Why do I get a different power series?

Question

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.

Answer 1

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.

Answer 2

The series $\displaystyle\sum_{n=0}^\infty(-1)^nx^n$ converges to $\dfrac1{1+x}$ when $\lvert x\rvert<1$, whereas the series $\displaystyle\sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}}(x-1)^n$ converges (also) to $\dfrac1{1+x}$ when $\left\lvert x-1\right\rvert<2$. There is no problem here.

More generally, if $a>0$, then $\displaystyle\sum_{n=0}^\infty\frac{(-1)^n}{a^{n+1}}(x-a)^n$ converges to $\dfrac1{1+x}$ when $\lvert x-a\rvert<a$.