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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.

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  • $\begingroup$ The geometric is the only needed tool to understand radius of convergence of power series. $\sum_{n=0}^\infty R^n z^n$ converges for $|z| < 1/R$. Thus if $|a_n| \le C R^{n}$ then $\sum_{n=0}^\infty a_n z^n$ converges for $|z| < 1/R$. Indeed this is a iff when $R$ is replaced by $R+\epsilon$. Finally sometimes it is easier to show $|\frac{a_n}{a_{n+1}}| \le R$, which implies $|a_n| \le C (R+\epsilon)^{n}$ and hence $\sum_{n=0}^\infty a_n z^n$ converges for $|z| < 1/R$. $\endgroup$
    – reuns
    Oct 5, 2017 at 1:50
  • $\begingroup$ $R$ is the limit, not the inequality. The equation says nothing about convergence – you must be quoting it out of context, there must be some words that go along with that equation. Anyway, if the inequality is true, then the series converges. But your last question leaves me speechless. Again, the context – presumably, we are talking about a sequence of coefficients, so of course there's an input $n$. $\endgroup$ Oct 5, 2017 at 3:07

1 Answer 1

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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 $x$.

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