MHB Root or Ratio Test: Interval of Convergence

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To find the interval of convergence for the series Σ(-x/10)^(2k), both the root and ratio tests confirm that the series converges when |x|<10. Expressing the series as a geometric series leads to the same conclusion, indicating convergence for |x|<10 and divergence for |x|>10. For x values of ±1, the series diverges. Therefore, the interval of convergence is determined to be (-10, 10). The analysis effectively demonstrates the convergence behavior of the series using multiple testing methods.
Fernando Revilla
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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.
 
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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 $(0,1).$
 
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