function at $[0, \infty]$.
I am trying to understand if $g(n) = \frac{f(n+1) - f(n)}{f(n)}$ is unbounded at $[0, \infty]$ (but bounded at $[0, a]$ for every positive $a$), Then $\sum_{n \in N}\frac{2^n}{f(n)}$ converges.
One such function satisfyes $g(n)$ can be defined as $f(0) = 2, f(n+1) = f(n)^2$.
Then we get:
$g(n) = \frac{f(n+1) - f(n)}{f(n)} = \frac{f(n)^2 - f(n)}{f(n)} =\frac{f(n)(f(n) - 1)}{f(n)} = f(n) -1 \rightarrow\infty $
And if we define:
$f(n) = 2^{2n}$
We get:
$g(n) = \frac{2^{2(n+1)} - 2^{2n}}{2^{2n}} = \frac{2^{2n+2} - 2^{2n}}{2^{2n}} = \frac{3*2^{2n}}{2^{2n}} = 3$ and $g$ is bounded.
$f(n) = 2^{2^n}$, however, satisfyes $g$.
It seems like the family of functions satisfy $g$ grows extremly fast.
Im not sure how to use any of the theorems to show convergence of the series.
Hints will be appericiated.
 A: Here is an answer for the case where $g(n)\to\infty$.
You have that
$$
g(n)=\frac{f(n+1)}{f(n)}-1
$$
is unbounded, so
$$
\frac{f(n+1)}{f(n)}
$$
is unbounded. Looking at the ratio test,
$$
\frac{\frac{2^{n+1}}{f(n+1)}}{\frac{2^n}{f(n)}}=\frac{2f(n)}{f(n+1)}\to0,
$$
so the series converges.
A: The statement is false if $g$ does not diverge to $\infty$ and we can even construct an increasing $f$ for a counterexample: We set $f(0):= 1$ and for nonnegative integers $n$ we define
$$f(n+1) \colon = \begin{cases} \sqrt{2}f(n) & \textrm{if }f(n)>2^{n}, \\\\
(n+2)f(n) & \textrm{otherwise.} \end{cases}$$
Between integers we can just interpolate linearly to get an increasing, positive function on $[0, \infty)$. Since $n!$ grows faster than $2^n$, this function will eventually satisfy $f(n) > 2^n$, at which point it starts to grow like $\sqrt{2}^n$ until it gets overtaken again by $2^n$, etc. This guarantees
$$g(n) = \frac{f(n+1) - f(n)}{f(n)} =  \frac{(n+1)f(n) - f(n)}{f(n)} = n$$
infinitely often, so $g$ is unbounded. But at the same time $f(n)< 2^n$ also happens infinitely often, so the series diverges because $\frac{2^n}{f(n)}$ doesn't even converge to $0$.
