Question

Question: Determine whether the series is convergent or divergent.

\(\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{{{\rm{n}}^{\rm{3}}}}}{{{{\rm{5}}^{\rm{n}}}}}} \)

Short answer

The series is perfectly convergent.

Step-by-step solution

Ratio Test.

Consider the Limit now \({\rm{L = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \left| {\frac{{{{\rm{a}}_{{\rm{n + 1}}}}}}{{{{\rm{a}}_{\rm{n}}}}}} \right|\)

The series is totally convergent if\({\rm{L}} \prec {\rm{1}}\).

The series is divergent if\({\rm{L}} \succ {\rm{1}}\).

If\({\rm{L = 1}}\), this test isn't applicable; instead, try something different.

Find out if the series is convergent or divergent.

In this case, \({{\rm{a}}_{\rm{n}}}{\rm{ = }}\frac{{{{{\rm{(n + 1)}}}^{\rm{3}}}}}{{{{\rm{5}}^{{\rm{n + 1}}}}}}\;\;\;{{\rm{a}}_{{\rm{n + 1}}}}{\rm{ = }}\frac{{{{\rm{n}}^{\rm{3}}}}}{{{{\rm{5}}^{\rm{n}}}}}\)

Therefore,

\(\begin{aligned}{c}{\rm{L = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \left| {\frac{{{{\rm{a}}_{{\rm{n + 1}}}}}}{{{{\rm{a}}_{\rm{n}}}}}} \right|\\{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \left| {\frac{{{{{\rm{(n + 1)}}}^{\rm{3}}}}}{{{{\rm{5}}^{{\rm{n + 1}}}}}}{\rm{ \div }}\frac{{{{\rm{n}}^{\rm{3}}}}}{{{{\rm{5}}^{\rm{n}}}}}} \right|{\rm{n}}\\{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \left| {\frac{{{{{\rm{(n + 1)}}}^{\rm{3}}}}}{{{{\rm{5}}^{{\rm{n + 1}}}}}}{\rm{ \times }}\frac{{{{\rm{5}}^{\rm{n}}}}}{{{{\rm{n}}^{\rm{3}}}}}} \right|{\rm{n}}\\{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{{{\rm{(n + 1)}}}^{\rm{3}}}}}{{{{\rm{n}}^{\rm{3}}}}}{\rm{ \times }}\frac{{{{\rm{5}}^{\rm{n}}}}}{{{{\rm{5}}^{{\rm{n + 1}}}}}}{\rm{n}}\\{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {\left( {\frac{{{\rm{n + 1}}}}{{\rm{n}}}} \right)^{\rm{3}}}{\rm{ \times }}\frac{{{{\rm{5}}^{\rm{n}}}}}{{{{\rm{5}}^{\rm{n}}}{\rm{ \times 5}}}}\\{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{n}}}} \right)^{\rm{3}}}{\rm{ \times }}\frac{{\rm{1}}}{{\rm{5}}}\\{\rm{ = (1 + 0}}{{\rm{)}}^{\rm{3}}}{\rm{ \times }}\frac{{\rm{1}}}{{\rm{5}}}{\rm{ < 1}}\end{aligned}\)

As a result, the series is perfectly convergent.