Is there any equivalent of Taylor/Maclaurin series of $\ln(1+x)$ for $|x| > 1$

Question

I happened to come across Taylor series and Maclaurin series recently, but everywhere I read about the expansion for $\ln(1+x)$, it was stated that the approximation is valid for $-1 < x < 1$.

I understand that the bounds for $x$ are because the series doesn't converge for $|x|>1$, but is there any equivalent of this series for the value of $|x|$ as greater than $1$?

Please note that I am not asking if we can compute for $|x|>1$ or not, as that can be done by computing for $\frac 1x$, which will then lie between $-1$ and $1$.

Also, I'm quite new to all this, so new that today was the day I read the name 'Maclaurin' for the first time. So any answers understandable with high school mathematics are highly appreciated.

Thanks!

Answer 1

We can actually develop a series that converges for all positive $x$.

Begin by rendering the Maclaurin series for $\ln(1+y)$ and $\ln(1-y)$:

$\ln(1+y)=\sum_{n=1}^\infty \dfrac{(-1)^{n-1}y^n}{n}$

$\ln(1-y)=\sum_{n=1}^\infty \dfrac{(-1)y^n}{n}$

Now subtract the second equation from the first, recognizing that (1) the difference between logarithms on the left side is now a logarithm of a quotient, (2) only the odd $n$ terms on the right will give a nonzero difference. Thereby

$\ln(\dfrac{1+y}{1-y})=2\sum_{m=1}^\infty {\dfrac{y^{2m-1}}{2m-1}}$

Now define $x=(1+y)/(1-y)$, which is solved for $y$ to give $y=(x-1)/(x+1)$. Note that this latter equation implies $|y|<1$ for all positive $x$ -- so, convergence of the series follows! The final, convergent series takes the form

$\ln(x)=2\sum_{m=1}^\infty {\dfrac{(\frac{x-1}{x+1})^{2m-1}}{2m-1}}$

Answer 2

$f(x)=\ln(1+x)$, If $x>1$, $$f(x)=\ln x+\ln\left(1+\frac{1}{x}\right)=\ln x+ \sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^{-k}}{k}.$$