12 INFINITE SEQUENCES AND SERIES

12.4 The Comparison Tests In this section, we will learn: How to find the value of a series by comparing it with a known series. INFINITE SEQUENCES AND SERIES

COMPARISON TESTS In the comparison tests, the idea is to compare a given series with one that is known to be convergent or divergent.

Consider the series  This reminds us of the series.  The latter is a geometric series with a = ½ and r = ½ and is therefore convergent. COMPARISON TESTS Series 1

As the series is similar to a convergent series, we have the feeling that it too must be convergent.  Indeed, it is. COMPARISON TESTS

The inequality shows that our given series has smaller terms than those of the geometric series.  Hence, all its partial sums are also smaller than 1 (the sum of the geometric series). COMPARISON TESTS

Thus,  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: COMPARISON TESTS

Similar reasoning can be used to prove the following test—which applies only to series whose terms are positive. COMPARISON TESTS

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. COMPARISON TESTS

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. COMPARISON TESTS

Suppose that Σ a n and Σ b n are series with positive terms. i.If Σ b n is convergent and a n ≤ b n for all n, then Σ a n is also convergent. ii.If Σ b n is divergent and a n ≥ b n for all n, then Σ a n is also divergent. THE COMPARISON TEST

Let  Since both series have positive terms, the sequences {s n } and {t n } are increasing (s n+1 = s n + a n+1 ≥ s n ).  Also, t n → t; so t n ≤ t for all n. THE COMPARISON TEST—PROOF Part i

Since a i ≤ b i, we have s n ≤ t n. Hence, s n ≤ t for all n.  This means that {s n } is increasing and bounded above.  So, it converges by the Monotonic Sequence Theorem.  Thus, Σ a n converges. THE COMPARISON TEST—PROOF Part i

If Σ b n is divergent, then t n → ∞ (since {t n } is increasing).  However, a i ≥ b i ; so s n ≥ t n.  Thus, s n → ∞; so Σ a n diverges. THE COMPARISON TEST—PROOF Part ii

It is important to keep in mind the distinction between a sequence and a series.  A sequence is a list of numbers.  A series is a sum. SEQUENCE VS. SERIES

With every series Σ a n, there are associated two sequences: 1.The sequence {a n } of terms 2.The sequence {s n } of partial sums SEQUENCE VS. SERIES

In using the Comparison Test, we must, of course, have some known series Σ b n for the purpose of comparison. COMPARISON TEST

Most of the time, we use one of these:  A p-series [Σ 1/n p converges if p > 1 and diverges if p ≤ 1]  A geometric series [Σ ar n–1 converges if |r| < 1 and diverges if |r| ≥ 1] COMPARISON TEST

Determine whether the given series converges or diverges: COMPARISON TEST Example 1

For large n, the dominant term in the denominator is 2n 2.  So, we compare the given series with the series Σ 5/(2n 2 ). COMPARISON TEST Example 1

Observe that since the left side has a bigger denominator.  In the notation of the Comparison Test, a n is the left side and b n is the right side. COMPARISON TEST Example 1

We know that is convergent because it’s a constant times a p-series with p = 2 > 1. COMPARISON TEST Example 1

Therefore, is convergent by part i of the Comparison Test. COMPARISON TEST Example 1

Although the condition a n ≤ b n or a n ≥ b n in the Comparison Test is given for all n, we need verify only that it holds for n ≥ N, where N is some fixed integer.  This is because the convergence of a series is not affected by a finite number of terms.  This is illustrated in the next example. NOTE 1

Test the given series for convergence or divergence: COMPARISON TEST Example 2

This series was tested (using the Integral Test) in Example 4 in Section 11.3  However, it is also possible to test it by comparing it with the harmonic series. COMPARISON TEST Example 2

Observe that l n n > 1 for n ≥ 3. So,  We know that Σ 1/n is divergent (p-series with p = 1).  Thus, the series is divergent by the Comparison Test. COMPARISON TEST Example 2

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, the Comparison Test doesn’t apply. NOTE 2

For instance, consider  The inequality is useless as far as the Comparison Test is concerned.  This is because Σ b n = Σ (½) n is convergent and a n > b n. NOTE 2

Nonetheless, we have the feeling that Σ1/(2 n -1) ought to be convergent because it is very similar to the convergent geometric series Σ (½) n.  In such cases, the following test can be used. NOTE 2

Suppose that Σ a n and Σ b n are series with positive terms. If where c is a finite number and c > 0, either both series converge or both diverge. LIMIT COMPARISON TEST

Let m and M be positive numbers such that m < c < M.  Since a n /b n is close to c for large n, there is an integer N such that LIMIT COMPARISON TEST—PROOF

If Σ b n converges, so does Σ Mb n.  Thus, Σ a n converges by part i of the Comparison Test. LIMIT COMPARISON TEST—PROOF

If Σ b n diverges, so does Σ mb n.  Thus, Σ a n diverges by part ii of the Comparison Test. LIMIT COMPARISON TEST—PROOF

Test the given series for convergence or divergence: COMPARISON TESTS Example 3

We use the Limit Comparison Test with: COMPARISON TESTS Example 3

We obtain: COMPARISON TESTS Example 3

This limit exists and Σ 1/2 n is a convergent geometric series.  Thus, the given series converges by the Limit Comparison Test. COMPARISON TESTS Example 3

Determine whether the given series converges or diverges: COMPARISON TESTS Example 4

The dominant part of the numerator is 2n 2. The dominant part of the denominator is n 5/2.  This suggests taking: COMPARISON TESTS Example 4

COMPARISON TESTS Example 4 We obtain:

Σ b n = 2 Σ 1/n 1/2 is divergent (p-series with p = ½ < 1).  Thus, the given series diverges by the Limit Comparison Test. COMPARISON TESTS Example 4

Notice that, in testing many series, we find a suitable comparison series Σ b n by keeping only the highest powers in the numerator and denominator. COMPARISON TESTS

ESTIMATING SUMS We have used the Comparison Test to show that a series Σ a n converges by comparison with a series Σ b n.  It follows that we may be able to estimate the sum Σ a n by comparing remainders.

As in Section 11.3, we consider the remainder R n = s – s n = a n+1 + a n+2 + …  For the comparison series Σ b n, we consider the corresponding remainder T n = t - t n = b n+1 + b n+2 + … ESTIMATING SUMS

As a n ≤ b n for all n, we have R n ≤ T n.  If Σ b n is a p-series, we can estimate its remainder T n as in Section 11.3  If Σ b n is a geometric series, then T n is the sum of a geometric series and we can sum it exactly (Exercises 35 and 36).  In either case, we know that R n is smaller than T n ESTIMATING SUMS

Use the sum of the first 100 terms to approximate the sum of the series Σ 1/(n 3 +1). Estimate the error involved in this approximation. ESTIMATING SUMS Example 5

Since the given series is convergent by the Comparison Test. ESTIMATING SUMS Example 5

The remainder T n for the comparison series Σ 1/n 3 was estimated in Example 5 in Section 11.3 (using the Remainder Estimate for the Integral Test).  We found that: ESTIMATING SUMS Example 5

Therefore, the remainder for the given series satisfies: R n ≤ T n ≤ 1/2n 2 ESTIMATING SUMS Example 5

With n = 100, we have:  With a programmable calculator or a computer, we find that with error less than ESTIMATING SUMS Example 5