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Chapter 1: Q13P (page 1)

Question: Use the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals.

13.鈭�1鈭�n2n3+1

Short Answer

Expert verified

The series鈭�1鈭�n2n3+1 is divergent.

Step by step solution

01

Definition of convergent and divergent

If the partial sumSnof an infinite series tend to a limit S, then the series is called convergent.

If the partial sumSn of an infinite series do not approach to a limit, the series is called divergent.

The limiting value S is called the sum of the series.

02

Apply the integral test

The given series is 鈭�1鈭�n2n3+1.

Use the integral in the given series as:

鈭�鈭�n2n3+1鈥�dn

Now, solve the integral is as follows:

Let, t=n3+1, dt=3n2鈥�dn.

03

Solve the integral

Substitute the value of t and dt into the integral and change the variable from n into t.

鈭�鈭�dt3t

Solve the integral is as follows:

鈭�鈭�dt3t=13lnt鈭�=13ln鈭�=鈭�

Here, the series approaches to infinite therefore the given series diverges.

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Most popular questions from this chapter

In(2-e-x)

Test the following series for convergence.

2.鈭�n=1鈭�(鈭�2)nn2

By computer or tables, find the exact sum of each of the following series.

(a)鈭�n=1鈭�n(4n2-1)2(b)鈭�n=1鈭�n3n!(c)鈭�n=1鈭�n(n+1)3n

1+ln(1+x)

Show that if pis a positive integer, then (pn)=0 when n>p,so (1+x)p=鈭�(pn)xnis just a sum of p+1terms, from n=0to n=p. For example, (1+x)2has 3terms, (1+x)3has 4terms, etc. This is just the familiar binomial theorem.
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