Question

I need to prove that ln(1-x) = - sum of x^n/n from 1 to infinity, but I don't understand where the negative comes from. Help?

Answer 1

Have you tried finding a pattern in the Taylor polynomial terms?

Answer 2

No, because we haven't learned that yet.

Answer 3

$$f(x)=\log(1-x)\\f'(x)=\frac{-1}{1-x}=-\frac1{1-x}\\f''(x)=-\frac{-1}{(1-x)^2}=-\frac1{(1-x)^2}\\f'''(x)=\frac{-2}{(1-x)^3}=-\frac2{(1-x)^3}\\f^{(4)}(x)=-\frac6{(1-x)^4}\\\dots$$At \(x=0\) observe these reduce to:$$f(0)=0\\f'(0)=-1\\f''(0)=-1\\f'''(0)=-2\\f^{(4)}(0)=-6\\\dots$$the key is to realize that the product rule of differentiation keeps multiplying the previous powers; e.g. for \(f^{(4)}\) our numerator \(6\) comes from \(3\times2\times1=3!\). In general, \(f^{(n)}(0)=(n-1)!\) by this pattern

Answer 4

rather \(f^{(n)}(0)=-(n-1)!\) **

Answer 5

From our Taylor expansion form at \(x=0\) it is obvious what happens next:$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=-\sum_{n=1}^\infty\frac{(n-1)!}{n!}x^n=-\sum_{n=1}^\infty\frac1nx^n$$

Answer 6

notice we don't include the \(n=0\) term since \(f(0)=0\)

Answer 7

Is there a way without using the Taylor expansion?

Answer 8

I think I have an idea. Did you learn about the derivative/integral of a sum?

Answer 9

Yes

Answer 10

The problem itself is about a Taylor expansion :-p but @SithsAndGiggles has a brilliant idea

Answer 11

tabby says he/she hasn't learned taylor expansions/series yet, though.

Answer 12

Good effort nonetheless @oldrin.bataku

Answer 13

Since \(f(x)=\ln(1-x)\), you have \(f'(x)=-\dfrac{1}{1-x}\).

Note that for \(|x|<1\),
\[-\frac{1}{1-x}=-\sum_{n=0}^\infty x^n\]
Take the integral of the sum and you get
\[f(x)=\ln(1-x)=\int\left(-\sum_{n=0}^\infty x^n\right)~dx\\
\ln(1-x)=\cdots\]

Answer 14

Btw, I'm a girl.

Answer 15

So what would the first few partial sums look like?

Answer 16

Partial sums? Just how rigorous is this proof supposed to be? I would think the integral of the series would suffice to show that the given series represents \(f(x)\).