Which function has this Taylor series: $ \sum _{n=1}^{\infty }\:\frac{x^n}{n} $ (no factorial)?

Question

I'm eager to know which functions has $ \sum _{n=1}^{\infty }\:\frac{x^n}{n} $ as a Taylor series. $ ln(x+1) $ comes close, but it has $ (-1)^n $ inside the summation.

[EDITED: changed n=0 to n=1]

Answer 1

Start with

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

The series has obviously $1$ as radius of convergence (ratio test)

Now in its domaine of convergence its derivative is

$$f’(x)=\sum_{n=1}^\infty x^{n-1}=\sum_{n=0}^\infty x^n={1\over 1-x}$$

Taking the anti derivative we get

$$f(x)=-\log(1-x)$$

Answer 2

$$f(x) = \sum\limits_{n = 1}^{\infty} \frac{x^n}{n}$$ $$f'(x) = \sum\limits_{n = 0}^{\infty} x^n$$ $$f'(x) = \frac{1}{1 - x}$$ $$f(x) = -\ln (1 - x) $$

I assumed that summation goes from $n = 1$ since you can't divide by $0$.

Answer 3

Consider for $|x| \lt 1$ $$\int\frac{1}{1-x}dx = \int\left(1 + x + x^2 + x^3 + ....\right)dx$$ $$\implies -\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots = \sum_{n=1}^\infty\frac{x^n}{n}$$

$$\implies 1 + \sum_{n=1}^\infty\frac{x^n}{n} = \boxed{1 - \ln(1-x)}$$

Answer 4

Yes,$$\log(x+1)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4-\cdots$$and therefore$$-\log(-x+1)=x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+\cdots$$