Is there any expansion for $\log(1+x)$ when $x\gt 1$? Is there any expansion for $\log(1+x)$ when $x\gt 1$ ?
 A: You can expand the function $\log (1+x)$ around any point at which it is defined. This means there exists an expansion of $\log (1+x)$ around the point $x=2$, for example, however it will be of the form 
$$\log(1+x) = \sum_{i=1}^\infty a_i (x-2)^i,$$
and the expansion will be valid for $x\in(-1,5)$.
However, I imagine you want the expansion to of the form
$$\log(1+x)=\sum_{i=1}^\infty a_i x^i,$$
for which you will have a problem. The problem is that any power series $$\sum_{i=1}^\infty a_i (x-x_0)^i$$
will converge on a interval symmetric around $x_0$ (meaning an interval of the type $(x_0-\delta, x_0+\delta)$). This means that if the series expansion for $\log(1+x)$ will converge for $x>1$, it will also converge for some $x<-1$ which is impossible.
A: You are looking for Laurent series for $|x|>1$
$$ \ln(1+x) = \ln( x(1+1/x) ) = \ln(x)+\ln(1+1/x) = \ln(x) + \sum_{k=1}^{\infty}\frac{(-1)^k}{k}x^{-k} $$
with the condition of convergence $ |x|>1 $.
A: I do not know if this is what you expect, so forgive me if I am wrong.
Change variable $1+x=\frac{1+y}{1-y}$, that is to say $x=-\frac{2 y}{y-1}$ and use $$\log(1+x)=\log\frac{1+y}{1-y}=2 \sum_{i=1}^\infty \frac{y^{2i-1}}{2i-1}$$ For example, using $x=9$ and adding ten terms gives a value of $2.29890$ instead of $2.30259$.
A: There is indeed another series expansion involving the natural logarithm, namely:
$$\ln\left(\sum_{r=0}^{n-1} x^r\right)=\sum_{k=0}^\infty {\left(\sum_{r=1}^{n-1} {\left(\frac{x^{nk+r}}{nk+r}-\frac{x^{nk+n}}{nk+n}\right)}\right)}$$
Note that the well-known series expansion of $\ln(1+x)$ is a special case of this formula (namely for $n=2$), so it can be seen as a quite natural generalization of the series expansion of $\ln(1+x)$. For $x=1$ it yields an interesting expansion of $\ln(n)$:
$$\ln(2)=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}...$$
$$\ln(3)=\frac{1}{1}+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{8}-\frac{2}{9}...$$
$$\ln(4)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}-\frac{3}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{3}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}-\frac{3}{12}...$$
It can be proven as follows:
Define the following function:
$$f_n(x):=\sum_{k=0}^\infty {\left(\sum_{r=1}^{n-1} {\left(\frac{x^{nk+r}}{nk+r}-\frac{x^{nk+n}}{nk+n}\right)}\right)}$$
With $n\ge2$. Then, since every $x$ in the series has an exponent greater or equal to $1$, we have $f_n^{}(0)=0$. Therefore:
$$f_n(x)=f_n(x)-f_n(0)=\int_{0}^{x} f_{n}^{´}(t) dt$$
$f_{n}^{´}(t)$ is given by:
$$f_{n}^{´}(t)=\sum_{k=0}^\infty {\left(\sum_{r=1}^{n-1} {\left(t^{nk+r-1}-t^{nk+n-1}\right)}\right)}=\sum_{r=1}^{n-1} {\left(\sum_{k=0}^\infty {\left(t^{nk+r-1}-t^{nk+n-1}\right)}\right)}=\sum_{r=1}^{n-1} {\left(\frac{t^{r-1}}{1-t^n}-\frac{t^{n-1}}{1-t^n}\right)}=\frac{\sum_{r=1}^{n-1} {\left(t^{r-1}-t^{n-1}\right)}}{1-t^n}=\frac{\sum_{r=1}^{n-1} {\left(t^{r-1}\cdot \left(1-t^{n-r}\right)\right)}}{1-t^n}=\frac{\sum_{r=1}^{n-1} {\left(t^{r-1}\cdot (1-t)\cdot \left(\sum_{k=0}^{n-r-1} t^k \right)\right)}}{1-t^n}=\frac{(1-t)\cdot \sum_{r=1}^{n-1} {\left(\sum_{k=0}^{n-r-1} t^{k+r-1} \right)}}{(1-t)\cdot \sum_{k=0}^{n-1} t^k}=\frac{\sum_{r=0}^{n-2} {\left(\sum_{k=0}^{n-r-2} t^{k+r} \right)}}{\sum_{k=0}^{n-1} t^k}$$
Furthermore, whe have:
$$\sum_{r=0}^{n-2} {\left(\sum_{k=0}^{n-r-2} t^{k+r} \right)}=(t^0+t^1+...+t^{n-2})+(t^1+t^2+...+t^{n-2})+...+(t^{n-1}+t^{n-2})+(t^{n-2})=1\cdot t^0+2\cdot t^1+...+(n-2)\cdot t^{n-3}+(n-1)\cdot t^{n-2}=\sum_{k=1}^{n-1} k\cdot t^{k-1}$$
And hence, if we define $g_n(t):=\sum_{k=0}^{n-1} t^k$ (so $g_{n}^{`}(t)=\sum_{k=0}^{n-1} k\cdot t^{k-1}=\sum_{k=1}^{n-1} k\cdot t^{k-1}$) we get:
$$f_{n}^{´}(t)=\frac{\sum_{k=1}^{n-1} k\cdot t^{k-1}}{\sum_{k=0}^{n-1} t^k}=\frac{g_{n}^{´}(t)}{g_n(t)}$$
Note that $g_n(0)=1$.
Therefore, we have:
$$f_n(x)=\int_{0}^{x} f_{n}^{´}(t) dt=\int_{0}^{x} \frac{g_{n}^{´}(t)}{g_n(t)} dt=\ln(g_n(x))-\ln(g_n(0))=\ln(g_n(x))=\ln\left(\sum_{r=0}^{n-1} x^r\right)$$
I hope, this is kind of what you wanted to see and I apologize for all the mistakes in my English, I'm not a native speaker. :)
