Binomial Series — Question 6

*Question 6 Derive the binomial series for $(1-x)^{-2}$ and show that it agrees with differentiating the geometric series. Determine the exact interval of convergence.

*Question 6 – Solution

For $n\ge0$, $$\binom{-2}{n}=\frac{(-2)(-3)\cdots(-n-1)}{n!}=(-1)^n(n+1).$$ Substitute $u=-x$ into $(1+u)^{-2}$: $$\binom{-2}{n}(-x)^n=(-1)^n(n+1)(-1)^n x^n=(n+1)x^n.$$ Thus $$\boxed{(1-x)^{-2}=\sum_{n=0}^{\infty}(n+1)x^n}.$$

$$\frac1{1-x}=\sum_{n=0}^{\infty}x^n\quad\Longrightarrow\quad \frac1{(1-x)^2}=\sum_{n=1}^{\infty}n x^{n-1}=\sum_{n=0}^{\infty}(n+1)x^n.$$

The radius is $1$. At $x=1$ terms are $n+1$; at $x=-1$ their magnitudes are $n+1$. Both endpoints diverge.

The interval is $\boxed{(-1,1)}$.