# Power Series Expansion for $\log(\frac{1}{1-x})$ [closed]

How can I prove that the power series expansion

$$\log{\big(\frac{1}{1 - x}\big)} = \sum_{m = 1}^{\infty}\frac{x^{m}}{m}$$

is valid for every real $0 \leq x < 1$?

## closed as off-topic by heropup, Davide Giraudo, user251257, Jack's wasted life, ShaileshNov 25 '16 at 0:11

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Hint. Note that $\frac{d}{dx}(-\log(1-x))=\frac{1}{1-x}$.

$$S=1+x+x^2+x^3+\dots$$

$$xS=x+x^2+x^3+\dots$$

Subtract the two as follows:

\begin{align}S&=1+x+x^2+x^3+\dots\\xS&=\quad\ \ \ x+x^2+x^3+\dots\\\hline(1-x)S&=1\end{align}

$$S=\frac1{1-x}=1+x+x^2+x^3+\dots$$

Can you figure out what the following is?

$$\int_0^t\frac1{1-x}dx=\int_0^t1+x+x^2+x^3+\dots dx$$