By definition, the logarithm log of base $b$ is given by $$\log_b(b^x)=x,$$ $$\log_b(1)=0.$$
In a similar fashion, I am trying to define a function that is always decreasing as its input is getting smaller. The problem is that my example is supposed to work with two variables.
What I want is $f_b$ such that $$f_b(\frac{n}{b^k}) > f_b(\frac{n}{b^{k+1}})$$ for every $b > 1$, $n$, and natural $k$. Also, it must hold for every allowed value that $$\sum_{k=0}^{\infty}f_b(\frac{n}{b^k}) = \infty.$$ I tried defining it as $$f_b(\frac{n}{b^k})=\frac{1}{n+k},$$ but picking $k=1$, $n=b$ gives $\frac{1}{b+1}$, and $n=1$, $k=0$ gives $1$, thus $b=0$.
Could you please let me know what a well-defined function for any $b$ could be? Is it even possible?
I think $f_b(x)=\frac{1}{1+\ln(1+\frac{1}{x})}$ works.
We get $f_b(\frac{n}{b^k}) > f_b(\frac{n}{b^{k+1}})$ because we have $f_b(\frac{n}{b^k})=\frac{1}{1-\ln n+\ln(n+b^k)}$ and $f_b(\frac{n}{b^{k+1}})=\frac{1}{1-\ln n+\ln(n+b^{k+1})}$.
We also have $$\sum_{k=0}^{\infty}f_b\bigg(\frac{n}{b^k}\bigg)=\infty$$ because $$\begin{align}\sum_{k=0}^{\infty}f_b\bigg(\frac{n}{b^k}\bigg)&=\sum_{k=0}^{\infty}\frac{1}{1-\ln n+\ln(n+b^k)} \\\\&\ge \sum_{k=0}^{\infty}\frac{1}{1-\ln n+\ln((n+1)b^k)} \\\\&=\sum_{k=0}^{\infty}\frac{1}{1-\ln n+\ln(n+1)+k\ln b} \\\\&\ge\sum_{k=0}^{\infty}\frac{1}{\left\lceil 1-\ln n+\ln(n+1)\right\rceil+k\left\lceil\ln b\right\rceil} \\\\&=\infty\end{align}$$