Radius of convergence about a different point

Question

Let $$\frac1{1+x^2}=\sum_{n=0}^\infty a_n(x-a)^n\ \ a\in\mathbb{R},$$then what is the radius of convergence of the power series?

I think it should be $\infty$. Is it right?

Answer 1

No. The radius of convergence of the series is $\sqrt{1+a^2}$, which turns out to be the distance from $a$ to $\pm i$.

Note that, if $a=0$, then you get the series $\displaystyle\sum_{n=0}^\infty(-1)^nx^{2n}$, whose radius of convergence is $1=\sqrt{1+0^2}$.

Answer 2

A good rule to look ratios of convergence is look for "circles of convergence" into the complex plane. Even $\frac1{1+x^2}$ is continuous over R isn't that over the complex plane, and the only converges inside the unitary circle, because the function don't exist for x=+i and x=-i. So, it can't "cross" the unitary circle, for complexes values of x. This fact is translate for the real domain as the intersection of the complex circle of convergence with the real axis. Then, the (real) radius of convergence is |x| < 1, and need to study apart the cases x=1 and x=-1.

Answer 3

The coefficients can be obtained by the Taylor formula developed around $x=a$. But the computation of the derivatives is uneasy, and it is better to use a change of variable $x^2=t$.

Then

$$\left((1+t)^{-1}\right)^{(n)}=(-1)^nn!(1+t)^{-1-n}$$

and the coefficients of the development around $a^2$ are

$$b_n:=(-1)^n(1+a^2)^{-1-n},$$

$$\frac1{1+t}=\frac1{1+a^2}-\frac t{(1+a^2)^2}+\frac{t^2}{(1+a^2)^3}-\cdots$$

By the ratio test, the radius of convergence for $t$ is

$$1+a^2$$ and the corresponding radius for $x$ is

$$\sqrt{1+a^2}.$$