3.1. Sequences and Their Limits Definition (3.1.1). A sequence of real numbers (or a sequence in R) is a function from N into R.


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1 CHAPTER 3 Sequences and Series 3.. Sequences and Their Limits Definition (3..). A sequence of real numbers (or a sequence in R) is a function from N into R. Notation. () The values of X : N! R are denoted as X(n) or x n, where X is the sequence. () (x n : n N) or simply (x n ) may denote a sequence this is not the same as {x n : n N}. (3) (x, x,..., x n,... ). Example. () (3n) = (3n : n N) = (3, 6, 9,..., 3n,... ). () () = ( : n N) = (,,,...,,... ). (3) ( ) n = ( ) n : n N = (4) + (, 4, 8,..., ( ) n,.... )n = + ( )n : n N = (0,, 0,,..., 0,,... ). 34
2 (5) n + ( )n = n + ( )n : n N =,,,,, 3,, 4,...,, n, SEQUENCES AND THEIR LIMITS 35 Sequences may also be defined inductively or recursively. Example. () x = 5, x n+ = x n 3 (n ) gives (5, 7,, 9, 35,... ). () Fibonacci sequence: x = x =, x n+ = x n + x n (n ) gives (,,, 3, 5, 8, 3,... ). Definition (3..3). A sequence X = (x n ) in R is said to converge to x R, or have limit x, if 8 > 0 9 K( ) N 3 8 n K( ), x n x <. We write this as lim X = x, lim(x n ) = x, lim n! x n = x, or x n! x as n!. A sequence that converges is called convergent, one that does not divergent. Example. lim = 0. n Proof. Let > 0 be given. By the Archimedean property, 9 K( ) N 3 <. Then, K( ) for n K( ), n apple, and so K( ) 0 = n n apple K( ) <. Thus lim = 0 by definition. n
3 36 3. SEQUENCES AND SERIES Calculator Visualisation lim n! n = lim = 0 if 8 > 0, x! x with ymin = 0 and ymax = 0 +, you can find K( ) N 3 xmin = K( ) and xmax = E99, if the graph only enters the screen from the left and exits from the right. Theorem (3..4 Uniqueness of Limits). A sequence in R can have at most one limit. Proof. [The technique.] Suppose lim(x n ) = x 0 and lim(x n ) = x 00. By Theorem..9, it su ces to show that x 0 x 00 < 8 > 0, for then x 0 x 00 = 0 =) x 0 = x 00. Let > 0 be given. Since lim(x n ) = x 0, 9 K 0 N 3 8 n K 0, x n x 0 <. Since lim(x n ) = x 00, 9 K 00 N 3 8 n K 00, x n x 00 <. Let K = max{k 0, K 00 }. Then n K =) n K 0 and n K 00 =) x 0 x 00 = x 0 x n + x {z n x 00 } smuggling apple x 0 x n + x n x 00 < + =.
4 3.. SEQUENCES AND THEIR LIMITS 37 Theorem (3..5). Let X = (x n ) be a sequence in R, and let x R. The following are equivalent: (a) X converges to x. (b) 8 > 0, 9 K N 3 8 n K, x n x <. (c) 8 > 0, 9 K N 3 8 n K, x < x n < x +. (d) 8 nbhd. V (x) of x, 9 K N 3 8 n Proof. (a) () (b) by definition. (b) () (c) () (d) since K, x n V (x). x n x < () < x n x < () x < x n < x + () x n V (x). Technique Given > 0. Produce or verify the existence of an integer K( ) so that n K( ) =) x n x <. Sometimes x n x < can be converted, with reversible steps, to an inequality of the form n > f( ). Take K( ) as the first integer greater than f( ) (by the Archimedean Property), K( ) = [f( )] +, for example. Then Example. n K( ) =) n > f( ) =) x n x <. () lim(c) = c, c R, i,e., x n = c 8n N. Proof. Given > 0. [To show 9 K( ) N 3 8 n K( ), c c <.] c c = 0 < 8 n N. Pick K( ) =. Then n K( ) =) c c <.
5 38 3. SEQUENCES AND SERIES () lim p n = 0. x n = p n here. Proof. Given > 0. h To show 9 K( ) N 3 n K( ) =) p i 0 <. n Now p 0 < () p < () n n < p n () < n. Pick K( ) = h i c (3) lim = 0, c R, p > 0. n p +. Then n K( ) =) n > =) p n 0 <. Proof. Case c = 0 was Example, so assume c 6= 0. Given > 0. c 0 < () c c c /p < () n p np < np () < n. apple c /p Take K = +. c /p c Then n K =) n > =) 0 <. n p Note. Thus x n = 3p n, x n = 5 n 5/4, and x n =, 000, 000! n all have limit 0.
