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