Prove $\sum _{n=1}^{\infty }\:\frac{x^n}{n}$ is equal to $-\ln\left(1-x\right)$, for $|x| < 1$

Question

Prove $$\sum _{n=1}^{\infty }\:\frac{x^n}{n} = -\ln\left(1-x\right),$$ for $|x| < 1$.

Pretty much title I've searched and seen some solutions using Taylor Series but isn't there a more intuative way to do this?

Answer 1

Since $|x|<1$, the series in the LHS is absolutely convergent by comparison with a geometric series.
Additionally, $\frac{x^n}{n}=\int_{0}^{x}y^{n-1}\,dy$, hence: $$ \sum_{n=1}^{N}\frac{x^n}{n} = \int_{0}^{x}\frac{1-y^N}{1-y}\,dy =-\log(1-x)-\int_{0}^{x}\frac{y^N}{1-y}\,dy.$$ by exchanging $\sum$ and $\int$. Now we just need to consider the limit as $N\to +\infty$, and that is just $-\log(1-x)$ since $|x|<1$ grants: $$ \left|\int_{0}^{x}\frac{y^N}{1-y}\,dy\right|\leq |x|^N \int_{0}^{x}\frac{dy}{1-y}\to 0.$$

Answer 2

Here is intuitive:

$$-\ln(1-x)=\int_0^x\frac1{1-t}\ dt=\int_0^x\left(1+t+t^2+t^3+\dots\ \right)\ dx\\=x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+\dots$$