Question

Determine whether the series is convergent or divergent \(\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{\rm{n}}}{{{{\rm{n}}^{\rm{3}}}{\rm{ + 1}}}}} .\)

Short answer

The series is convergent.

Step-by-step solution

the denominator is reduced,

The fraction becomes larger when reducing the denominator.

Therefore,

\(\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{\rm{n}}}{{{{\rm{n}}^{\rm{3}}}{\rm{ + 1}}}}} \le \sum\limits_{{\rm{n = 1}}}^\infty {\frac{{\rm{n}}}{{{{\rm{n}}^{\rm{3}}}}}} {\rm{ = }}\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{\rm{1}}}{{{{\rm{n}}^{\rm{2}}}}}} \)

The blue series is known as the \({\rm{p}}\)-series, it is convergent because \({\rm{p = 2 > 1}}\)

Note that:

(1) All the terms of the given series are positive.

(2) It is smaller than a convergent series.

Use the Direct Comparison Test to conclude that it is also convergent.

the Direct Comparison Test,

Part of the Direct Comparison Test:

Consider two series \({{\rm{a}}_{\rm{n}}}\) and \({{\rm{b}}_{\rm{n}}}\) such that

\((1)\) Every term \({{\rm{a}}_{\rm{n}}}\) is positive.

\((2)\) \({{\rm{b}}_{\rm{n}}}\) is convergent.

\((3)\)\({{\rm{a}}_{\rm{n}}}{\rm{ < }}\;{{\rm{b}}_{\rm{n}}}\)

If above the three conditions are satisfied, then \({{\rm{a}}_{\rm{n}}}\) is also convergent.

Therefore, the series is convergent.