Is my reasoning on this power series problem correct?

Question

$Problem:$

Let $$\sum_{n=0}^\infty c_nx^n$$

be convergent when $x=-4$ and divergent when $x=6$. What can we say about the convergence or divergence of the following series?

$a)$ $\sum_{n=0}^\infty c_n$

$b)$ $\sum_{n=0}^\infty c_n 8^n$

$c)$ $\sum_{n=0}^\infty c_n (-3)^n$

$d)$ $\sum_{n=0}^\infty (-1)^nc_n9^n$

$Solution$:

Let $I = \lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|<1$, with $a_n=c_nx^n$, be the convergence interval of the series, perhaps including or perhaps excluding the extremes of the interval.

We know that $-4 \in I$ and $6 \notin I $. Because this interval is defined by the ratio test, which involves an absolute value, we can conclude that $I=(-\lim_{n\to\infty}\frac{a_{n+1}}{a_n}, \lim_{n\to\infty}\frac{a_{n+1}}{a_n})$ is at least $[-4, 4)$ and at most $[-6, 6)$.

$a)$ This series has $x=1$, and $1\in I$. Therefore, the series converges.

$b)$ This series has $x=8$, and $8\notin I$. Therefore, the series diverges.

$c)$ This series has $x=-3$, and $-3\in I$. Therefore, the series converges.

$d$) See that $\sum_{n=0}^\infty (-1)^nc_n9^n=\sum_{n=0}^\infty c_n(-9)^n$ This series has $x=-9$, and $-9\notin I$. Therefore, the series diverges.

I'm very new to power series and I'm not confident that this reasoning is correct. If it's not, could someone point out my mistakes, or what would be the right approach? Thank you.

Answer 1

Yes, the solutions you provided itself are correct. Well done!

Maybe one remark can be useful here: $I= |a_{n+1}/a_n| < 1$ doesn't make much sense, because $I$ is an interval (not depending on $n$). This happens later in your proof again.

Here is how I would write the proof:

Denote the interval of convergence with $I:= \left]-R,R\right[$ (where we possibly also have convergence on boundary points). Since we have convergence for $x = -4$, it follows that $R \geq 4$ and since we have divergence for $x=6$, we have $R\leq 6$, thus $4 \leq R \leq 6$ and then we follow your reasoning to conclude.

Answer 2

The ratio test doesn't always work to determine the radius of convergence. That it exists is clear, because we can define it as the supremum of all positive numbers $x$ such that the series converges absolutely. If this supremum $R$ is $\infty$, the series converges everywhere; if it is zero the series converges only at zero; if $R>0$ is finite, then the series converges for $|x|<R$ and doesn't converge for $|x|>R$. Nothing general can be said at $x=\pm R$.

The Hadamard formula says that $R^{-1}=\limsup_{n\to\infty}\sqrt[n]{|a_n|}$ (with obvious conventions about zero and infinity), but you don't need it.

What you know is that the radius of convergence $R$ satisfies $4\le R\le 6$.

Therefore the series 1 and 3 converge, whereas 2 and 4 don't converge. Your argument is good, except for the mention of the ratio test.