By applying term-wise differentiation and integration find the sum of the series $\sum_{k=1}^{\infty}\frac{x^k}{k}$

Question

I need to find the sum of the following series: $$\sum_{k=1}^{\infty}\frac{x^k}{k}$$ on the interval $x\in[a,b], -1<a<0<b<1$ using term-wise differentiation and integration.

Can anyone give me a hint on how to start this using the method described because it seems fairly simple but I'm stumped.

Answer 1

$$f(x)=\sum_{k=1}^{\infty}\frac{x^k}{k}=x+\frac{x^2}2+\frac{x^3}3+\ldots$$ By differentiang w.r.t. $x$ $$f'(x)=\sum_{k=1}^{\infty} x^{k-1}=1+x+x^2+\cdots$$

for$|x|\lt1$we have

$$f'(x)=1+x+x^2+\cdots=\frac{1}{1-x}$$ by integrating we get $$f(x)=-\ln (1-x)+C$$ and can determine the unique value of $C$ to get

$$\sum_{k=1}^{\infty}\frac{x^k}{k}=-\ln (1-x)$$