Find the appropriate power series for $f(x) = \frac{1}{x + 2} $
First method : $\frac{1}{x + 2} = \frac{1}{2(1 - (-\frac{x}{2}))} = \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}x^n$ , $|x|\lt2$
Second method : $\frac{1}{x + 2} = \frac{1}{1 - (-x-1)} = \sum_{n=0}^\infty (-x-1)^n$ , $-2 \lt x \lt 0$
Is there any problem in my answers ? Why these power series have different radii of convergence ?
On
Your computations are correct, but the radii of convergence aren't the same. It's $2$ for the first power series and $1$ for the second one.
On
Convergence is doomed to fail at $x=-2$, hence the convergence radius when developing around $a$ cannot be larger than the distance from $a$ to $-2$, i.e., $|a+2|$ (and for higher reasons, this is indeed the radius of convergence). So when developing around $0$, the radius is $2$, when developing around $-1$, it is $1$ (and when developing around $42$ it would be $44$)
There is no problem (in principle; I haven't checked the details). Note that one power series has general term $\frac{(-1)^n}{2^n}x^n$ and the other series has general term $(-1)^n(x+1)^n$.
A power series with general term $a_n(x-a)^n$ is said to be centered around $a$. For meromorphic functions (like rational functions), the radius of convergence for a power series turns out to be the distance from the center to the nearest pole.
$\frac{1}{x+2}$ has a pole at $x=-2$. Your first series is centered around $0$, and therefore has radius of convergence $2$, while your second series is centered at $-1$, and therefore has radius of convergence $-1$.
In fact, one can use this to gain a hint of the geometry of the complex plane using just the real numbers. The series of $\frac{1}{x^2+1}$ centered around $a\in\Bbb R$ has radius of convergence $\sqrt{a^2+1}$, hinting of a pole one unit away from $0$, perpendicular to the real line. We know its two poles as $\pm i$, of course.