   Chapter 11.2, Problem 43E

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.43. ∑ n = 2 ∞ 2 n 2 − 1

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=22n21.

Here, an=2n21.

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, n=22n21.

n=22n21=n=22(n1)(n+1) (1)

Obtain the partial fraction of 2(n1)(n+1).

2(n1)(n+1)=a0n1+a1n+1 (2)

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

2(n1)(n+1)(n1)(n+1)=a0(n1)(n+1)n1+a1(n1)(n+1)n+12=a0(n+1)+a1(n1)2=a0n+a0+a1na12=(a0+a1)n+(a0a1)

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

a0+a1=0a0a1=2

Solve the above equations and obtain a0=1 and a1=1.

Substitute 1 for a0 and -1 for a1 in equation (3),

2(n1)(n+1)=1n1+1n+1=1n11n+1

Therefore, the partial fraction is 2(n1)(n+1)=1n11n+1. (3)

Substitute equation (3) in equation (1), the series is n=21n(n+1)=n=2(1n11n+1).

Let sn be the nth partial sum of the series n=2(1n11n+1)

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