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For each of the sequences in Exercises 23–52 determine whether the sequence is monotonic or eventually monotonic and whether the sequence is bounded above and/or below. If the sequence converges, give the limit.

1+1kk

Short Answer

Expert verified

The given sequence is monotonic, bounded and convergent.

The limit of the sequence is e.

Step by step solution

01

Step 1. Given Information  

We are given the sequence 1+1kkand we need to find if the sequence is monotonic, bounded and the limit if it is convergent.

02

Step 2. Finding monotonic 

The general term is ak=1+1kk.

The ratio

ak+1ak=1+1k+1k+11+1kk=k+2k+1k+1k+1k+1k+1kk+1=k+2k+1.kk+1k.k+2k+1>1(k>0)∴ak+1>ak

The sequence is strictly increasing so it is monotonic.

03

Step 3. Finding bounded  

The sequence 1+1kkis bounded below because 2<ak. As k>1,ak≤3. The decreasing sequence has a lower bound and is 2and an upper bound 3.

Therefore, the given sequence is bounded.

04

Step 4. Finding the limit 

The monotonic decreasing sequence is bounded below and hence convergent.

limk→∞ak=limk→∞1+1kk=e

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Most popular questions from this chapter

True/False:

Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: If ak→0, then ∑k=1∞akconverges.

(b) True or False: If ∑k=1∞akconverges, then ak→0.

(c) True or False: The improper integral ∫1∞f(x)dxconverges if and only if the series ∑k=1∞f(k)converges.

(d) True or False: The harmonic series converges.

(e) True or False: If p>1, the series ∑k=1∞k-pconverges.

(f) True or False: If f(x)→0as x→∞, then ∑k=1∞f(k) converges.

(g) True or False: If ∑k=1∞f(k)converges, then f(x)→0as x→∞.

(h) True or False: If ∑k=1∞ak=Land {Sn}is the sequence of partial sums for the series, then the sequence of remainders {L-Sn}converges to 0.

Find the values of x for which the series ∑K=0∞sinxkconverges.

Examples: Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading.

(a) A divergent series ∑k=1∞akin which ak→0.

(b) A divergent p-series.

(c) A convergent p-series.

Prove Theorem 7.31. That is, show that if a function a is continuous, positive, and decreasing, and if the improper integral ∫1∞a(x)dxconverges, then the nth remainder, Rn, for the series∑k=1∞a(k) is bounded by0≤Rn=∑k=n+1∞a(k)≤∫n∞a(x)dx

Given that a0=-3,a1=5,a2=-4,a3=2and ∑akk=2∞=7, find the value ofrole="math" localid="1648828282417" ∑akk=1∞.

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