Taylor Series/Radius of Convergence - I just need a hint

[DivGradCurl]

Consider the following:

$$f(x) = 1 + x + x^2 = 7 + 5 (x-2) + (x-2)^2$$

which is a Taylor series centered at 2. My question is: what is the radius of convergence? The answer in my book is $$R=\infty$$, but take a look at this:

$$f(x) = 7 + 5 (x-2) + (x-2)^2 = \sum _{n=0} ^{\infty} b_n (x-2)^n \Longrightarrow \left| x-2 \right| < 1$$

Then, I get

$$1 \leq x \leq 3 \Longrightarrow R = \frac{3-1}{2}=1 \neq \infty$$

In other words, I'm a bit confused!

Thanks

 

[Tide]

Your confusion is that the $b_n$ are ALL 0 for n > 2. There's nothing to test since the series is perfectly well defined for all x. You might try, for example, comparing the terms in your series with the terms in a series you know converges for all x such as $e^{-x^2}$. Clearly, for n > 2 each term of your series is smaller than the corresponding term in the latter expansion.

 

[Hurkyl]

It looks like you tried to use the ratio test, but you can't because you need to find the limit of 0/0.

However, you can apply the n-th root test, which will, indeed, say that it converges for all x. (as it should because the function is a polynomial which exists for all x!)

 

[DivGradCurl]

It makes sense now.

Thanks.