Ratio test for power series with arbitrary terms?

Question

im goin through Petersons, Kelleys Difference equations textbook and in one exercise following idea appeared.

$a_k$ is coefficient of power series.

“Returning to the difference equation, we have

$\frac{a_{k+3}}{a_k}=\frac{1}{(k+3)(k+2)}$

which goes to 0 as $k$ approaches infty, and the ratio test implies the power series converges for all values of $x$.”

If i understand it correctly, the text suggests that since $\frac{a_{k+3}}{a_k}$ goes to zero, then $\frac{a_{k+1}}{a_k}$ goes to zero, hence the radius of convergence is $\infty$. Does that mean that i don’t have to look for ratio of the immediate succesor term, but a arbitrary one in this example? I struggle to find the formula for $\frac{a_{k+1}}{a_k}$.

Answer 1

let $x_k=\frac{a_{k+1}}{{a_k}}$ for $k \in \mathbb{N}$

$$x_k \space x_{k+1}\space x_{k+2} =\frac{a_{k+1}}{a_k}\times \frac{a_{k+3}}{a_{k+2}}\times \frac{a_{k+2}}{a_{k+1}}=\frac{1}{(k+3)(k+2)}$$

Answer 2

If i understand it correctly, the text suggests that since $\frac{a_{k+3}}{a_k}$ goes to zero, then $\frac{a_{k+1}}{a_k}$ goes to zero, hence the radius of convergence is $\infty$.

This argument doesn't work. For simplicity, I'll consider the ratios $\frac{a_{k+2}}{a_k}$, but the idea are the same for the ratios $\frac{a_{k+3}}{a_k}$.

You can have a small ratio $\frac{a_{k+2}}{a_k}$ with $\frac{a_{k+1}}{a_k}$ "large" and $\frac{a_{k+2}}{a_{k+1}}$ "very small", so that their ratio is still small. For instance, take

$$a_k = \frac{1}{k!} \text{ if } k \text{ is even,}$$

$$a_k = \frac{1}{(k-1)!} \text{ if } k \text{ is odd.}$$

Then $\frac{a_{k+1}}{a_k} = 1$ if $k$ is even, and $\frac{a_{k+1}}{a_k} = \frac{1}{k (k+1)}$ if $k$ is odd. As a consequence, $\frac{a_{k+2}}{a_k} \sim \frac{1}{k^2}$, while $\frac{a_{k+1}}{a_k}$ does not converge to $0$.

However, the convergence of $\frac{a_{k+3}}{a_k}$ to $0$ is still enough to conclude that the power series converges everywhere. The trick is to split the sums. Let $z \in \mathbb{C}$. Then

$$\sum_{k=0}^{+ \infty} a_k z^k = \sum_{k=0}^{+ \infty} a_{3k} (z^k)^3 + z \sum_{k=0}^{+ \infty} a_{3k+1} (z^k)^3 + z^2 \sum_{k=0}^{+ \infty} a_{3k+2} (z^k)^3.$$

In addition, the sequences $\frac{a_{3(k+1)}}{a_{3k}}$, $\frac{a_{3(k+1)+1}}{a_{3k+1}}$ and $\frac{a_{3(k+1)+2}}{a_{3k+2}}$ converge to $0$. By the ratio test, all the sums on the right hand-side converge, so $\sum_{k=0}^{+ \infty} a_k z^k$ converge.

This can be generalized: it is enough to prove that $\frac{a_{k+n}}{a_k}$ converges to $0$ for some $n \geq 1$ to conclude that the generating series converges everywhere.