In this video I go further into the wonderful world of infinite sequences and series and this time look at using comparisons with known series to test if a given series is convergent or divergent. The standard Comparison Test simply shows that if the terms of a series are less than the terms of a known convergent series, such as a p-series or geometric series, than the given series is also convergent; and vice versa if the terms are greater than the terms of a divergent series so too is the given series divergent. I go over many examples and exercises illustrating this test as well as go over a similar Limit Comparison Test. This particular test involves taking the limit as the number of terms approach infinity of the ratio of a given series to a known series; if the ratio approaches a number greater than 0 than both series are either both divergent or both convergent. In later exercises I show that if the ratio approaches 0 and the known series is convergent then so too is the given series; but if the ratio approaches infinity and the known series is divergent than so too is the given series. Many more examples and details are covered in this video so hope you enjoy!
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Note that I mainly follow along the following calculus book:
In comparison tests the idea is to compare a given series with a series that is known to be convergent or divergent.
For instance, the series:
reminds us of the series ∑∞n=1 1/2n, which is a geometric series with a = 1/2 and r = 1/2 and is therefore convergent.
Recall the geometric series from my earlier video on Infinite Series.
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Retrieved: 27 July 2019
Archive: http://archive.fo/FGAyJ
Because the series is so similar to a convergent series, we have the feeling that it too must be convergent and indeed, it is.
The inequality:
shows that our given series has smaller terms than those of the geometric series and therefore all its partials sums are also smaller than 1 (the sum of the geometric series).
This means that its partial sums form a bounded increasing sequence, which is convergent.
It also follows that the sum of the series is less than the sum of the geometric series:
Similar reasoning can be used to prove the following test, which applies only to series whose terms are positive.
The first part says that if we have a series whose terms are smaller than those of a known convergent series, then our series is also convergent.
The second part says that if we start with a series whose terms are larger than those of a known divergent series, then it too is divergent.
Suppose that ∑ an and ∑ bn are series with positive terms.
(i) If ∑ bn is convergent and an ≤ bn for all n, then ∑ an is also convergent.
(ii) If ∑ bn is divergent and an ≥ bn for all n, then ∑ an is also divergent.
Proof
Firstly, it is important to keep in mind the distinction between a sequence and a series.
A sequence is a list of numbers, whereas a series is a sum.
With every series ∑ an there are associated two sequences: the sequence {an} of terms and the sequence {sn} of partial sums.
Since both series have positive terms, the sequence {sn} and {tn} are increasing.
This means that {sn} is increasing and bounded above and therefore converges by the Monotonic Sequence Theorem.
Thus ∑ an converges.
Recall the Monotonic Sequence Theorem from my earlier video on Infinite Sequences.
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Retrieved: 25 June 2019
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Essentially, if a sequence is always increasing (or decreasing), hence monotonic, and is bounded above (or below) then it will approach a number or limit L ≤ M (or L ≥ m if the sequence is decreasing).
(ii) If ∑ bn is divergent, then tn → ∞ (since {tn} is increasing).
Therefore ∑ an diverges.
In using the Comparison Test we must, of course, have some known series ∑ bn for the purpose of comparison.
Most of the time we use one of these series:
These are the standard series for use with the Comparison Test.
Both the p-series and the geometric series are detailed in my earlier video on Infinite Series.
Determine whether the following series converges or diverges.
Solution:
For large n, the dominant term in the denominator is 2n2 so we compute the given series with the series ∑ 5/(2n2).
Observe that:
because the left side has a bigger denominator.
In the notation of the Comparison Test, an is the left side and bn is the right side.
We know that:
is convergent because it's a constant times a p-series with p = 2 > 1.
Therefore:
is convergent by part (i) of the Comparison Test.
Although the condition an ≤ bn or an ≥ bn in the Comparison Test is given for all n, we need to verify only that it holds for n ≥ N, where N is some fixed integer, because the convergence of a series is not affected by a finite number of terms.
This is illustrated in the next example.
Test the following series for convergence or divergence.
Solution:
Note that this series was tested (using the Integral Test) in my earlier video on the Integral Test.
@mes/infinite-sequences-and-series-the-integral-test-and-estimate-of-sums
Retrieved: 28 July 2019
Archive: http://archive.fo/3YMxr
But it is also possible to test it by comparing it with the harmonic series.
Observe that ln n > 1 for n ≥ 3:
We know that the harmonic series ∑ 1/n is divergent (p-series with p = 1).
Thus the given series is divergent by the Comparison Test.
The terms of the series being tested must be smaller than those of a convergent series or larger than those of a divergent series.
If the terms are larger than the terms of a convergent series or smaller than those of a divergent series, then the Comparison Test doesn't apply.
Consider, for instance, the series:
The inequality:
is useless as far as the Comparison Test is concerned because ∑ bn = ∑ (1/2)n is convergent and an > bn.
Nonetheless, we have the feeling that ∑ 1/(2n - 1) ought to be convergent because it is very similar to the convergent geometric series ∑ (1/2)n.
