   Chapter 11.2, Problem 48E

Chapter
Section
Textbook Problem

# Determine whether the series is convergent or divergent by expressing sn as a telescoping sum (as in Example 8). If it is convergent, find its sum.48. ∑ n = 2 ∞ 1 n 3 − n

To determine

Whether the series is convergent or divergent and obtain the series sum if the series is convergent.

Explanation

Given:

The series is n=21n3n.

Here, an=1n3n.

Result used:

If the limit of the partial sums exists and limnsn=L, then the series convergent and its sum is n=1an=L.

Calculation:

Consider the series, n=21n3n. (1)

Obtain the partial fraction of 1n3n.

1n3n=1n(n21)=1n(n+1)(n1)

=a1n(n1)+a2n(n+1) (2)

Multiply the equation by (n+1)(n1) and simplify the terms,

(n+1)(n1)n3n=a1(n+1)(n1)n(n1)+a2(n+1)(n1)n(n+1)1n=a1(n+1)n+a2(n1)n1=a1(n+1)+a2(n1)1=(a1+a2)n+(a1a2)

Equate the coefficient of n and the constant term on both sides,

a1+a2=0a1a2=1

Solve the above equations and obtain a1=12and a2=12.

Substitute the values a1=12and a2=12 in equation (2),

1n3n=12n(n1)+12n(n+1)=12n(n1)12n(n+1)=12(1n(n1)1n(n+1))

Therefore, the partial fraction is 1n3n=12(1n(n1)1n(n+1)). (3)

Substitute equation (3) in equation (1), then the series is

n=21n3n=n=212(1n(n1)1n(n+1))=12n=2(1n(n1)1n(n+1))=12[n=2(1n(n1)1n(n+1))]

Therefore, the series is n=21n3n=12[n=2(1n(n1)1n(n+1))]

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