Infinite Series Lecture Notes PDF

Lecture Notes on Infinite Series Laurel Benn July 9, 2009 Definition 1 An Infinite Series is an expression of the form ∞  an = a1 + a2 +... + an +... n=1 Where a1 , a2 , a3 ,..., an ,... are called the t...

Lecture Notes on Infinite Series Laurel Benn July 9, 2009 Definition 1 An Infinite Series is an expression of the form ∞  an = a1 + a2 +... + an +... n=1 Where a1 , a2 , a3 ,..., an ,... are called the terms of the series. If we let Sn be the sum of the first n terms of the series then we have the following: S1 = a1 S 2 = a1 + a2 S 3 = a1 + a2 + a 3... n  Sn = a1 + a2 + a3 +... + an = an n=1 We’ll call Sn the nth partial sum of the series. The partial sums form {Sn }+∞ n=1 - the sequence of partial sums. ∞  Definition 2 Let {Sn } be a sequence of partial sums of an. n=1 If {Sn } converges to a limit S, then the series also converges and S is called the sum of the series. ∞  S= an n=1 If {Sn } diverges then the series is said to diverge. A divergent series has no sum. Example 1 Determine if the series 1 − 1 + 1 − 1 + 1 − 1 +... converges or diverges. 1 Now, S1 = 1, S2 = 1 − 1 = 0, S3 = 1, S4 = 0, e.t.c 1, 0, 1, 0, 1, 0,... is the sequence of partial sums. This sequence is divergent ⇒ the given series is divergent. We’ll now look at a class of series called a geometric series. Definition 3 A geometric series is a series of the form a + ar + ar2 + ar3 +... + arn +....Where a = 0 and r is a real number called the ratio of the series. Theorem 1 A geometric series converges if |r| < 1 and diverges if |r| ≥ 1. a When the series converges the sum is 1−r Example 2 5 + 54 + 452 + 453 + 454 +... + 4k−1 5 +... 1 is a geometric series with a = 5,and r = 4. ⇒ the series converges with sum = 1−5 1 = 20 3 4 ∞  1 Example 3 Determine if converges or diverges. If it converges find k=1 5k the sum. Well I’ll leave this one for you. Just identify it as a geometric series and do what’s needed. ∞  1 Example 4 Determine if the series converges or diverges. If k=1 k(k + 1) it converges find its sum. n 1 1 1 1 1 Here, Sn = = + + +... + k=1 k(k + 1) 1·2 2·3 3·4 n(n + 1) Using partial fractions we see that 1 k(k+1) = k1 − k+1 1 n   1 1 1 1 1 1 1 This implies that Sn = − = 1 − + − +... + − k=1 k k+1 2 2 3 n n+1 1 =1− n+1   1 1 So Sn = 1 − n+1 and n→∞ lim 1 − lim Sn = n→∞ =1 n+1 ∞ 1 ⇒ =1 k=1 k(k + 1) 2 Which means that the series converges with sum 1. The series in Example 4 is an example of what we call a Telescoping series. ∞  1 Example 5 Determine if converges or diverges. If it k=1 (k + 2)(k + 3) converges find the sum. This is another one that I would like you to try for me. We now come to an important theorem that allows us to quickly decide if a series diverges or not. ∞  Theorem 2 (Divergence Test) If lim ak = 0 then ak diverges. k→∞ k=1 ∞  k Example 6 diverges since k=1 k + 1 k 1 lim = lim 1 = 1 = 0 k→∞ k+1 k→∞ 1+ k Theorem 3 (Properties of Infinite Series) ∞  ∞  1. can = c an n=1 n=1 ∞  ∞  ∞  2. (an ± bn ) = an ± bn n=1 n=1 n=1 3. Convergence and Divergence are unaffected by deleting a finite number of terms from the beginning of a series. From (1) we see that if a series is convergent then a scalar times that series is also convergent. Similarly, if a series diverges then a scalar times that series also diverges. From (2) it is obvious that the sum or difference of 2 convergent series also converges. 