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

Question

I am having trouble understainding why

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

I understand how I got

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

but when I substitute $x$ with $x+1$ and multiply $-1$, I get

$$-\ln(x+1)=-x+x^2/2-x^3/3+... $$

which is off by the alternating negative sign. . .

Can someone help me out?

Answer 1

The reason is that the statement in your question is incorrect.

The correct statement is $$\sum_{k = 1}^\infty \frac{x^k}k = -\ln(1 - x).$$

Answer 2

Take IGP as $$\frac{1}{1+x}=1-x+x^2-x^3+... +\frac{x^k}{k}+..., |x|<1.~~~~(1)$$ Integrate w,r,t, $x$ both sides, we get $$\ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^k}{k}+C~~~~(2)$$ $C$ is constant of integration which can be obtained to be 0 by putting $x=0$ in )2). So $C=0$. let $1+x=y$ in (2),we get $$\ln y=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{(y-1)^k}{k}$$ Your Eq.(1), the factor $(-1)^{k-1}$.

Answer 3

$\log (1-x)=\int_{1}^{1-x}\frac{1}{t}dt=$

$-\int_{0}^{x}\frac{1}{1-t}dt=-\int_{0}^{x}\sum_{i=0}^{\infty}t^i dt=$

$-\sum_{0}^{\infty}\int_{0}^{x}t^i dt=-\sum_{0}^{\infty}\frac{x^{i+1}}{i+1}=$

$-\sum_{1}^{\infty}\frac{x^i}{i}.$