Question

Determine whether the series is convergent or divergent.

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

Short answer

The series is convergent.

Step-by-step solution

To use the comparison test.

The \({\rm{p - }}\)series \(\sum\nolimits_{{\rm{n = 1}}}^\infty {{\textstyle{{\rm{1}} \over {{{\rm{n}}^{\rm{p}}}}}}} \)is convergent if \({\rm{p}} \succ {\rm{1}}\)and divergent if\({\rm{p}} \le {\rm{1}}\).

Find out if the series is convergent or divergent.

\(\begin{array}{c}{{\rm{a}}_{\rm{n}}}{\rm{ = }}\frac{{\sqrt {{\rm{n + 1}}} {\rm{ - }}\sqrt {{\rm{n - 1}}} }}{{\rm{n}}}\\{\rm{ = }}\frac{{{\rm{(}}\sqrt {{\rm{n + 1}}} {\rm{ - }}\sqrt {{\rm{n - 1}}} {\rm{)(}}\sqrt {{\rm{n + 1}}} {\rm{ + }}\sqrt {{\rm{n - 1}}} {\rm{)}}}}{{{\rm{n(}}\sqrt {{\rm{n + 1}}} {\rm{ + }}\sqrt {{\rm{n - 1}}} {\rm{)}}}}\\{\rm{ = }}\frac{{{\rm{(n + 1) - (n - 1)}}}}{{{\rm{n(}}\sqrt {{\rm{n + 1}}} {\rm{ + }}\sqrt {{\rm{n - 1}}} {\rm{)}}}}\\{\rm{ = }}\frac{{\rm{2}}}{{{\rm{n(}}\sqrt {{\rm{n + 1}}} {\rm{ + }}\sqrt {{\rm{n - 1}}} {\rm{)}}}}\\{\rm{(1) < }}\frac{{\rm{2}}}{{{\rm{n}}\sqrt {{\rm{n + 1}}} }}{\rm{ }}\\{\rm{(2) < }}\frac{{\rm{2}}}{{{\rm{n}}\sqrt {\rm{n}} }}{\rm{ = }}\frac{{\rm{2}}}{{{{\rm{n}}^{{\rm{3/2}}}}}}\end{array}\)

The series is thus convergent.