Divergence of harmonic-like series $f^{*}(n):=2f(n)-f(2n)$
$f:\mathbb{N}\rightarrow\mathbb{R}^+\;$ is such that $\;\forall n\in\mathbb{N}\;\;f^{*}(n)>0$
Is it true that $\;\sum_\limits{n=1}^\infty \frac1{f^{*}(n)}\;$ always diverges?
To be honest, I have no idea if this statement is true, and there is no context of this problem, just a curiosity.
 A: I found this confusing to visualise what is going on, but I believe I have found a counter-example nonetheless.
Start by considering a function $\ f_2\ $ that maps $\ A_2=\{2,4,8,16\ \ldots\}\to\mathbb{R}, $ defined as $\ f_2(n) = \frac{n}{\ln(n)}\ (>0\ \forall n\geq2).\ $ Then $\  {f_2}^{*}(n) = 2f_2(n) - f_2(2n) = 2n\left( \frac{1}{\ln(n)} - \frac{1}{\ln(n)+\ln(2)} \right)\ (>0\ \forall n\geq 2),\ $ and so, using the substitution $\ n = 2^k,$
$$ \sum_{n\in A_2} \frac{1}{{f_2}^*(n)} = \sum_{n\in A_2} \frac{1}{2n\left( \frac{1}{\ln(n)} - \frac{1}{\ln(n)+\ln(2)} \right)} = \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{2^k \left( \frac{1}{k\ln(2)} - \frac{1}{k\ln(2)+\ln(2)} \right)} $$
$$= \frac{\ln(2)}{2} \sum_{k=1}^{\infty} \frac{1}{2^k \left( \frac{1}{k} - \frac{1}{k+1} \right)} = \frac{\ln(2)}{2} \sum_{k=1}^{\infty} \frac{\left(k^2+k\right)}{2^k},$$
which converges due to the ratio test or root test. Let's say it converges to $\ \alpha>0.\ $
Next: $\ 3\ $ is the next number after $\ 2\ $ that wasn't in our domain, so let $\ A_3 = \{ 3,6,12,24,\ldots\},\ $ and let $\ f_3: A_3\to\mathbb{R}\ $ be defined as $\ c_3 \frac{n}{\ln(n)}\ $ where is $\ c_3>0\ $ is some constant to be discussed in a moment.
$$\sum_{n\in A_3} \frac{1}{{f_3}^*(n)} $$
converges for the same reason that  $ \sum_{n\in A_2} \frac{1}{{f_2}^*(n)}\ $ converges in the calculation above. Now we may freely choose $\ c_3\ $ so that $\sum_{n\in A_3} \frac{1}{{f_3}^*(n)}\ $ converges to $\ \frac{\alpha}{2}.$
Then, we do a similar thing by defining for $\ A_5\ $ and $\ f_5,\ $ then choosing $\ c_5\ $ so that $\sum_{n\in A_5} \frac{1}{{f_5}^*(n)}\ $ converges to $\ \frac{\alpha}{4},\ $ then we define $\ A_7\ $ and $\ f_7,\ $  then choose $\ c_7\ $ so that $\sum_{n\in A_7} \frac{1}{{f_7}^*(n)}\ $ converges to $\ \frac{\alpha}{8},\ $  then we define $\ A_9\ $ and $\ f_9,\ $  then choose $\ c_9\ $ so that $\sum_{n\in A_9} \frac{1}{{f_9}^*(n)}\ $ converges to $\ \frac{\alpha}{16},\ $  etc, and continuing indefinitely with this process whereby we do not skip any numbers as we go, we end up with: $\;\sum_\limits{n=1}^\infty \frac1{f^{*}(n)}\;$ converging to $\ \alpha + \frac{\alpha}{2} + \frac{\alpha}{4} + \frac{\alpha}{8} + \ldots = 2\alpha.$
