While showing the existence of radius of convergence, standard proofs make use of the following theorem
Theorem: Suppose we have a series $\sum_{n=0}^\infty a_ny^n$ which is convergent. Then any series of the form $\sum_{n=0}^\infty a_nx^n$ is absolutely convergent as long as $0<|x|<|y|$.
Proof: As $\sum_{n=0}^\infty a_ny^n$ is convergent, $a_n|y|^n < 1$ for sufficiently large $n$ and for those values of $n$, $a_n|x|^n = a_n|y|^n \times |x/y|^n < |x/y|^n$. As $a_n|x|^n$ is sandwiched between $0$ and $|x/y|^n$ with $|0<|x/y|<1|$, the series $\sum_{n=0}^\infty a_n|x|^n$ is convergent.
Even though the hypothesis is that $\sum_{n=0}^\infty a_ny^n$ is convergent, the proof just makes use of a weaker fact (and I suppose the theorem could have used a weaker hypothesis) that the sequence $(a_ny^n)$ is a null sequence, as in it tends to $0$. This made me wonder about the following problem.
My Question: If a power series $\sum_{n=0}^\infty a_ny^n$ is such that the sequence is a $(a_ny^n)$ is a null sequence, then what can we say about the convergence of the power series?
I know that in general, a series $\sum_{n=0}^\infty a_n$ formed from a null sequence $(a_n)$ need not converge, as is the case with $a_n=1/n$. However, this time, it is a null sequence of the form $(a_ny^n)$. Does that make any difference at all when we have to say something about the convergence of its corresponding power series $\sum_{n=0}^\infty a_ny^n$?
Best Answer
Since every null sequence $(a_n)$ can be written as the terms of a power series — we can trivially take $y = 1$, and if we want it a little less trivial set $b_n = a_ny^{-n}$ for some $y \neq 0$ to have $a_n = b_ny^n$ — it makes no difference, from $a_ny^n \to 0$ we cannot deduce that the power series converges at $y$.
You are right, the theorem can use a weaker hypothesis than convergence, even weaker than your $a_ny^n \to 0$, it suffices that $(a_ny^n)$ is bounded. If $\lvert a_n y^n\rvert \leqslant M$ for all $n$, then we can majorise $$\lvert a_n x^n\rvert \leqslant M\cdot \biggl\lvert \frac{x}{y}\biggr\rvert^n\,.$$ For every $x$ with $\lvert x\rvert < \lvert y\rvert$ the terms on the right are the terms of a convergent geometric series, whence $\sum a_nx^n$ converges absolutely for $\lvert x\rvert < \lvert y\rvert$.
We thus can characterise the radius of convergence $R$ of the power series as $$R = \sup\: \{ r \geqslant 0 : a_nr^n \to 0\} = \sup\: \{ r \geqslant 0 : \lvert a_n\rvert r^n \text{ is bounded}\}\,.$$ These are often easier to find than $\limsup \lvert a_n\rvert^{1/n}$ as used in the Cauchy–Hadamard formula.