INFINITE SEQUENCES AND SERIES 11 INFINITE SEQUENCES AND SERIES

11.2 Series In this section, we will learn about: INFINITE SEQUENCES AND SERIES 11.2 Series In this section, we will learn about: Various types of series.

SERIES Series 1 If we try to add the terms of an infinite sequence we get an expression of the form a1 + a2 + a3 + ··· + an + ∙·∙

This is called an infinite series (or just a series). It is denoted, for short, by the symbol

INFINITE SERIES However, does it make sense to talk about the sum of infinitely many terms?

It would be impossible to find a finite sum for the series INFINITE SERIES It would be impossible to find a finite sum for the series 1 + 2 + 3 + 4 + 5 + ∙∙∙ + n + ··· If we start adding the terms, we get the cumulative sums 1, 3, 6, 10, 15, 21, . . . After the nth term, we get n(n + 1)/2, which becomes very large as n increases.

However, if we start to add the terms of the series INFINITE SERIES However, if we start to add the terms of the series we get:

INFINITE SERIES The table shows that, as we add more and more terms, these partial sums become closer and closer to 1. In fact, by adding sufficiently many terms of the series, we can make the partial sums as close as we like to 1.

INFINITE SERIES So, it seems reasonable to say that the sum of this infinite series is 1 and to write:

INFINITE SERIES We use a similar idea to determine whether or not a general series (Series 1) has a sum.

We consider the partial sums s1 = a1 s2 = a1 + a2 s3 = a1 + a2 + a3 INFINITE SERIES We consider the partial sums s1 = a1 s2 = a1 + a2 s3 = a1 + a2 + a3 s3 = a1 + a2 + a3 + a4 In general,

INFINITE SERIES These partial sums form a new sequence {sn}, which may or may not have a limit.

SUM OF INFINITE SERIES If exists (as a finite number), then, as in the preceding example, we call it the sum of the infinite series Σ an.

let sn denote its nth partial sum: SUM OF INFINITE SERIES Definition 2 Given a series let sn denote its nth partial sum:

SUM OF INFINITE SERIES Definition 2 If the sequence {sn} is convergent and exists as a real number, then the series Σ an is called convergent and we write: The number s is called the sum of the series. Otherwise, the series is called divergent.

SUM OF INFINITE SERIES Thus, the sum of a series is the limit of the sequence of partial sums. So, when we write , we mean that, by adding sufficiently many terms of the series, we can get as close as we like to the number s.

SUM OF INFINITE SERIES Notice that:

SUM OF INFINITE SERIES VS. IMPROPER INTEGRALS Compare with the improper integral To find this integral, we integrate from 1 to t and then let t → ∞. For a series, we sum from 1 to n and then let n → ∞.

An important example of an infinite series is the geometric series

GEOMETRIC SERIES Example 1 Each term is obtained from the preceding one by multiplying it by the common ratio r. We have already considered the special case where a = ½ and r = ½ earlier in the section.

If r = 1, then sn = a + a + ∙∙∙ + a = na → ±∞ GEOMETRIC SERIES Example 1 If r = 1, then sn = a + a + ∙∙∙ + a = na → ±∞ Since doesn’t exist, the geometric series diverges in this case.

rsn = ar + ar2 + ∙∙∙ +ar n–1 + ar n GEOMETRIC SERIES Example 1 If r ≠ 1, we have sn = a + ar + ar2 + ∙∙∙ + ar n–1 and rsn = ar + ar2 + ∙∙∙ +ar n–1 + ar n

Subtracting these equations, we get: GEOMETRIC SERIES E. g. 1—Equation 3 Subtracting these equations, we get: sn – rsn = a – ar n

GEOMETRIC SERIES Example 1 If –1 < r < 1, we know from Result 9 in Section 11.1 that r n → 0 as n → ∞. So, Thus, when |r | < 1, the series is convergent and its sum is a/(1 – r).

So, by Equation 3, does not exist. GEOMETRIC SERIES Example 1 If r ≤ –1 or r > 1, the sequence {r n} is divergent by Result 9 in Section 11.1 So, by Equation 3, does not exist. Hence, the series diverges in those cases.

GEOMETRIC SERIES The figure provides a geometric demonstration of the result in Example 1.

If s is the sum of the series, then, by similar triangles, GEOMETRIC SERIES If s is the sum of the series, then, by similar triangles, So,

We summarize the results of Example 1 as follows. GEOMETRIC SERIES We summarize the results of Example 1 as follows.

is convergent if |r | < 1. GEOMETRIC SERIES Result 4 The geometric series is convergent if |r | < 1.

The sum of the series is: GEOMETRIC SERIES Result 4 The sum of the series is: If |r | ≥ 1, the series is divergent.

Find the sum of the geometric series Example 2 Find the sum of the geometric series The first term is a = 5 and the common ratio is r = –2/3

GEOMETRIC SERIES Example 2 Since |r | = 2/3 < 1, the series is convergent by Result 4 and its sum is:

GEOMETRIC SERIES What do we really mean when we say that the sum of the series in Example 2 is 3? Of course, we can’t literally add an infinite number of terms, one by one.

GEOMETRIC SERIES However, according to Definition 2, the total sum is the limit of the sequence of partial sums. So, by taking the sum of sufficiently many terms, we can get as close as we like to the number 3.

The table shows the first ten partial sums sn. GEOMETRIC SERIES The table shows the first ten partial sums sn. The graph shows how the sequence of partial sums approaches 3.

Is the series convergent or divergent? GEOMETRIC SERIES Example 3 Is the series convergent or divergent?

