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Can someone help me finding a power series around $a \in \mathbb R \backslash \{0\}$ which converges to $\frac 1 {1-x}$ for $|x-a| < R$ and some $R > 0$ ?

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Around $0$, $$\frac1{1-x}=1+x+x^2+x^3+x^4+\ldots$$ absolutely convergent if $|x| <1$.

You can use this, let $x=y+a$, then $$\frac1{1-x}=\frac1{1-(y+a)}=\frac1{(1-a)-y} = \frac1{1-a}\cdot\frac1{1-\frac y{1-a}} . $$

So, use the above for $z:=\frac y{1-a}\ $ for $a\ne 1$. We have a convergent powerseries for $|z|<1$, i.e.\ $|y|<|1-a|=:R$.

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Yes. Thanks. I get $$ \sum_{n=0}^\infty (1-a)^{-n-1}(x-a)^n$$ for $x \in (a-1,a+1)$ –  André Jan 22 '13 at 15:30
    
Yes, but rather $x\in (1,2a-1)$ in case $a>1$.. –  Berci Jan 23 '13 at 12:50

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