Power series and shifting index

Question

First I have to find the power series represantion for the following function:

$$\ f(x) = \ln(1+x)$$

I tried the following:

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

Therefore,

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

Up to this point I think my answer is correct, but I have to shift the index by one so it starts from

n = 1 and I get:

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

I need some help in this last part if everything above is correct

Answer 1

You have to write $$(-x)^n=(-1)^{n}\cdot x^n$$ If you integrate that you have $${(-1)}^n\cdot \frac{x^{n+1}}{n+1}$$ Shifting gives $${(-1)}^{n-1}\cdot \frac{x^n}{n}=(-1)^{n+1}\cdot \frac{x^n}{n}$$ So your approach was correct, but you forgot to seperate the sign before integrating.

Answer 2

Index shifting is simple:

$$\sum_{n=a}^b f(n)$$

can be rewritten with a variable substitution, for instance $n\to m+s$

$$\sum_{m+s=a}^b f(m+s)$$ or

$$\sum_{m=a-s}^{b-s} f(m+s).$$

If you want a specific starting value, choose the suitable $s$.