Taylor Series Representation of $\ln(x+1) - \ln(x)$

Question

Hello I am trying to find the
Taylor Series Representation of

$\ln(x+1) - \ln(x)$

So I know

$$\ln(1+x) =\sum_{n \ge 0}\frac{(-1)^nx^{n+1}}{n+1}\\$$

But for $\ln(x)$ I 'm told nothing about where it's centered and I don't believe theres a common taylor series representation for ln(x)

My teacher and TA are out of town. Is there a way I can do this without knowing where ln(x) is centered at?

Answer 1

There's no such thing as the Taylor series representation.

The function you have is (real) analytic on its domain, which is $(0,\infty)$, which means it can be represented as a Taylor series at each point of the domain.

Choose $x=1/2$ as the center; it's simpler if you set $x=t+1/2$, so you get $$ \ln(x+1)-\ln x= g(t)=\ln\left(t+\frac{3}{2}\right)-\ln\left(t+\frac{1}{2}\right)= \ln\frac{3}{2}-\ln\frac{1}{2}+\ln\left(1+\frac{2}{3}t\right)-\ln(1+2t) $$ Now we can expand both parts: $$ g(t)= \ln3+\sum_{n\ge1}(-1)^{n+1}\frac{2^n}{n3^n}t^n-\sum_{n\ge1}(-1)^{n+1}\frac{2^n}{n}t^n =\ln3+\sum_{n\ge1}(-1)^{n+1}\frac{2^n}{n}(1+3^{-n})t^n $$ and finally $$ \ln(x+1)-\ln x=\ln3+\sum_{n\ge1}(-1)^{n+1}\frac{2^n}{n}(1+3^{-n})\left(x-\frac{1}{2}\right)^{\!n} $$ Note that the radius of convergence is $1/2$.

Answer 2

The series you have for $\ln(x+1)$ works for $|x|<1$, so the series $$\ln(x)=\sum_{n\geq0}(-1)^{n}\frac{(x-1)^{n+1}}{n+1}$$ works for $x\in(0,2)$. Thus the series $$\ln(x+1)-\ln(x)=\sum_{n\geq0}\frac{(-1)^n}{n+1}\left[x^{n+1}-(x-1)^{n+1}\right]$$ works for $x\in(0,1)$.

Answer 3

$\begin{array}\\ \ln(x+1)-\ln(x) &=\ln(1+1/x)\\ &=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{nx^n}\\ \end{array} $

Converges for $|1/x| < 1$ or $|x| > 1$.