In such cases the following test can be used.
Suppose that ∑ an and ∑ bn are series with positive terms.
If:
where c is a finite number and c > 0, then either both series converge or both diverge.
Proof:
Let m and M be positive numbers such that m < c < M.
Because an/bn is close to c for large n, there is an integer N such that:
If ∑ bn converges, so does ∑ Mbn.
Thus ∑ an converges by part (i) of the Comparison Test.
If ∑ bn diverges, so does ∑ mbn and part (ii) of the Comparison Test shows that ∑ an diverges.
Note that Exercises 1 and 2 at the end of this video deal with the cases c = 0 and c = ∞.
Test the following series for convergence or divergence.
Solution:
We use the Limit Comparison Test with:
Since this limit exists and ∑ 1/2n is a convergent geometric series, the given series converges by the Limit Comparison Test.
Determine whether the following series converges or diverges.
Solution:
The dominant part of the numerator is 2n2 and the dominant part of the denominator is n5/2.
This suggests taking:
Since ∑ bn = 2 ∑ 1/n1/2 is divergent (p-series with p = 1/2 < 1), the given series diverges by the Limit Comparison Test.
Notice that in testing many series we find a suitable comparison series ∑ bn by keeping only the highest powers in the numerator and denominator.
If we used the Comparison Test to show that a series ∑ an converges by comparison with a series ∑ bn then we may be able to estimate the sum ∑ an by comparing remainders.
As in my earlier video on the Integral Test and Estimate of Sums, we consider the remainder:
For the comparison series ∑ bn we consider the corresponding remainder:
Since an ≤ bn for all n, we have Rn ≤ Tn.
If ∑ bn is a p-series, we can estimate its remainder Tn as in my earlier video but replace Rn with Tn in the Remainder Estimate for the Integral Test.
@mes/infinite-sequences-and-series-the-integral-test-and-estimate-of-sums
If ∑ bn is a geometric series, then Tn is the sum of a geometric series and we can sum it exactly (see Exercises 3 and 4 at the end of this video).
In either case we know that Rn is smaller than Tn.
Use the sum of the first 100 terms to approximate the sum of the series:
Estimate the error involved in this approximation.
Solution:
the given series is convergent by the Comparison Test.
The remainder Tn for the comparison series ∑ 1/n3 was estimated in my earlier video using the Remainder Estimate for the Integral Test.
There we found that:
Therefore the remainder Rn for the given series satisfies:
With n = 100 we have:
Calculation Check #1: 1/(2*100^2) = 0.0001 which is rounded up.
Calculation Check #2: 1/100^3 = 0.0 which is rounded down.
Calculation Check #3: 1/100^2 = 0.0001 which is right.
Manual Calculation Check: Thus 0.0001 / 2 = 0.00005 which is right.
Using a spreadsheet we can determine the partial sum s100.
https://1drv.ms/x/s!As32ynv0LoaIh_ESqUCeCXooqaAdqg
Retrieved: 29 July 2019
Archive: Not Available
with error less than 0.00005.
The following 4 exercises are referenced in this section of my calculus book so I thought it would be good to go over them too.
(a) Suppose that ∑ an and ∑ bn are series with positive terms and ∑ bn is convergent.
Prove that if:
then ∑ an is also convergent.
(b) Use part (a) to show that the following series converges:
Solution:
(a) Since:
there is a number N > 0 such that:
And since an and bn are both positive, then an < bn.
Thus, since ∑ bn converges, so does ∑ an by the Comparison Test.
Thus ∑ 1/en is a geometric series with ratio r = 1/e and hence convergent since |r| < 1.
Thus, ∑ an converges by part (a).
(a) Suppose that ∑ an and ∑ bn are series with positive terms and ∑ bn is divergent.
Prove that if:
then ∑ an is also divergent.
(b) Use part (a) to show that the following series diverge.
Solution:
(a) Since:
then there is an integer N such that:
Essentially we are just taking the definition M = 1 from the definition of infinite sequences approaching infinity in my earlier video on Infinite Sequences.
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Retrieved: 29 July 2019
Archive: http://archive.fo/s7iYZ
And since ∑ bn is divergent then ∑ an is also divergent by the Comparison Test.
Use the sum of the first 10 terms to approximate the sum of the following series and estimate the error.
Solution:
Calculation Check:
Thus, the error is:
Calculation Check:
https://duckduckgo.com/?q=1%2F2%5E10&t=h_&ia=calculator
Retrieved: 29 July 2019
Archive: http://archive.fo/kT89d
Use the sum of the first 10 terms to approximate the sum of the following series and estimate the error.
Solution:
Calculation Check:
Since I don't have the solutions manual for this particular exercise, I have double-checked the calculation using the same spreadsheet from earlier.
Thus, the error is:
Calculation Check:
https://duckduckgo.com/?q=1%2F2*1%2F3%5E10%3D&t=h_&ia=calculator
Retrieved: 29 July 2019
Archive: http://archive.fo/6jXmT
