Question

Determine whether the series is conditionally convergent, absolutely convergent, or divergent.

\(\sum\limits_{{\rm{n = 1}}}^\infty {{{{\rm{( - 1)}}}^{{\rm{n - 1}}}}{{\rm{n}}^{{\rm{ - 3}}}}} \)

Short answer

The provided series is convergent in all directions.

Step-by-step solution

check whether the series is absolutely convergent.

We'll start by seeing if the series is absolutely convergent.

Because if a series is absolutely convergent, it must be convergent.

However, if the series isn't absolutely convergent. After that, we'll see if the series is convergent or divergent.

Remember that if \(\sum {\left| {{{\rm{a}}_{\rm{ - }}}{\rm{n}}} \right|} \)is convergent, \(\sum {{{\rm{a}}_{\rm{ - }}}} {\rm{n}}\)is said to be absolutely convergent.

As a result, when the following series converges, the supplied series will be absolutely convergent.

\(\begin{array}{c}\sum\limits_{{\rm{n = 1}}}^\infty {\left| {{{{\rm{( - 1)}}}^{{\rm{n - 1}}}}{{\rm{n}}^{{\rm{ - 3}}}}} \right|} {\rm{ = }}\sum\limits_{{\rm{n = 1}}}^\infty {{{\rm{n}}^{{\rm{ - 3}}}}} \\{\rm{ = }}\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{\rm{1}}}{{{{\rm{n}}^{\rm{3}}}}}} \end{array}\)

This is a \({\rm{p}}\)-series with\({\rm{p = 3 > 1}}\).

As a result, the series converges.

Furthermore, the original series in the question is absolutely convergent.

Result.

The series \(\sum\limits_{{\rm{n = k}}}^\infty {\frac{{\rm{1}}}{{{{\rm{n}}^{\rm{p}}}}}} \)is called \({\rm{p}}\)-series

A \({\rm{p}}\)-series diverges when \({\rm{p}} \le {\rm{1}}\)

And converges when \({\rm{p > 1}}\)

Therefore, the provided series is convergent in all directions.