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Determine whether the series is convergent or divergent \(\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{{{\rm{n}}^{\rm{2}}}{\rm{ + 1}}}}{{{{\rm{n}}^{\rm{3}}}{\rm{ + 1}}}}} \).

Short Answer

Expert verified

The series is perfectly Diverges because the result is positive\({\rm{1}}\).

Step by step solution

01

Limit comparison test (LCT).

The limit comparison test (LCT) (in contrast with the related direct comparison test) is a method of testing for the convergence of aninfinite series. Suppose that in a two series\({{\rm{\Sigma }}_{\rm{n}}}{{\rm{a}}_{\rm{n}}}\)and\({{\rm{\Sigma }}_{\rm{n}}}{{\rm{b}}_{\rm{n}}}\)with\({{\rm{a}}_{\rm{n}}} \ge {\rm{0,}}{{\rm{b}}_{\rm{n}}}{\rm{ > 0}}\)for all\({\rm{n}}\).Then if\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{{\rm{a}}_{\rm{n}}}}}{{{{\rm{b}}_{\rm{n}}}}}{\rm{ = c}}\)with\({\rm{0 < c < }}\infty \), then either both series converge or both series diverge.

02

Find out if the series is convergent or divergent.

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

This series looks like Limit comparison test (LCT)\(\frac{{\rm{1}}}{{\rm{n}}}\)

So,

\(\begin{array}{c}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{\frac{{{{\rm{n}}^{\rm{2}}}{\rm{ + 1}}}}{{{{\rm{n}}^{\rm{3}}}{\rm{ + 1}}}}}}{{\frac{{\rm{1}}}{{\rm{n}}}}}{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{{\rm{n}}^{\rm{2}}}{\rm{ + 1}}}}{{{{\rm{n}}^{\rm{3}}}{\rm{ + 1}}}}{\rm{ \times n}}\\{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{{\rm{n}}^{\rm{3}}}{\rm{ + n}}}}{{{{\rm{n}}^{\rm{3}}}{\rm{ + 1}}}}{\rm{ \times }}\frac{{\frac{{\rm{1}}}{{{{\rm{n}}^{\rm{3}}}}}}}{{\frac{{\rm{1}}}{{{{\rm{n}}^{\rm{3}}}}}}}\\{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{\rm{1 + }}\frac{{\rm{1}}}{{{{\rm{n}}^{\rm{2}}}}}}}{{{\rm{1 + }}\frac{{\rm{1}}}{{{{\rm{n}}^{\rm{3}}}}}}}\\{\rm{ = }}\frac{{{\rm{1 + 0}}}}{{{\rm{1 + 0}}}}\\\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{\frac{{{{\rm{n}}^{\rm{2}}}{\rm{ + 1}}}}{{{{\rm{n}}^{\rm{3}}}{\rm{ + 1}}}}}}{{\frac{{\rm{1}}}{{\rm{n}}}}}{\rm{ = 1}}\end{array}\)

As the result is positive integer \({\rm{1}}\) so, that the series is diverges.

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Most popular questions from this chapter

(a)Show that if \(\mathop {\lim }\limits_{n \to \infty } {a_2}_n = L\)and \(\mathop {\lim }\limits_{n \to \infty } {a_{2n + 1}} = L,\) then {\({a_n}\)} is convergent and \(\mathop {\lim }\limits_{n \to \infty } {a_n} = L\).

(a) If \({a_1} = 1\) and

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