My question pertains to what one can deduce from the following relation, taken by trying to compute the ratio test of a generic power series:
$$|x-x_0| < \lim_{n\to \infty} | \frac{a_n}{a_{n+1}}| = R$$
This is how it was displayed in my lecture notes, and doesn't address the questions I'm about to ask.
- Does R equal that whole equation? That seems to be a relation with an inequality. I'm not sure how that can equate to some number R.
- What does that equation say? That if the ratio of $a_n$ to $a_{n+1}$ is greater than $|x-x_0|$, the series will diverge?
- Why do we do about the limit operator on those two entries $a_n$ and $a_{n+1}$? How do we treat them if they have no input $n$?
Any help is sought for earnestly.
Let $\sum_{n=0}^{\infty}a_n(x-x_0)^n$ be a power series with $a_n \ne 0$ for all $n$ and put $b_n(x)=a_n(x-x_0)^n$.
Furthermore let $\lim_{n\to \infty} | \frac{a_n}{a_{n+1}}| = R>0$
For $x \ne x_0$ we then have
$|\frac{b_{n+1}(x)}{b_n(x)}|=|\frac{a_{n+1}}{a_n}||x-x_0| \to \frac{1}{R}|x-x_0|$.
The ratio test now gives:
If $\frac{1}{R}|x-x_0|<1$, then the power series converges in $x$.
If $\frac{1}{R}|x-x_0|>1$, then the power series diverges in $x$.
Conclusion:
If $|x-x_0|<R$, then the power series converges in $x$.
If $|x-x_0|>R$, then the power series diverges in $x$.