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Chapter 3: Infinite Series

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Chapter 3: Infinite Series Chapter 3: Infinite series Introduction Definition: Let (an : n ∈ N) be a sequence. For each n ∈ N, we define the nth partial sum of (an : n ∈ N) to be: sn := a1 + a2 + . an. By the series generated by (an : n ∈ N) we mean the sequence (sn : n ∈ N) of partial sums of (an : n ∈ N) (ie: a series is a sequence). We say that the series (sn : n ∈ N) generated by a sequence (an : n ∈ N) is convergent if the sequence (sn : n ∈ N) of partial sums converges. If the series (sn : n ∈ N) is convergent then we call its limit the sum of the series and we write: P∞ lim sn = an. In detail: n→∞ n=1 s1 = a1 s2 = a1 + a + 2 = s1 + a2 s3 = a + 1 + a + 2 + a3 = s2 + a3 ... = ... sn = a + 1 + ... + an = sn−1 + an. Definition: A series that is not convergent is called divergent, ie: each series is ei- P∞ ther convergent or divergent. It has become traditional to use the notation n=1 an to represent both the series (sn : n ∈ N) generated by the sequence (an : n ∈ N) and the sum limn→∞ sn. However, this ambiguity in the notation should not lead to any confu- sion, provided that it is always made clear that the convergence of the series must be established. Important: Although traditionally series have been (and still are) represented by expres- P∞ sions such as n=1 an they are, technically, sequences of partial sums. So if somebody comes up to you in a supermarket and asks you what a series is - you answer “it is a P∞ sequence of partial sums” not “it is an expression of the form: n=1 an.” n−1 Example: (1) For 0 < r < ∞, we may define the sequence (an : n ∈ N) by, an := r for each n ∈ N. 2 n−1 (2) Then the partial sums of this sequence are: sn := 1 + r + r + . r . Therefore, 2 3 n r · sn = r + r + r + . r and so n (1 − r)sn = sn − rsn = 1 − r . 15 n (3) Hence if r 6= 1 then, sn = (1 − r )/(1 − r). (4) Moreover, for 0 < r < 1, (sn : n ∈ N) is monotonely increasing and bounded above by n 1/(1 − r) since, (1 − r ) < 1 for all n ∈ N. Hence the limit of the sequence (sn : n ∈ N) exists. (4) Furthermore, since sn+1 = r · sn + 1 for all n ∈ N we have that L = rL + 1 ⇒ L = 1/(1 − r) , where L := lim sn. 2 n→∞ Definition: The series in the above example is called the geometric series with ratio r. p Example: For each 0 < p < ∞, let (an : n ∈ N) be the sequence defined by, an := 1/n for all n ∈ N. Then the series generated by (an : n ∈ N) is called a p-series. It will be shown later that a p-series is convergent if, and only if, 1 < p < ∞. The divergent case p = 1 has special significance and is called the harmonic series. n Exercise: Let (an : n ∈ N) be the sequence defined by, an := (−1) for all n ∈ N. Show that the series generated by (an : n ∈ N) is divergent. Hint: show that sn is 0 if n is even and 1 if n is odd. ∞ X 1 Example: Calculate the sum of the series, . n(n + 1)(n + 2) n=1 Answer: By using partial fractions we see that, n X 1 s = n j(j + 1)(j + 2) j=1 n 1 X 1 1 = − 2 j(j + 1) (j + 1)(j + 2) j=1 1 1 1 1 1 1 1 = − + − + ··· + − 2 2 6 6 12 n(n + 1) (n + 1)(n + 2) 1 1 1 = − 2 2 (n + 1)(n + 2) ∞ X 1 and so lim sn = = 1/4. 2 n→∞ n(n + 1)(n + 2) n=1 P∞ Any series of the form: n=1(an − an+1), for some sequence (an : n ∈ N) is called a telescoping series and so we see that the previous example is in fact a telescoping series. P∞ Exercise: Let (an : n ∈ N) be a sequence. Show that the series n=1(an−an+1) converges if, and only if, the sequence (an : n ∈ N) converges. Hint: Show that the nth-partial sum of this series equals: a1 − an+1. ∞ X 1 Exercise: Express the sum of the series, in terms of a. (n + a)(n + a + 1) n=1 16 ∞ P Theorem: If an is a convergent series then lim an = 0. n=1 n→∞ Proof: Let sn denote the nth partial sum of the series. Then sn −sn−1 = an for all n ∈ N and so by taking the limit of both sides of this equation as, n → ∞, we see that, ∞ ∞ ! X X 0 = an − an = lim sn − lim sn−1 = lim (sn − sn−1) = lim an 2 n→∞ n→∞ n→∞ n→∞ n=1 n=1 Note: (sn−1) is a “subsequence” of (sn). If sn → L then it can be shown that every subsequence converges to L also. This concept is very useful, but will not be developed here. ∞ P Warning: The fact that lim an = 0 does not guarantee that an is convergent. n→∞ n=1 ∞ X Example: The harmonic series, 1/n is