Infinite series $\sum_{n=1}^{+\infty} a_n x^n$ is convergent when $x=-3$, and it is divergent when $x=20$ Assume that the series $\displaystyle\sum_{n=1}^{+\infty} a_n x^n$ is convergent when $x=-3$, and it is divergent when $x=20$
Is the series $\displaystyle\sum_{n=1}^{+\infty} a_n (-2)^n$ divergent?
Is the series  $\displaystyle\sum_{n=1}^{+\infty} a_n 24^n$ convergent?
I figure I could use a comparison test for the second one, but the first one isn't positive, so I'm not sure.
 A: Remember that the collection of all values of $x$ at which a power series around $0$ (such as this one) converges must be one of:


*

*Just the number $0$; or

*Every real number; or

*An interval that goes from $-R$ to $R$ (for some positive real number $R$), possibly including either (or both) endpoints.


Now, you know this series converges at $x=-3$, so we are not in the first situation.
You also know that it diverges at $x=20$, so we are not in the second situation.
So we must be in the third situation: the interval of convergence goes from some $-R$ to $R$. It is one of: $(-R,R)$, $[-R,R)$, $(-R,R]$, or $[-R,R]$. Everything strictly between $-R$ and $R$ is a value where the series definitely converges; everything outside is a value where it definitely diverges.
What do you know? You know that the series converges at $-3$; that means that $-3$ must be either inside the interval, or at an endpoint of that interval. So $-R\leq -3$, or equivalently, $R\geq 3$.
You know it diverges when $x=20$, so $20$ is outside the interval. That means that you must have $20\leq R$.
So we know that, whatever the value of $R$ (the radius of convergence) is, it is something between 3 and 20. 
Definitely, anything that is closer to $0$ than $-3$ is somewhere that the series converges; and anything further away from $0$ than $20$ is something where the series diverges.
That should be enough to answer this problem.
For a bonus question: What, if anything, can you say about $\sum\limits_{n=1}^{\infty}a_n(-5)^n$ ?
