Consider the alternating series $ \sum_{n=1}^{\infty} (-1)^n \frac{5}{n^3} x^n $.

Find the radius of convergence of it.


$a_n=(-1)^n \frac{5}{n^3} $

Root test:

$ \lim_{n \to \infty} \sqrt[n]{|a_n|}=\lim_{n \to \infty} \sqrt[n]{\frac{5}{n^3}}=5$

Ratio test:

$ \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|=\lim_{n \to \infty} |\frac{(n+1)^3}{n^3}|=1 $

So both limits are not equal.

But we know that if limits exists , then they will be equal.

So I am doing something wrong.

Help me

  • 2
    $\begingroup$ You're doing the root test wrongly, it should be $1$ $\endgroup$ – LucaMac Sep 7 '18 at 15:40
  • $\begingroup$ @LucaMac, clarify my mistake please $\endgroup$ – timebound Sep 7 '18 at 15:41
  • 1
    $\begingroup$ $\sqrt[n]{5} \to 1$ $\endgroup$ – LucaMac Sep 7 '18 at 15:45

By roots test we have

$$ \lim_{n \to \infty} \sqrt[n]{|a_n|}=\lim_{n \to \infty} \sqrt[n]{\frac{5}{n^3}}=1$$

then the radius of convergence is $1$.

To check for the convergence we need to consider separately the cases $x=1$ and $x=-1$.

  • 1
    $\begingroup$ You don't need to check $x=1$ and $x=-1$ if the question is just about the radius of convergence. $\endgroup$ – Robert Israel Sep 7 '18 at 15:46
  • 1
    $\begingroup$ Yes you are absolutely right, I fix it! Thanks $\endgroup$ – gimusi Sep 7 '18 at 15:49

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