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 neglected in the identity1 a-b+ 1 a-2b+ 1…

Given,

$$\frac{1}{a-b}+\frac{1}{a-2b}+\frac{1}{a-3b}+\ldots .+\frac{1}{a-nb}$$a−b1​+a−2b1​+a−3b1​+….+a−nb1​

$$={(a-b)}^{-1}+{(a-2b)}^{-1}+\ldots .+{(a-nb)}^{-1}$$=(a−b)−1+(a−2b)−1+….+(a−nb)−1

$$=\frac{1}{a}\sum _{r=1}^{n}{(1-\frac{rb}{a})}^{-1}$$=a1​r=1∑n​(1−arb​)−1

$$=\frac{1}{a}\sum _{r=1}^{n}{(1+\frac{rb}{a}+\frac{{r}^{2}{b}^{2}}{{a}^{2}})+(\mathrm{terms}\mathrm{to}\mathrm{be}\mathrm{neglected})}$$=a1​r=1∑n​(1+arb​+a2r2b2​)+(termstobeneglected)

$$=\frac{1}{a}[n+\frac{n(n+1)}{2}\cdot \frac{b}{a}+\frac{n(n+1)(2n+1)}{6}\cdot \frac{{b}^{2}}{{a}^{2}}]$$=a1​[n+2n(n+1)​⋅ab​+6n(n+1)(2n+1)​⋅a2b2​]

$$=\frac{1}{a}[n+\frac{{n}^{2}+n}{2}\cdot \frac{b}{a}+\frac{2{n}^{3}+3{n}^{2}+n}{6}\cdot \frac{{b}^{2}}{{a}^{2}}]$$=a1​[n+2n2+n​⋅ab​+62n3+3n2+n​⋅a2b2​]

So $\gamma =\frac{{b}^{2}}{3{a}^{3}}$γ=3a3b2​