Let’s rewrite the nth term of the series in the form ar n-1: GEOMETRIC SERIES Example 3 Let’s rewrite the nth term of the series in the form ar n-1: We recognize this series as a geometric series with a = 4 and r = 4/3. Since r > 1, the series diverges by Result 4.

Write the number as a ratio of integers. GEOMETRIC SERIES Example 4 Write the number as a ratio of integers. 2.3171717… After the first term, we have a geometric series with a = 17/103 and r = 1/102.

GEOMETRIC SERIES Example 4 Therefore,

Find the sum of the series where |x| < 1. GEOMETRIC SERIES Example 5 Find the sum of the series where |x| < 1. Notice that this series starts with n = 0. So, the first term is x0 = 1. With series, we adopt the convention that x0 = 1 even when x = 0.

Thus, This is a geometric series with a = 1 and r = x. Example 5 Thus, This is a geometric series with a = 1 and r = x.

Since |r | = |x| < 1, it converges, and Result 4 gives: GEOMETRIC SERIES E. g. 5—Equation 5 Since |r | = |x| < 1, it converges, and Result 4 gives:

is convergent, and find its sum. SERIES Example 6 Show that the series is convergent, and find its sum.

This is not a geometric series. Example 6 This is not a geometric series. So, we go back to the definition of a convergent series and compute the partial sums:

SERIES Example 6 We can simplify this expression if we use the partial fraction decomposition. See Section 7.4

SERIES Example 6 Thus, we have:

SERIES Example 6 Thus, Hence, the given series is convergent and

SERIES The figure illustrates Example 6 by showing the graphs of the sequence of terms an =1/[n(n + 1)] and the sequence {sn} of partial sums. Notice that an → 0 and sn → 1.

Show that the harmonic series Example 7 Show that the harmonic series is divergent.

HARMONIC SERIES Example 7 For this particular series it’s convenient to consider the partial sums s2, s4, s8, s16, s32, … and show that they become large.

HARMONIC SERIES Example 7 Similarly,

HARMONIC SERIES Example 7 Similarly,

Similarly, s32 > 1 + 5/2, s64 > 1 + 6/2, and, in general, HARMONIC SERIES Example 7 Similarly, s32 > 1 + 5/2, s64 > 1 + 6/2, and, in general, This shows that s2n → ∞ as n → ∞, and so {sn} is divergent. Therefore, the harmonic series diverges.

HARMONIC SERIES The method used in Example 7 for showing that the harmonic series diverges is due to the French scholar Nicole Oresme (1323–1382).

If the series is convergent, then Theorem 6 If the series is convergent, then

Let sn = a1 + a2 + ∙∙∙ + an Then, an = sn – sn–1 SERIES Theorem 6—Proof Let sn = a1 + a2 + ∙∙∙ + an Then, an = sn – sn–1 Since Σ an is convergent, the sequence {sn} is convergent.

Since n – 1 → ∞ as n → ∞, we also have: SERIES Theorem 6—Proof Let Since n – 1 → ∞ as n → ∞, we also have:

SERIES Theorem 6—Proof Therefore,

With any series Σ an we associate two sequences: Note 1 With any series Σ an we associate two sequences: The sequence {sn} of its partial sums The sequence {an} of its terms

If Σ an is convergent, then SERIES Note 1 If Σ an is convergent, then The limit of the sequence {sn} is s (the sum of the series). The limit of the sequence {an}, as Theorem 6 asserts, is 0.

The converse of Theorem 6 is not true in general. SERIES Note 2 The converse of Theorem 6 is not true in general. If , we cannot conclude that Σ an is convergent.

SERIES Note 2 Observe that, for the harmonic series Σ 1/n, we have an = 1/n → 0 as n → ∞. However, we showed in Example 7 that Σ 1/n is divergent.

THE TEST FOR DIVERGENCE If does not exist or if , then the series is divergent.

The Test for Divergence follows from Theorem 6. If the series is not divergent, then it is convergent. Thus,

Show that the series diverges. TEST FOR DIVERGENCE Example 8 Show that the series diverges. = So, the series diverges by the Test for Divergence.

If we find that , we know that Σ an is divergent. SERIES Note 3 If we find that , we know that Σ an is divergent. If we find that , we know nothing about the convergence or divergence of Σ an.

Remember the warning in Note 2: SERIES Note 3 Remember the warning in Note 2: If , the series Σ an might converge or diverge.

SERIES Theorem 8 If Σ an and Σ bn are convergent series, then so are the series Σ can (where c is a constant), Σ (an + bn), and Σ (an – bn), and

SERIES These properties of convergent series follow from the corresponding Limit Laws for Sequences in Section 11.1 For instance, we prove part ii of Theorem 8 as follows.

THEOREM 8 ii—PROOF Let

The nth partial sum for the series Σ (an + bn) is: THEOREM 8 ii—PROOF The nth partial sum for the series Σ (an + bn) is:

Using Equation 10 in Section 5.2, we have: THEOREM 8 ii—PROOF Using Equation 10 in Section 5.2, we have:

Hence, Σ (an + bn) is convergent, and its sum is: THEOREM 8 ii—PROOF Hence, Σ (an + bn) is convergent, and its sum is:

Find the sum of the series Example 9 Find the sum of the series The series Σ 1/2n is a geometric series with a = ½ and r = ½. Hence,

In Example 6, we found that: SERIES Example 9 In Example 6, we found that: So, by Theorem 8, the given series is convergent and

SERIES Note 4 A finite number of terms doesn’t affect the convergence or divergence of a series.

SERIES Note 4 For instance, suppose that we were able to show that the series is convergent. Since it follows that the entire series is convergent.

SERIES Note 4 Similarly, if it is known that the series converges, then the full series is also convergent.