Radius of convergence for a given sum

Question

What is a brief description of the radius of convergence? How do you find the radius of convergence for $$\sum_{i=1}^{\infty}2^i\cdot x^{-3(i-1)}$$

score 5 · Answer 1 · answered Jun 01 '14 at 06:59

5

I would say that it is a geometric series. Quotient = $\frac{2}{x^3}$. Konvergens when $\frac{2}{|x^3|}<1$.

answered Jun 01 '14 at 06:59

georg

2,749

1

more clever approach (+1) – S L Jun 01 '14 at 07:01

score 1 · Accepted Answer · answered Jun 01 '14 at 06:50

1

Root test gives $$\lim_{n\to\infty}\sqrt[n]{\left|2^n x^{-3(n-1)}\right |} < 1$$ which can be simplified into $$2 |x^{-3}| < 1$$ from where you get $$|x|>\sqrt[3]2$$

answered Jun 01 '14 at 06:50

S L

11,731

Could you please provide a brief description of what the radius of convergence is? – user117520 Jun 01 '14 at 06:57
@user117520 wikipedia is a very good source to learn it also here it is explained as well as there are few examples. – S L Jun 01 '14 at 06:59
Thanks! That's very helpful – user117520 Jun 01 '14 at 06:59
It looks like the radius of convergence is usually a number that |x| has to be less than. Why is it that |x| > r allows the series to converge in this instance? – user117520 Jun 01 '14 at 07:02
@user117520 I would say radius of convergence is an interval where value of $x$ gives converging value. It could be $|x| < r$ or $|x|>r$ but 99% of examples are $|x|>r$ – S L Jun 01 '14 at 07:04

score 1 · Answer 3 · answered Jun 01 '14 at 07:00

Hint

Start rewriting $$S=\sum_{i=1}^{\infty}2^i\cdot x^{-3(i-1)}=x^3~\sum_{i=1}^{\infty} \Big(\frac{2}{x^3}\Big)^i$$ in which you recognize the sum of a geometric progression. Using the standard formula, you then have $$S=\frac{2 x^3}{x^3-2}$$ I am sure that you can take from here.

Radius of convergence for a given sum

3 Answers3