Power Series and Functions — Question 2

Question 2 – Solution

Step 1: Make the sign substitution.

Replace $x$ by $-x$ in $1/(1-x)=\sum x^n$: $$\boxed{\frac1{1+x}=\sum_{n=0}^{\infty}(-x)^n=\sum_{n=0}^{\infty}(-1)^n x^n}.$$

Step 2: Translate the restriction.

The geometric condition is $|-x|<1$, equivalent to $|x|<1$, so $R=1$.

Step 3: Test the endpoints.

• At $x=1$, the series is $\sum(-1)^n$, which diverges by the nth-term test.

• At $x=-1$, it is $\sum1$, which diverges.

Conclusion.

The interval is $\boxed{(-1,1)}$. Although $1/(1+x)$ is finite at $x=1$, its Taylor series about $0$ does not converge there.