Binomial Series — Question 2

Question 2 – Solution

Step 1: Compute the generalized coefficients.

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

Step 2: Connect with geometry.

This is exactly $$\frac1{1-(-x)}=\sum_{n=0}^{\infty}(-x)^n,$$ whose common ratio is $-x$. It converges when $|-x|<1$, so $R=1$.

Step 3: Test endpoints.

At $x=1$, terms are $(-1)^n$; at $x=-1$, terms are $1$. Neither tends to zero.

Conclusion.

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