3  ∞    3 2 Example 7 Find the sum of the series −. k=1 4k 5k−1 ∞    ∞  ∞ 3 2 3 2 From the above theorem − = − k=1 4k 5k−1 k=1 4k k=1 5 k−1 3 4 2 = 1 − 1 1− 4 1− 5 5 3 =1− =− 2 2 ∞  2 Example 8 Find the sum of k=1 5k ∞  ∞  2 1 From (1) in Theorem 3, we have = 2 k=1 5k k=1 5k ∞  1 Well is a series you already dealt with in Example 3, so you know 5k k=1 what to do. ∞  k Example 9 Determine if converges or diverges. k=10 k + 1 ∞ ∞  k k From Example 6 diverges, therefore also diverges since k=1 k + 1 k=10 k + 1 ∞ k it is with the first nine terms taken out and according to (3) from k=1 k + 1 Theorem 3 such a series must also diverge. ∞  Theorem 4 (Integral Test) Let an be a series with positive terms , n=1 and let f (x) be the function such that f (n) = an. If f is decreasing and continuous for x ≥ 1 , then ∞   ∞ an and f (x) dx both converge or both diverge. n=1 1 ∞  1 Example 10 Determine if converges or diverges. n=1 n2 1 f (x) = x2 4   ∞  M dx dx 2 = lim 1 x M →∞ 1 x2   1 M = lim − M →∞ x 1   1 = lim 1 − =1 M →∞ M We have just shown that the improper integral converges,therefore the series converges. ∞  1 Example 11 Show that diverges using the integral test. n n=1 I’ll leave this one to you. You just need to set up an improper integral like the one I set up in Example 1. Then show that the integral diverges. ∞  n Example 12 Determine if the series converges or diverges. n=1 en2 x 2 Here we’ll let f (x) = ex 2 = xe−x then 2 f  (x) = e−x (1 − 2x2 ) ≤ 0 This implies that f is decreasing for x ≥ 1 and since all the terms of the series are positive we can go ahead and use the integral test.  ∞  M −x2 2 xe dx = lim xe−x dx 1 M →∞ 1 M   1 2 = lim − e−x M →∞ 2 1   1 1 −M 2 = lim − e M →∞ 2e 2 1 = 2e This implies that the improper integral converges and therefore the series converges. The Integral Test leads us to the following theorem. 5  ∞  1 1 1 Theorem 5 = 1+ + p +..., where p > 0 converges if p > 1 and k=1 kp 2p 3 diverges if 0 < p ≤ 1. The above series is called a p - series. ∞ 1 When p = 1 we get the series which is called the harmonic series and k=1 k which is of course divergent. ∞  1 Example 13 Determine if the series √3 converges. k=1 k 1 Now √ 3 k = 11 This means that the series is a p-series with p = 13. From k3 the last theorem we know that a p - series converges if p > 1 and diverges if 0 < p ≤ 1. Therefore the given series diverges. ∞  Theorem 6 (Ratio Test) Let an be a series with non-zero terms.And   k=1  an+1  let ρ = n→∞  lim    an 1. The series converges if ρ < 1 2. The series diverges if ρ > 1 3. The test is inconclusive if ρ = 1 ∞  2n Example 14 Determine if converges or diverges. n=1 n! 2n+1 n an+1 = (n+1)! , an = 2n!    n+1   an+1   a  n =  (n+1)! 2 · 2n!n  = n+1 2 2 ρ = n→∞ lim =0 n+1 Therefore the series converges by the ratio test. Theorem 7 (Alternating Series Test) An alternating series a1 − a2 + a3 − a4 +... + (−1)k+1 ak +... or −a1 + a2 − a3 +... + (−1)k ak +..., all ak > 0 converges if the following conditions are met: 6  1. a1 ≥ a2 ≥ a3 ≥... ≥ ak ≥... 