<s>[INST] Why do I get a different power series? My question revolves around the power series of the following function:
$$ f(x) = \frac{1}{1+x}$$
Now, it is almost immediate to have the power series of this function by substituting $-x$ into the geometric series to get:
$$\frac{1}{1 + x} = \frac{1}{1-(-x)} = \sum_{k=0}^\infty (-x)^k = \sum_{k=0}^\infty (-1)^kx^k = 1 - x + x^2 - x^3 + \ldots$$
with radius of convergence $1$. Now what got me thinking is why if I do a slight manipulation, I didn't arrive at the same result. Here was what I did:
$$\frac{1}{1+x} = \frac{1}{2 + x - 1} = \frac{1}{2}\left(\frac{1}{1 +\frac{x}{2} - \frac{1}{2}}\right) = \frac{1}{2}\left(\dfrac{1}{1-\{ -\frac{1}{2}(x-1)\}}\right)$$
And using the definition of the geometric series:
$$\frac{1}{1+x} = \frac{1}{2}\sum_{k=0}^\infty \left(-\frac{1}{2}(x-1)\right)^k = \sum_{k=0}^\infty (-1)^k\frac{1}{2^{k+1}}(x-1)^k$$
When I try to evaluate them on small values, this series doesn't look equal to the above series. Why do I get a different answer? Or are they identically the same? If so, can someone please guide/show me? Thanks.  
 [/INST] Note that the center of the  $$\frac{1}{1+x} = \frac{1}{2}\sum_{k=0}^\infty \left(-\frac{1}{2}(x-1)\right)^k = \sum_{k=0}^\infty (-1)^k\frac{1}{2^{k+1}}(x-1)^k$$
is the point $x=1$ while the center of $$ \frac{1}{1 + x} = \frac{1}{1-(-x)} = \sum_{k=0}^\infty (-x)^k = \sum_{k=0}^\infty (-1)^kx^k = 1 - x + x^2 - x^3 + \ldots$$
is the point $x=0$
The series approximate the function about the center which in this case are two  different points.
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