Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Expert verified

The series is perfectly convergent.

Step by step solution

01

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.

02

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,

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

As a result, the series is perfectly convergent.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A certain ball has the property that each time it falls from a height \(h\)\(\) onto a hard, level surface, it rebounds to a height \(rh\), where \(0 < r < 1\). Suppose that the ball is dropped from an initial height of \(H\) meters.

(a) Assuming that the ball continues to bounce indefinitely, find the total distance that it travels.

(b) Calculate the total time that the ball travels. (Use the fact that the ball falls \(\frac{1}{2}g{t^2}\) meters in \({t^{}}\)seconds.)

(c) Suppose that each time the ball strikes the surface with velocity \(v\) it rebounds with velocity \( - kv\) , where \(0 < k < 1\). How long will it take for the ball to come to rest?

Use definition 2 directly to prove that \(\mathop {\lim }\limits_{n \to \infty } {(0.8)^n} = 0\)(from ? with, \(r = 0.8\)). Use logarithms to determine how large n has to be so that \((0.8) < 0.000001\).

We have seen that the harmonic series is a divergent series whose terms approach 0. Show that \(\sum\limits_{n = 1}^\infty {\ln (1 + \frac{1}{n})} \) is another series with this property.

Determine whether the series is convergent or divergent. If its convergent, find its sum.

\(\sum\limits_{n = 1}^\infty {\frac{{{3^n}}}{{{e^{(n - 1)}}}}} \)

Let S be the sum of a series of \(\sum {{a_n}} \) that has shown to be convergent by the Integral Test and let f(x) be the function in that test. The remainder after n terms is

\({R_n} = S - {S_n} = {a_{n + 1}} + {a_{n + 2}} + {a_{n + 3}} + .......\)

Thus, Rn is the error made when Sn , the sum of the first n terms is used as an approximation to the total sum S.

(a) By comparing areas in a diagram like figures 3 and 4 (but with x โ‰ฅ n), show that

\(\int\limits_{n + 1}^\infty {f(x)dx \le {R_n} \le \int\limits_n^\infty {f(x)dx} } \)

(b) Deduce from part (a) that

\({S_n} + \int\limits_{n + 1}^\infty {f(x)dx \le S \le {S_n} + \int\limits_n^\infty {f(x)dx} } \)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free