Closed formula for the power series

Question

I have no clue how to attempt this problem.

consider the power series: $$\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}$$ Find the closed form formula for the function $f(x)$ to which the power series converges.

Answer 1

$$f(x)=\sum_{n=1}^{\infty} (-1)^n\frac{x^{n+1}}{n+1}$$ $$f'(x)=\sum_{n=1}^{\infty} (-x)^n=\frac1{x+1}$$ $$\therefore f(x)=\int \frac1{x+1} dx=\ln (1+x)+C$$ $$f(0)=0 \rightarrow C=0$$

Answer 2

One way forward is to write for $-1<x\le 1$

$$\begin{align} \sum_{n=0}^\infty \frac{(-1)^nx^{n+1}}{n+1}&=\sum_{n=0}^\infty (-1)^n\int_0^x u^n\,du\\\\ &=\lim_{N\to \infty}\int_0^x \sum_{n=0}^N(-u)^n\,du\\\\ &=\lim_{N\to \infty}\int_0^x \left(\frac{1-(-u)^{N+1}}{1+u}\right)\,du\\\\ &=\log(1+x)-\lim_{N\to \infty}\int_0^x \left(\frac{(-u)^{N+1}}{1+u}\right)\,du \tag 1\\\\ &=\log(1+x)-\int_0^x\lim_{N\to \infty} \left(\frac{(-u)^{N+1}}{1+u}\right)\,du \tag 2\\\\ &=\log(1+x) \end{align}$$

where the Dominated Convergence Theorem justifies going from $(1)$ to $(2)$