# Find the radius of convergence of the alternating power series

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

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

$$\lim_{n \to \infty} \sqrt[n]{|a_n|}=\lim_{n \to \infty} \sqrt[n]{\frac{5}{n^3}}=1$$
then the radius of convergence is $1$.
To check for the convergence we need to consider separately the cases $x=1$ and $x=-1$.
• You don't need to check $x=1$ and $x=-1$ if the question is just about the radius of convergence. – Robert Israel Sep 7 '18 at 15:46