Infinite Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics

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1 Infinite Series MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2018

2 Background Consider the repeating decimal form of 2/ = = = 6(0.1) + 6(0.1) 2 + 6(0.1) 3 + 6(0.1) 4 + = 6(0.1) k

3 Background Consider the repeating decimal form of 2/ = = = 6(0.1) + 6(0.1) 2 + 6(0.1) 3 + 6(0.1) 4 + = 6(0.1) k Question: what does it mean to perform a summation of an infinite number of terms?

4 Infinite Series Definition An infinite series is an expression of the form a 1 + a 2 + a a k + = A finite summation of the form S n = a 1 + a 2 + a a n = a k. n is called the n th partial sum of the series. The sequence {S n } n=1 is called the sequence of partial sums of the series. a k

5 Convergence and Divergence Definition If {S n } n=1 is the sequence of partial sums of the series and if lim S n = S n where S is finite, then S is called the sum of the series, we say the series converges and we can write S = a k. If S is infinite or does not exist then we say the series diverges. a k

6 Previous Example (1 of 2) Consider 6(0.1) k. What is the sequence of partial sums of the series?

7 Previous Example (1 of 2) Consider the series? 6(0.1) k. What is the sequence of partial sums of S 1 = S 2 = S 3 = 1 6(0.1) k = 6(0.1) 1 = (0.1) k = 6(0.1) 1 + 6(0.1) 2 = (0.1) k = S n = }{{ 66 } n digits

8 Previous Example (2 of 2) Consider 6(0.1) k. Does the series converge and, if so, to what number?

9 Previous Example (2 of 2) Consider 6(0.1) k. Does the series converge and, if so, to what number? The series converges to 2/3. 6(0.1) k = 2 3

10 Examples Determine if the following series converge or diverge k k 1 k 1 k(k + 1) ( 1) k

11 Solution: 21 k S 1 = S 2 = S 3 =. S N = k = = k = = k = = 7 4 N 2 1 k = N 1 S. = ( lim N = lim 2 1 ) N N 2 N 1 = 2 The infinite series converges to S = 2.

12 Solution: k 1 k S 1 = S 2 = S 3 =. S N = N k 1 k k 1 k k 1 k k 1 k = = 0 = = 1 2 = = 7 6 = S N 1 + N 1 N The infinite series diverges.

13 Solution: 1 k(k+1) Using partial fraction decomposition S N = = = N N 1 k(k + 1) ( = 1 + ( 1 k 1 ) + ( = 1 1 N + 1 ) k + 1 ( ) 3 ) + S = lim N S N = lim N 1 k(k + 1) = 1 k 1 k + 1. Thus ( ( N 1 ) N + 1 ) + + ( 1 1 ) = 1 N + 1 The infinite series converges to S = 1. ( 1 N + 1 N ) 1 N + 1

14 Solution: ( 1)k S 1 = S 2 = S 3 = S N = 1 ( 1) k = ( 1) 1 = 1 2 ( 1) k = = 0 3 ( 1) k = = 1 { 1 if N is odd, 0 if N is even. The infinite series diverges.

15 Geometric Series Theorem For a 0, the geometric series a r k converges to k=0 a 1 r if r < 1 and diverges if r 1. The term r is called the ratio.

16 Proof S N = r S N = r N a r k = a + ar + ar 2 + ar ar N k=0 N a r k = ar + ar 2 + ar ar N+1 k=0 S N r S N = a r N+1 S N (1 r) = a r N+1 S N = a r N+1 1 r lim S N = lim N N S = a 1 r a r N+1 1 r if r < 1.

17 Examples (1 of 3) Find the sum of the series

18 Examples (1 of 3) Find the sum of the series = 5( ) ( [ 2 ] 0 [ ] 2 1 [ 2 = [ ] 2 k = 5 3 = k=0 5 1 [ ] = ] 2 + [ ] )

19 Examples (2 of 3) Is the series 2 2k 3 1 k convergent or divergent?

20 Examples (2 of 3) Is the series 2 2k 3 1 k convergent or divergent? 2 2k 3 1 k = 4 k k = [ ] 4 k 3 3 Since r = 4 3 > 1, the infinite series diverges.

21 Examples (3 of 3) Write the repeating decimal as a ratio of integers.

22 Examples (3 of 3) Write the repeating decimal as a ratio of integers = = = ( ) = / = = =

23 Issues of Convergence and Divergence Theorem If a k converges, then lim a k = 0. k

24 Issues of Convergence and Divergence Theorem If a k converges, then lim a k = 0. k Proof. Let the sum of the series be L and let S N be the nth partial sum. S N S N 1 = N N 1 a k S N S N 1 = a N lim N S N 1 ) N = lim N N L L = 0 = lim N N a k = (a a N 1 + a N ) (a a N 1 )

25 k th -Term Test for Divergence Theorem (k th -Term Test for Divergence) If lim a k 0, then a k diverges. k

26 k th -Term Test for Divergence Theorem (k th -Term Test for Divergence) If lim a k 0, then a k diverges. k Example Show that the series k 2 5k diverges.