divergent. n=1 Proof: We prove this result by induction. Let, 1 1 1 1 s := 1 + + + ··· + + and P (n) := “s n 1 + n/2” n 2 3 (n − 1) n 2 > Step 1. P (1) = “s21 > 1 + 1/2”; which is true. Therefore P (n) is true for the case n = 1. Step 2. Suppose that P (k) is true, ie: suppose that s2k > 1 + k/2, then, 1 1 1 s k+1 = 1 + + + ··· + 2 2 3 2k+1 1 1 1 1 1 = 1 + + + ··· + + ··· + 2 3 2k 2k + 1 2k+1 1 1 1 (1 + k/2) + + ··· + > 2k+1 2k+1 2k+1 = (1 + k/2) + 1/2 = 1 + (k + 1)/2. Therefore P (k + 1) is true. Step 3. Hence by induction, P (n) is true for all n ∈ N, ie: s2n > 1 + n/2 for all n ∈ N. Now (sn : n ∈ N) is a monotonely inceasing sequence (why ?) that is not bounded above. ∞ P Hence, 1/n = lim sn diverges to ∞. Note however that, lim 1/n = 0. 2 n=1 n→∞ n→∞ The following theorem enables us to show that certain series converge without having to resort to using ε’s and N’s. ∞ ∞ X X Theorem: Let an and bn be convergent series then; n=1 n=1 ∞ ∞ ∞ X X X (i) (an + bn) = an + bn; n=1 n=1 n=1 17 ∞ ∞ ! X X (ii) can = c · an . n=1 n=1 Proof: Exercise. ∞ ∞ ∞ X X X Note: If an and bn converge then the product series anbn may or may not n=1 n=1 n=1 n √ converge. E.g. an = bn = (−1) / n. Series of non-negative terms Definition: If (an : n ∈ N) is a sequence with non-negative terms (ie: 0 6 an for all P∞ n ∈ N) then the series n=1 an is called a series of non-negative terms. Importantly Pn for such series the partial sums sn := j=1 aj are monotonely increasing, since sn+1 −sn = Pn+1 Pn j=1 aj − j=1 aj = an+1 > 0. Hence to show that a series of non-negative terms is convergent it is sufficient (and nec- essary) to show that the series is bounded above. We now look at some tests for convergence of series with non-negative terms. The first test we consider is the integral test. Theorem: (Integral Test) Let (an : n ∈ N) be a sequence of non-negative terms and let f : [1, ∞) → R be a monotonely decreasing continuous function such that f(n) = an for P∞ R ∞ all n ∈ N. Then the series n=1 an converges if, and only if, the integral 1 f(x)dx = R N limN→∞ 1 f(x)dx converges. Proof: In the next figure, the area of the rectangles is: a2+a3+···+an−1+an and is clearly Z n no more than the area under the curve and so, a2 + a3 + ··· + an−1 + an 6 f(x)dx. 1 y . N .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ... ... ... ... ... ... ... ... ... ... ... ... .... .... ..... ..... ..... ...... ...... ...... ...... ....... ....... ....... ........ ........ ......... .......... ........... ............ .............. ................ ................... ....................... .............................. ....................................... ........................................................ ........................................................................................... y = f(x) a2 a3 a4 an−2 an−1 an I x 1 2 3 4 n−3 n−2 n−1 n 18 Z n Thus, sn 6 a1 + f(x)dx. 1 In the next figure we see that the area of the rectangles is: a1 + a2 + a3 + ··· + an−1 + an Z n+1 and so f(x)dx 6 a1 + a2 + ··· + an−1 + an 1 y . N .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ... ... ... ... ... ... ... ... ... ... ... ... .... .... ..... ..... ..... ...... ...... ...... ...... ....... ....... ....... ........ ........ ......... .......... ........... ............ .............. ................ ................... ....................... .............................. ....................................... ........................................................ ........................................................................................... y = f(x) a1 a2 a3 an−2 an−1 an I x 1 2 3 4 n−2 n−1 n n+1 Z n+1 Z n Hence, f(x)dx 6 sn 6 a1 + f(x)dx. 1 1 By taking the limit as, n → ∞, we see that that the series and improper integral are either both convergent or both divergent. Note also that the last inequality gives us both an upper and lower bound for the partial sum sn. 2 ∞ X Example: Show that the p-series, 1/np is convergent if, and only if, 1 < p < ∞. n=1 Answer: Suppose that 0 < p < ∞. Then the function f : [1, ∞) → R defined by, f(x) := 1/xp satisfies the hypotheses of the integral test. Moreover, for p 6= 1 we have that: Z ∞ Z n 1−p n 1−p ( 1 1 −p x n − 1 p > 1, dx = lim x dx = lim = lim = p − 1 xp n→∞ n→∞ 1 − p n→∞ 1 − p 1 1 1 ∞ 0 < p < 1.
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