I would like to ask for your help for the following exercise:
"Let the power Series $\sum a_nx^n$, where the formula of the sequence an is not given. It is given that the power series converges if $x=2$ and diverges if $x=-3 \ $. We would like to prove that the power series converges if $x=1$ and diverges if $x=4$".
I was thinking to use D' Alembert Criterion to prove it, but the formula of $a_n$ is not given. Could you please guide me?
Thank you very much in advance.
On
Suppose that $\sum a_n c^n$ diverges and that $\vert x \vert > \vert c \vert$. If $\sum \vert a_n \vert \vert x \vert^n$ would converges, its general term would be bounded, let say by $M>0$. Which means $\vert a_n \vert \vert x \vert ^n \le M$ for all $n \in \mathbb N$. But then $\vert a_n \vert \vert c \vert^n \le M \left\vert \frac{c}{x}\right\vert^n$. A contradiction as the last series is absolutely convergent being a geometric series with ratio less than one.
You can have a similar proof to prove that $\sum a_n x^n$ converges if $\sum a_n y^n$ converges with $\vert x \vert < \vert y \vert$.
You can apply that to the specific cases of your question.
Whenever a power series $\sum_{n=0}^\infty a_n(x-x_0)^n$ converges at a point $x$, then it converges at any point $y$ such that $\lvert y-x_0\rvert<\lvert x-x_0\rvert$. And if it diverges at a point $x$, then it diverges at any point $y$ such that $\lvert y-x_0\rvert>\lvert x-x_0\rvert$.
So, since your series is centered at $0$ and converges at $2$, then it converges at $1$ too, since $\lvert1\rvert<\lvert2\rvert$. And, since it diverges at $-3$, it diverges at $4$, since $\lvert4\rvert>\lvert-3\rvert$.