Question

Determine whether the series is convergent or divergent.

\(\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{{{\left( {{\rm{ - 1}}} \right)}^{\rm{n}}}}}{{\sqrt {{\rm{n + 1}}} }}} \)

Short answer

In the Alternating Series Test, the provided series is convergent.

Step-by-step solution

Alternating series test.

Such that \({{\rm{a}}_{\rm{n}}}{\rm{ = ( - 1}}{{\rm{)}}^{\rm{n}}}{{\rm{b}}_{\rm{n}}}\;\;\;\)or \(\;{a_n} = {( - 1)^{n + 1}}{b_n},\;\;\;\)were \({{\rm{b}}_{\rm{n}}} \ge {\rm{0}}\)for all\({\rm{n}}\).

The series is convergent if the following two requirements are satisfied.

1. \(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{b}}_{\rm{n}}}{\rm{ = 0}}\)
2. \({{\rm{b}}_{\rm{n}}}\) is a decreasing sequence

Find out if the series is convergent or divergent.

\(\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{{{{\rm{( - 1)}}}^{\rm{n}}}}}{{\sqrt {{\rm{n + 1}}} }}} \)

The series is convergent if the following two requirements are satisfied.

\({{\rm{b}}_{\rm{n}}}{\rm{ = }}\frac{{\rm{1}}}{{\sqrt {{\rm{n + 1}}} }}\)

Which satisfies both the Alternating series conditions.

Note: \({{\rm{b}}_{\rm{n}}}{\rm{ = }}\frac{{\rm{1}}}{{\sqrt {{\rm{n + 1}}} }}\)is decreasing because \(\sqrt {{\rm{n + 1}}} \)is increasing. Also\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{b}}_{\rm{n}}}{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{\rm{1}}}{{\sqrt {{\rm{n + 1}}} }}\).

As a result of the Alternating Series Test, the provided series is convergent.