<s>[INST] Problem with a simple convergent power series Create a power series around $x=0$ and find its radius. What is $f(1)$?
$$f(x)=\frac{1}{2+x}$$
Well according to me it is easy to see that the radius is $\left|x\right|<2$ and $f(1)=\frac{1}{3}$...right?
If I try to manipulate it to the geometric serie $f(x)=\frac{1}{1-x}$ I get $f(x)=\frac{1}{1-(-1-x)} =\frac{1}{2+x} $ which would be $\sum (-1-x)^k$ because $f(x)=\frac{1}{1-x}=\sum x^k $ but $\sum (-1-x)^k$ diverge when $x=1$. Why doesnt that work when $f(x)=\frac{1}{2}\frac{1}{1+x/2)}=\frac{1}{2}\sum (\frac{-x}{2})^k$ seems to do fine.$f(1) =1/3$
Do not use maclaurin to prove the series.
I am doing something fundamentally wrong here, please help me out.
 [/INST] I start with
$$\frac{1}{t+1}=\sum _{k=0}^{\infty } (-1)^k t^k$$
Now I substitude
$$t=\frac{x}{2}$$
It gives me
$$\frac{1}{\frac{x}{2}+1}=\sum _{k=0}^{\infty } (-1)^k \left(\frac{x}{2}\right)^k$$
Multiply with $\frac{1}{2}$ I get
$$\frac{1}{2 \left(\frac{x}{2}+1\right)}=\frac{1}{2} \sum _{k=0}^{\infty } (-1)^k \left(\frac{x}{2}\right)^k=\sum _{k=0}^{\infty } \frac{(-1)^k x^k}{2^{k+1}}$$
And this is
$$\frac{1}{x+2}=\sum _{k=0}^{\infty } \frac{(-1)^k x^k}{2^{k+1}}$$
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