2. lim ak = 0 k→∞ Definition 4 (Power Series) An infinite series of the form ∞  an xn = a0 + a1 x + a2 x2 +... + an xn +... n=0 is called a power series in x. ∞  An infinite series of the form an (x − c)n = a0 + a1 (x − c) + a2 (x − c)2 + n=0... + an (x − c)n +... is called a power series centered at c. Theorem 8 (Convergence of a Power Series) For a power series cen- tered at c only one of the following is true. 1. The series converges only at x = c. 2. The series converges for all x. 3. There exists a positive real number R such that the series converges for |x − c| < R and diverges for |x − c| > R In the third case the series converges in the interval (c − R, C + R) and diverges in intervals (−∞, c − R) and (c + R, ∞). We would still need to check the endpoints c − R and c + R for convergence. The interval in which the series converges is called the interval of convergence. Definition 5 (Radius of Convergence) The radius of convergence of a power series  centered  at c is  a   n  lim  R = n→∞ , 0 ≤ R ≤ ∞. an+1  ∞  xn Example 15 Find the radius of convergence of.     n=0 n!  a   1/n!   n   lim  R = n→∞ = lim   an+1  n→∞  1/(n + 1)!     (n + 1)!    lim  = n→∞  n!  7  lim (n + 1) = ∞ = n→∞ A radius of convergence of infinity means that the power series converges for all real values of x. ∞  (−1)n (x + 1)n Example 16 Find the radius of convergence of.     n=0  2n  a   (−1)n /2n  2   n+1   n    lim  R = n→∞ = lim  = lim  = lim 2 = 2 an+1  n→∞  (−1)n+1 /2n+1  n→∞  2n  n→∞ Since the center of the series is c = −1, we conclude that the series converges in the interval (−1−2, −1+2) = (−3, 1). In fact if we check for convergence at the endpoints we find that the series diverges at the endpoints and (−3, 1) is in fact the interval of convergence. We now look at an important type of power series called the Taylor series. Here we’ll show how to use derivatives of a function to write the power series for that function. Definition 6 (Taylor Series) If f (x) has derivatives of all orders at c, then the power series for f (x) centered at c is called the Taylor series for f (x) centered at c and is given by ∞  f (n) (c) f  (c) f (n) (c) (x−c)n = f (c)+f  (c)(x−c)+ (x−c)2 +...+ (x−c)n +... n=0 n! 2! n! If c = 0 then the Taylor series is called a Maclaurin series. Example 17 Find the Maclaurin series for f (x) = ex. Now f (0) = e0 = 1 and since f  (x) = ex and all higher derivatives of f also equal ex. This implies that f (n) (0) = 1 for all n. Now by the definition of the Maclaurin series,  2  3 ex = f (0) + f  (0)x + f (0)x 2! + f (0)x 3! +... x2 x3 = 1 + x + 2! + 3! +... ∞  xn =. n=0 n! Example 18 Find the Taylor series for f (x) = 1/x, centered at 1. f (x) = x−1 ⇒ f (1) = 1 f  (x) = −x−2 ⇒ f  (1) = −1 f  (x) = 2x−3 ⇒ f  (1) = 2 8 f  (x) = −6x−4 ⇒ f  (1) = −6 f (4) (x) = 24x−5 ⇒ f (4) (1) = 24  2  3 (4) 4 ⇒ f (x) = x1 = f (1) + f  (1)(x − 1) + f (1)(x−1) 2! + f (1)(x−1) 3! + f (1)(x−1) 4! +... 2(x−1)2 6(x−1)3 24(x−1)4 = 1 − (x − 1) + 2! − 3! + 4! −... = 1 − (x − 1) + (x − 1)2 − (x − 1)3 + (x − 1)4 −... ∞  = (−1)n (x − 1)n. n=0 Which is the Taylor series we wanted. 9