Show that $\ln(1/1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$ for $-1

Question

Show that $\ln(1/1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$ for $-1<x<1$ using the power series $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$

I did this using term by term integration

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

LHS: $\ln (1) - \ln (1-x) = \ln \frac{1}{1-x} + C$

What do I have to do with the right hand side of the equation? And afterwards, what is the reason for substituting $x=0$ to find the value of $C$ ? Where did the $x=0$ came from?

$\ln(1/1-x)$ should be written $\ln(1/(1-x))$ with parentheses. — Jean Marie, Nov 18 '22 at 11:25
If you have an equation which is true for any value of x, then you can substitute in any value that you like. So if it's useful to set $x=0$ then you are free to do so. — Blitzer, Nov 18 '22 at 11:35

score 3 · Answer 1 · answered Nov 18 '22 at 11:37

You are on the right track. For $ |x|<1$ let

$$f(x)= \sum_{n=1}^{\infty}\frac{x^n}{n}.$$

Then

$$f'(x)= \sum_{n=1}^{\infty}x^{n-1}= \sum_{n=0}^{\infty}x^{n}= \frac{1}{1-x}.$$

Hence

$$f(x)= - \ln (1-x)+c.$$

With $x=0$ we see that $c=0.$

Therefore

$$f(x)= - \ln (1-x)= \ln (1)-\ln (1-x)= \ln ( \frac{1}{1-x}).$$

Show that $\ln(1/1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$ for $-1

1 Answers1