If b is very small as compared to the value of a , so that the cube and other higher powers of b a can be…

Given,

$\frac{1}{a-b}+\frac{1}{a-2b}+\frac{1}{a-3b}+\dots .+\frac{1}{a-nb}$

$={\left(a-b\right)}^{-1}+{\left(a-2b\right)}^{-1}+\dots .+{\left(a-nb\right)}^{-1}$

$=\frac{1}{a}\sum _{r=1}^n{\left(1-\frac{rb}{a}\right)}^{-1}$

$=\frac{1}{a}\sum _{r=1}^n\left\{\left(1+\frac{rb}{a}+\frac{r^2{b}^2}{a^2}\right)+\left(\mathrm{terms} \mathrm{to} \mathrm{be} \mathrm{neglected}\right)\right\}$

$=\frac{1}{a}\left[n+\frac{n(n+1)}{2}·\frac{b}{a}+\frac{n(n+1)(2n+1)}{6}·\frac{b^2}{a^2}\right]$

$=\frac{1}{a}\left[n+\frac{n^2+n}{2}·\frac{b}{a}+\frac{2n^3+3n^2+n}{6}·\frac{b^2}{a^2}\right]$

So $\gamma =\frac{b^2}{3a^3}$