Find the following sum

Question

$$x-\frac{x^2}2+\frac{x^3}3-...+\frac{(-1)^{n+1}x^{n}}n+...$$

---

All I've found out is that is is similar to $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}=\ln2 $$

Answer 1

Let $f(x):=\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n}$ and let $R>0$ be its ratio of convergence. Since $f$ converges uniformly in $[-R,R]$, you can take the derivative of $f$, which is $$f’(x)=\sum_{n=1}^{\infty} (-1)^{n+1}x^{n-1}=\sum_{n=0}^{\infty} (-1)^n x^n=\frac{1}{1+x},$$ and hence $$f(x)=\int_0^x \frac{1}{1+t}dt=\ln(1+x).$$ In fact, observe that $$\ln(2)=f(1)=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n},$$ as you mentioned.