27 The Harmonic Series Remark: just because lim a k = 0 we cannot conclude that k a k converges. Lemma The harmonic series 1 k diverges.

28 Divergence of the Harmonic Series S 1 = 1 S 2 = = 3 2 S 4 = > 2 S 8 > 5 2 S 16 > 3 S 2 n > 1 + n 2 lim 2 n n =.

29 Properties of Convergent Series (1 of 2) Theorem If a k and b k are convergent series with sums A and B respectively, then 1. (a k + b k ) converges to A + B. 2. (a k b k ) converges to A B. 3. If c is a constant then (ca k ) converges to ca.

30 Properties of Convergent Series (2 of 2) Theorem If a k converges and b k diverges, then 1. (a k + b k ) diverges. 2. If c 0 is a constant then (cb k ) diverges.

31 Sierpinksi Gasket (1 of 3) Remove the center 1/9 of a square of side 1, then remove the center ninths of the eight remaining squares, and continue. Find the total area of all the removed squares.

32 Sierpinksi Gasket (2 of 3) Remove the center 1/9 of a square of side 1, then remove the center ninths of the eight remaining squares, and continue. Find the total area of all the removed squares.

33 Sierpinksi Gasket (3 of 3) Area removed: A = = 1 ( ) 81 + ( ) = 1 9 = 1/ = [ 8 9] 2 +

34 Half Circles Assume the largest circle has a radius of 1 and find the sum of the shaded areas in the figure.

35 Solution The largest circle has radius r = 1, the next r = 1/2, the next r = 1/4, and so on. A = = = k=0 k=0 π 2 π 2 π/2 1 1/4 ( (1 ) ) k 2 2 ( ) 1 k 4 = 2π 3

36 Triangles Assume the square has sides of length 1 and find the sum of the shaded areas in the figure.

37 Solution The square has area 1, so the largest triangle has area A 1 = 1/4. The second largest triangle is one quarter the area of a square whose sides are 1/2 in length, thus A 2 = (1/4)(1/4) = 1/16. The third largest triangle will have area A 3 = 1/64 and so on. A = ( = ( 1 4) 2 + ) = = k= /4 1 1/4 ( ) 1 k 4 = 1 3

38 Right Triangles b θ 1. Express the total length of the dotted line in the triangle in terms of b and θ. 2. Compute the sum of the length of the dotted line. 3. What happens as θ π/2?

39 Solution The right-most right triangle has a hypotenuse of length b and a height of length b sin θ. The third right-most triangle has a hypotenuse of length b sin 2 θ and a height of length n sin 3 θ. L = b + b sin θ + b sin 2 θ + b sin 3 θ + L = = b(1 + sin θ + sin 2 θ + sin 3 θ + ) = b(sin θ) k k=0 b 1 sin θ lim L = lim θ π/2 θ π/2 b 1 sin θ =

40 Stacking Blocks Suppose you have a large supply of rectangular blocks all of the same size and you stack them at the edge of a table with each block extending farther beyond the edge of the table than the one beneath it. Show that the top block can extend arbtrarily far beyond the edge of the table.

41 Stacking Blocks: Illustration 1/2 1/4 1/2 1/6 1/4 1/2 x 1/6 1/4 1/2

42 Stacking Blocks: Solution When there are n blocks, the stack will balance if exactly half the weight of the blocks extends beyond the edge of the table. Thus when n = 1 the block may overhang the edge of the table a distance s = 1/2. When n = 2, the top block overhangs the bottom block by 1/2 and the bottom block overhangs the edge of the table by x. A total weight of 2/2 = 1 extends beyond the edge of the table. 1 = x + (x + 1/2) = x = 1/4 Thus the total overhang is s = 1/2 + 1/4 = 3/4.

43 Stacking Blocks: Solution When n = 3, the top block overhangs the middle block by 1/2, the middle block overhangs the bottom block by 1/4, and the bottom block overhangs the edge of the table by x. A total weight of 3/2 extends beyond the edge of the table. 3 = x + (x + 1/4) + (x + 1/2 + 1/4) = x = 1/6 2 The total overhang is s = 3/4 + 1/6 = 11/12. Thus when there are n blocks the total overhang is s = 1 ( ). n

44 Homework Read Section 11.2 Exercises: WebAssign/D2L