Power Series and Functions — Question 4

Question 4 – Solution

Step 1: Expand the derivative.

For $|t|<1$, $$\frac1{1+t}=\sum_{n=0}^{\infty}(-1)^n t^n.$$

Step 2: Integrate from $0$ to $x$.

Term-by-term integration is valid when $|x|<1$: $$\begin{align*} \log(1+x)-\log1&=\int_0^x\frac{dt}{1+t}\\&=\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+1}}{n+1}. \end{align*}$$ Thus $$\boxed{\log(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n}}.$$ The lower limit makes the constant $0$.

Step 3: Test endpoints.

At $x=1$, the alternating harmonic series converges to $\log2$. At $x=-1$, the series is $-\sum1/n$ and diverges.

Conclusion.

$R=1$ and the interval is $\boxed{(-1,1]}$.