# Is there any expansion for log(1+x) when x>1?

Is there any expansion for $\mathrm{log}\left(1+x\right)$ when $x>1$?
Is there any expansion for $\mathrm{log}\left(1+x\right)$ when $x>1$?
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bargeolonakc
You can expand the function $\mathrm{log}\left(1+x\right)$ around any point at which it is defined. This means there exists an expansion of $\mathrm{log}\left(1+x\right)$ around the point $x=2$, for example, however it will be of the form
$\mathrm{log}\left(1+x\right)=\sum _{i=1}^{\mathrm{\infty }}{a}_{i}\left(x-2{\right)}^{i},$
and the expansion will be valid for $x\in \left(-1,5\right)$
However, I imagine you want the expansion to of the form
$\mathrm{log}\left(1+x\right)=\sum _{i=1}^{\mathrm{\infty }}{a}_{i}{x}^{i},$
for which you will have a problem. The problem is that any power series
$\sum _{i=1}^{\mathrm{\infty }}{a}_{i}\left(x-{x}_{0}{\right)}^{i}$
will converge on a interval symmetric around ${x}_{0}$ (meaning an interval of the type $\left({x}_{0}-\delta ,{x}_{0}+\delta \right)$). This means that if the series expansion for $\mathrm{log}\left(1+x\right)$ will converge for $x>1$, it will also converge for some $x<-1$ which is impossible.
###### Did you like this example?
You are looking for Laurent series for $|x|>1$
$\mathrm{ln}\left(1+x\right)=\mathrm{ln}\left(x\left(1+1/x\right)\right)=\mathrm{ln}\left(x\right)+\mathrm{ln}\left(1+1/x\right)=\mathrm{ln}\left(x\right)+\sum _{k=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{k}}{k}{x}^{-k}$
with the condition of convergence $|x|>1$