   Chapter 11.1, Problem 36E

Chapter
Section
Textbook Problem

# Determine whether the sequence converges or diverges. If it converges, find the limit.36. a n ( − 1 ) n + 1 n n + n

To determine

Whether the sequence converges or diverges and obtain the limit if the sequence converges.

Explanation

Given:

The sequence is an=(1)n+1nn+n .

Definition used:

If an is a sequence and limnan exists, then the sequence an is said to be converges otherwise it diverges.

Results used:

(1) If f(x) and g(x) are two functions, then, limxa[f(x)g(x)]=limxaf(x)limxag(x) and limxag(x)0 .

(2) If f(x) is a function and c is any constant, then, limxa[cf(x)]=climxaf(x) .

(3) If f(x) is a functions then, limxa[f(x)]b=[limxaf(x)]b .

Calculation:

Obtain the limit of the sequence to investigate whether the sequence converges or diverges.

Compute limnan=limn(1)n+1nn+n .

Use the subsequence n=2m in limn(1)n+1nn+n .

limn(1)n+1nn+n=limm(1)2m+1(2m)2m+2m                    =limm(1)1(1)2m(2m)2m+2m                    =limm((1)2)m(2m)2m+2m                    =limm2m2m+2m

Divide numerator and the denominator by the highest power.

limm(1)2m+1(2m)2m+2m=limm2mm2m+2mm=limm22mm+2mm=limm22+2m=limm(2)limm(2)+limm(2m)   [By Result (1) and Result (2)]

Apply the infinity property limn(kna)=0 and simplify the expressions.

limm(1)2m+1(2m)2m+2m=22+limm(2m)12=22+1=22+0=1

Therefore, limn(1)n+1nn+n=1 when n=2m .

Use the subsequence n=2m+1 in limn(1)n+1nn+n

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