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Determine whether the series is convergent or divergent.

\(\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{{{\left( {{\rm{ - 5}}} \right)}^{{\rm{2n}}}}}}{{{{\rm{n}}^{\rm{2}}}{{\rm{9}}^{\rm{n}}}}}} \)

Short Answer

Expert verified

The Root Test shows that the provided series is divergent.

Step by step solution

01

To use the Root Test.

Because \({\rm{2n}}\)is always positive, this is not an alternating series. Alternatively, you can use exponent rules to make it more evident.

\(\begin{array}{c}\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{{{{\rm{( - 5)}}}^{{\rm{2n}}}}}}{{{{\rm{n}}^{\rm{2}}}{{\rm{9}}^{\rm{n}}}}}} {\rm{ = }}\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{{{\left[ {{{{\rm{( - 5)}}}^{\rm{2}}}} \right]}^{\rm{n}}}}}{{{{\rm{n}}^{\rm{2}}}{{\rm{9}}^{\rm{n}}}}}} \\{\rm{ = }}\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{{\rm{2}}{{\rm{5}}^{\rm{n}}}}}{{{{\rm{n}}^{\rm{2}}}{{\rm{9}}^{\rm{n}}}}}} \end{array}\)

Root Test:

\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \sqrt[{\rm{n}}]{{\left| {\frac{{{\rm{2}}{{\rm{5}}^{\rm{n}}}}}{{{{\rm{n}}^{\rm{2}}}{{\rm{9}}^{\rm{n}}}}}} \right|}}{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{\rm{25}}}}{{{{\rm{n}}^{{\rm{2/n}}}}{\rm{9}}}}{\rm{ - - - - - - - - - - - - (1)}}\)

02

Find out if the series is convergent or divergent.

Separately check the \({\rm{n}}\)component.

\(\begin{array}{c}{\rm{y = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{n}}^{{\rm{2/n}}}}\\{\rm{lny = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {\rm{ln}}{{\rm{n}}^{{\rm{2/n}}}}\\{\rm{lny = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{\rm{2}}}{{\rm{n}}}{\rm{lnn}}\\{\rm{lny = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{\rm{2}}\frac{{\rm{1}}}{{\rm{n}}}}}{{\rm{1}}}\\{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{\rm{2}}}{{\rm{n}}}\\{\rm{ = 0y}}\\{\rm{ = }}{{\rm{e}}^{\rm{0}}}\\{\rm{ = 1}}\end{array}\)

Substitute in equation \({\rm{(1)}}\)

\(\begin{array}{c}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{\rm{25}}}}{{{{\rm{n}}^{{\rm{2/n}}}}{\rm{9}}}}{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{\rm{25}}}}{{{\rm{(1)9}}}}\\{\rm{ = }}\frac{{{\rm{25}}}}{{\rm{9}}}\end{array}\)

As a result, the Root Test shows that the provided series is divergent.

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