I have been trying to find upper bound of following thing with no success. Is there any inequality s.t following holds .
assume that $a \geq ,b \geq 1$ then find $f(a,b) \rightarrow \mathbb{R}^{+}$ s.t forall $n\geq 1$
$$\frac{an}{n-b} \leq f(a,b)\cdot n $$
Assuming that $n$ is a natural number, you should divide the inequality by $n$ as @RossMillikan mentioned in the comments. Then you get $\frac{a}{n-b} \leq f(a,b)$ for all $n \geq 1$. Now it is easy to show that $\frac{a}{n-b}>\frac{a}{n+1-b}$ if $n>b$ and $\frac{a}{n-b}<0$ for all $n<b$. Thus in the case $b \notin \mathbb{N}$ the function $f(a,b)=\frac{a}{\lceil b \rceil -b}$ works, since it holds for the smallest $n>b$ and is positive. In the case $b \in \mathbb{N}$ the same argumentation holds for $f(a,b)=a$.