# Finding the radius of convergence of a cubic Taylor series

I'm trying to find the Taylor series and radius of convergence for $x-x^3$ centered at $a = -2$.

I've found the series to be: $$f(x) = 6-11\cdot(x+2)+6\cdot(x+2)^2-(x+2)^3$$

However, I'm having trouble figuring out how to find the radius of convergence for the series.

If it were written as a normal summation, I'd just use the absolute ratio test with $a_n$ and $a_{n+1}$, but since I don't have the series in this form I'm not sure what to do. Can anyone point me in the right direction?

• All the terms in this series from degree $4$ on have coefficient $0$. It's a finite sum and so converges everywhere. Just think before you jump to formalism and convergence tests... – Ethan Bolker Feb 19 '18 at 15:19
• @EthanBolker That was my intuition, but you've given me words to describe it. Thank you! – Alex Johnson Feb 19 '18 at 15:31
• You're welcome. You can answer your own question here so it doesn't remain on the unanswered queue. – Ethan Bolker Feb 19 '18 at 15:36
• A polynomial is its own Taylor series in disguise, whatever the center. – Christian Blatter Feb 19 '18 at 19:44

A series is convergent if the sequence of its partial sums $\left \{ S_1,\ S_2,\ S_3,\dots \right \}$ tends to a limit; that means that the partial sums become closer and closer to a given number when the number of their terms increases.
As all the terms in the series greater than degree $4$ have a coefficient of $0$, the partial sums will remain unchanged as the number of terms increases beyond 4. Because of this, the series will converge for all values of $x$.