Welcome to our community

Be a part of something great, join today!

Root or ratio test

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I quote a question from Yahoo! Answers

Σ(-x/10)^(2k) how do I find the interval of convergence using the root or ratio test?
I have given a link to the topic there so the OP can see my response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
We can express $\displaystyle\sum_{k=0}^{\infty}\left(\frac{-x}{10}\right)^{2k}=\sum_{k=0}^{\infty}\left(\frac{x^2}{100}\right)^{k}.$ Then,

$(a)$ Considering this series as a geometric series:
$$\left| \frac{x^2}{100} \right|<1\Leftrightarrow x^2<100\Leftrightarrow |x|<10$$
and the series is convergent iff $|x|<10.$

$(b)$ Using the ratio test:
$$\lim_{k\to \infty}\;\left| \left(\frac{x^2}{100}\right)^{k+1} \left(\frac{100}{x^2}\right)^{k} \right|=\frac{x^2}{100}<1\Leftrightarrow |x|<10$$
So, the series is convergent if $|x|<10$ and divergent if $|x|>10.$ If $x=\pm 1$ we get $\displaystyle\sum_{k=0}^{\infty}1=1+1+\ldots$ (divergent).

$(c)$ Using the root test:
$$\lim_{k\to \infty}\;\left| \left(\frac{x^2}{100}\right)^{k} \right|^{1/k}=\frac{x^2}{100}<1\Leftrightarrow |x|<10$$
So, the series is convergent if $|x|<10$ and divergent if $|x|>10.$

The interval of convergence is $(0,1).$