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

Question

If $b$ is very small compared to $a$, so that the cube and higher powers of $b/a$ can be neglected in the identity

$$\frac{1}{a-b}+\frac{1}{a-2b}+\frac{1}{a-3b}+\cdots+\frac{1}{a-nb}=\alpha n+\beta n^2+\gamma n^3$$

then the value of $\gamma$ is

Accepted Answer

$$(a-rb)^{-1}=\frac1a\left(1-\frac{rb}{a}\right)^{-1}\approx\frac1a\left(1+\frac{rb}{a}+\frac{r^2b^2}{a^2}\right)=\frac1a+\frac{rb}{a^2}+\frac{r^2b^2}{a^3}$$

Summing $r=1$ to $n$:

$$S=\frac{n}{a}+\frac{n(n+1)b}{2a^2}+\frac{n(n+1)(2n+1)b^2}{6a^3}$$

Coefficient of $n^3$: $\dfrac{2b^2}{6a^3}=\dfrac{b^2}{3a^3}$

Solutions · 1