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Determine whether the series is convergent or divergent.

\(\sum\limits_{{\rm{n = 1}}}^\infty {{\rm{ln}}\left( {\frac{{\rm{n}}}{{{\rm{3n + 1}}}}} \right)} \)

Short Answer

Expert verified

The series is divergent.

Step by step solution

01

Check the limit.

\({{\rm{a}}_{\rm{n}}}{\rm{ = ln}}\frac{{\rm{n}}}{{{\rm{3n + 1}}}}\)

Use Divergence Test

\(\begin{array}{c}\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to \infty } {\rm{ln}}\left( {\frac{{\rm{n}}}{{{\rm{3n + 1}}}}{\rm{ \times }}\frac{{{\textstyle{{\rm{1}} \over {\rm{n}}}}}}{{{\textstyle{{\rm{1}} \over {\rm{n}}}}}}} \right){\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to \infty } {\rm{ln}}\frac{{\rm{1}}}{{{\rm{3 + }}{\textstyle{{\rm{1}} \over {\rm{n}}}}}}\\{\rm{ = ln}}\frac{{\rm{1}}}{{{\rm{3 + 0}}}}\\{\rm{ = ln}}\frac{{\rm{1}}}{{\rm{3}}}\end{array}\)

02

Result.

\(\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to \infty } {\rm{ln}}\left( {\frac{{\rm{n}}}{{{\rm{3n + 1}}}}{\rm{ \times }}\frac{{{\textstyle{{\rm{1}} \over {\rm{n}}}}}}{{{\textstyle{{\rm{1}} \over {\rm{n}}}}}}} \right){\rm{ = ln}}\frac{{\rm{1}}}{{\rm{3}}}\)

Therefore, the series is divergent.

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