Absolute and conditional convergence

 Objective: Determine whether a series is absolutely convergent or conditionally 

Theorem

Example. Determine whether the following series converges absolutely, conditionally or diverges.

Solution

Method

1) We start by calculating the series of the absolute value of the general term of the given series

2) We find a new series of which we study the convergence or divergence

3) If this series is convergent, the series of the absolute value of the general term is convergent. Therefore the given series converges absolutely

4) If this series diverges as in the example above, the series of the absolute value of the general term diverges. The series doesn't converge absolutely.  In this case we study the convergence of the given series. We conclude that the series converges conditionally.

Practice

Determine if the following series converges absolutely, conditionally or diverges.

The alternating series test

 Objectives:

1) Define the alternating series

2) Use the alternating series test for convergence

Introduction

Any series whose terms alternate in sign is called an alternating series.

Examples:

Definition

An alternate series is a series whose terms alternate between positive and negative values. An alternating series can be written in either form where bₙ is positive for all positive integers n

Theorem

Examples

Solution

a. Let's apply the condition above. First, let's see if the first condition is verified.

We have : bₙ = 1/n². Therefore bₙ₊₁ = 1/(n+1)² = 1/n² + 2n +1 >0 for all n≥1 and bₙ.₊₁≤ bₙ The first condition is verified. Let's see if the second condition is verified:

b.  In this problem we have bₙ₊₁>bₙ. We can prove that by calculating bₙ₊₁ -bₙ. The first condition not being verified, therefore we cannot apply the alternating series test. Instead let's find the limit of the sequence of the partial sum or the limit of the general term of the partial sum. We have:

 

The series is then divergent.

Practice

The comparison test for infinite series

 Objectives:

1) Know what the comparison test is about

2) Show a procedure to use the comparison test

3) Deduct a general method for the comparison test

Comparison test

The comparison test consists in comparing a given series to a known series  in order to determine its convergence or divergence. Since the convergence and the divergence of geometric series are known, it's easy to compare a series to those series.

Procedure

The general method for determining the convergence or divergence of a series consists in determining if the sequence of partial sums is convergent or divergent. In the comparison test, we try to determine the convergence or divergence of the sequence of partial sums from the convergence or divergence of the sequence of the partial sums of the known series.

We follow these steps:

1) Comparing the general terms of the 2 series

2) Deduct the convergence or divergence of the given series from the comparison of those two terms

Example 1.

Find out if the following series is convergent.

Solution

Let's compare this series to the series 

This series is convergent by the integral test and also as a p-series

Since all the terms in both series are positive the series are monotone increasing

In comparing the general terms of both series, we have:

0<1/n² + 1<1/n²

Let's compare each term of the partial sums of the given series to the corresponding term of the sequence of the second series.

Each term of the sequence of the partial sum of the given series is less than the corresponding term of the other series. Therefore in taking into account this observation and the inequality above, we can write:

The series on the right is convergent therefore the sequence of its partial sum is also convergent. A convergent  sequence is also bounded above. The sequence Sₖ is then bounded above, Since the sequence of the partial sum is monotone increasing. the sequence Sₖ is monotone increasing above. The Monotone convergence theorem states that if a sequence is monotone increasing  above, it is therefore convergent. The series Sₖ is then convergent. The given series is then convergent.

Conclusion 1

In the example above, we note that the general term of the given series is less than the general term of the series 1/n₂. If we name the given series aₙ and the harmonic series by bₙ, we can write 

aₙ <bₙ,. This condition is sufficient to say that the series aₙ is convergent provided that  bₙ is convergent and we know it's already convergent

Example 2

Find out if the following series is convergent or divergent

Solution

This series is similar to the series

Let's compare the example series to that series

The sequence of partial sum for each series is monotone increasing. In comparing the general terms of both series we have:

Let's compare each term of the partial sums of the given series to the corresponding term of the sequence of the second series.

Each term of the partial sum of the given series is more than the corresponding partial sum of the other series. Therefore in taking into account this observation and the inequality above, we can write:

Since the series 

is divergent, the sequence of partial sum 

is divergent. Therefore the sequence of partial sum is unbounded. Based on the inequality above, the sequence Sₖ is unbounded and therefore divergent, The example series is then divergent. 

Conclusion 2

In the example above, we note that the general term of the given series is greater than the general term of the harmonic series. If we name the given series aₙ and the harmonic series by bₙ, we can write 

aₙ > bₙ,. This condition is sufficient to say that the series aₙ is divergent provided that  bₙ is divergent and we know it's already divergent.

Final Conclusion

 Conclusions 1 and 2 can be combined into one:

One comparing two series  aₙ and bₙ  , we compare the general terms.

 

Let's do some examples based on the theorem above

Use the comparison test to determine whether the following series are convergent or divergent:

Solution

a. Let's compare the given series to 

This series is a p-series. A p-series is convergent if p>1 and divergent if p≤ 1. Here p = 3>1. The series is convergent

Let's compare the general terms:

1/n³ + 3n + 1 < 1/n³

According to the theorem of the comparison test, if  0≤ aₙ ≤bₙ, and the series bₙ convergent, then the series  aₙ is convergent, we conclude that the series  is convergent.

b. Let's compare the given series to the series 1/2ⁿ or the series (1/2)ⁿ. This series is a geometric series. We can write it in explicit form as:

Here we have r = 1/2 < 1. Then the series is convergent

Let's compare the general terms. We have:

1/2ⁿ + 1 < 1/2ⁿ

Since the series 1/2ⁿ is convergent the series 1/2ⁿ + 1 is convergent.

Practice 

Use the convergence test to determine if the following series is convergent or divergent

The comparison test limit

 The objectives of this post are: 

a) Define the limit comparison test

2) Use it to find the convergence or divergence of a series

The comparison test limit works effectively if it's possible to find another series that  satisfies the hypothesis of the test. Let's consider the series:

Let's compare it to the series:

 

This series is convergent a a p-series: p = 2>.1. Let's compare the general terms of both series.. We have:

1/n²-1>1/n²

We can't say anything about the given series meaning we can't say if it's convergent or divergent. In order for the comparison test to work here we should have 1/n²-1<1/n².

Let's find a theory to solve this problem. 

These results can be summarized in the following theorem:

Examples

For each of the following series, use the limit comparison test to determine whether the series converges or diverges. If the test doesn't apply, say so.

How to estimate the value of a series ?

 Let's estimate the value of a convergent series :

We can write this series as:

From the equation above, the reminder can be written as:

Let's use the integral test to estimate the remainder. We are going to compare the reminder to the improper integral of a function f. The remainder can be written as a sum of areas of rectangles as follows:

The length of each rectangle is considered to be 1.

We consider two scenarios. The first scenario is considered in the following figure:

In this case the sum of the areas of the rectangles that represents the reminder Rₙ is inferior to the improper integral whose limits are N and +∞. We write:

Let's consider the second scenario in the following figure:

 

The reminder Rₙ is superior to the improper integral and we can write:

We can combine the two comparisons of  Rₙ by writing:

The preceding inequality represents an approximation of the remainder.

Example

Solution

a) S₁₀  = 1/1³ + 1/2³ + 1/3³ + 1/4⁴ + .....                 + 1/10³≃ 1.19753

Estimate the error means estimate the remainder. According to the formula for the estimation of the remainder:

b) Estimating the series within 0.001 means the error or the remainder should be less than 0.001.

Then 1/2N²  < 0.001

2N² > 1000⇒ N² > 500 ⇒ N>10⎷5 ⇒N>22.36

By rounding N to the nearest hundreth, we can choose the least value of N as N = 23.  

Practice