6 (4) lim = 0. n Proof. Given > 0. n 0 < () n < () < n () 3.. SEQUENCES AND THEIR LIMITS 39 ln < ln n () ln < n ln () ln ln < n. h ln i Take K = max, +. ln Then n K =) n > ln ln =) n 0 <. (5) Let x n = + ( ) n. X = (0,, 0,,... ). lim(x n ) does not exist. Proof. [We use contradiction.] Suppose lim(x n ) = x. Then, 8 > 0, 9 K N 3 8 n K, x n x <. In particular, for =, 9 K N 3 8 n K, x n x <. ( 0 x < for n odd But x < for n even, so = x + x apple x + x apple x + x 0 < + =, a contradiction. Thus lim(x n ) does not exist.
7 40 3. SEQUENCES AND SERIES (6) Let x n = p n. lim(x n ) does not exist. Proof. [We again use contradiction.] Suppose lim( p n) = x. Then, 8 > 0, 9 K N 3 8 n K, p n x < or, equivalently, < p n x < or x < p n < x +. Then, for =, 8 n K(), p n < x + or n < (x + ), contradicting the Archimedean Property. Thus lim( p n) does not exist. Homework Pages 66 #5b,5d (Do not use Theorem 3..0 with these work from the definition) Extra Problem: Prove ( ) n diverges. (Hint: This is a translation of Example 5 watch your inequalities, though.) Note. Sometimes it is awkward or impossible to solve x n x < for n. In such cases, it may be possible to establish an inequality of the form where C > 0 and lim a n = 0. x n x apple C a n
8 3.. SEQUENCES AND THEIR LIMITS 4 Theorem (3..0). Let (a n ) and (x n ) be sequences in R, lim(a n ) = 0, and x R. If for some C > 0 and some m N we have then lim(x n ) = x. x n x apple C a n 8 n m, Proof. Let > 0 be given. Since lim(a n ) = 0, 9 K a N 3 8 n K a, a n 0 < C C C. Let K x ( ) = max m, K a. Then C 8 n K x ( ), x n x apple {z} n m C a n < {z} n K a C C C =. Thus lim(x n ) = x. Example. + ( ) n (7) lim = 0. n h Proof. X = 0,, 0,, 0, 3,..., 0, i n, ( ) n 0 = + ( )n apple + n n n = n! 0 by Example 3. The result follows from Theorem 3..0.
9 4 3. SEQUENCES AND SERIES n + (8) lim = 3n + 3. h Proof. X = 5, 3 8, 4, 5 4, 6 7, 7 0,..., n + i 3n +,.... n + 3n + 3 = 3n + 3 3n 3(3n + ) by Example 3. The result follows from Theorem = 3(3n + ) apple 3(3n) = 9 n! 0 n + (9) lim n 3 + p = 0. n h 3 Proof. X =, 8 + p, p 3, 5 66, p 5,..., n + i n 3 + p n,.... n + n 3 + p n 0 = n + n 3 + p n apple n + n 3 apple n + n by Example 3. The result follows from Theorem n 3 = n n 3 = n! 0
10 3.. SEQUENCES AND THEIR LIMITS 43 n (0) lim = n 3 n. h Proof. X = 4, 4, 0 7, 67 4, 45,..., n i n 3 n,.... n n 3 n = n3 + 6 n 3 + n (n 3 n) = n + 6 (n 3 n) apple {z} n 6 n + n (n 3 n 3 ) = n n = 3 n! 0 by Example 3. The result follows from Theorem Ultimate Behavior Definition (3..8). If X = (x, x,..., x n,... ) is a sequence in R and if m N, the mtail of X is the sequence X m = (x m+n : n N) = (x m+, x m+,..., x m+n,... ). Example. The 4tail of,, 3,..., n,... is X 4 = 5, 6, 7,..., 4 + n,.... Theorem (3..9). Let X = (x n : n N) be a sequence and let m N. Then the mtail X m = (x m+n : n N) converges () X converges. In this case, lim X m = lim X. Proof. Read in text it is just a translation argument. Homework Pages 66 #6a, 6c, 0, Extra: Prove lim( p n + p n) = 0. Hint for #0: Look at Theorem 3..5(c) and pick the right.
11 44 3. SEQUENCES AND SERIES 3.. Limit Theorems Definition (3..). A sequence (x n ) is bounded if 9 M > 0 3 x n apple M 8 n N. Theorem (3..). If (x n ) converges, then (x n ) is bounded. Proof. Suppose lim(x n ) = x and =. Then 9 K() N 3 8 n K(), x n x <. Then, for n K(), x n = x n x + x apple x n x + x < + x. Let M = sup x, x,..., x K(), + x. Then x n apple M 8 n N. Example. ( ) n is bounded since ( ) n apple 8 n N, but does not converge. Thus, bounded 6=) convergent. Example. ( n ) diverges. Proof. If ( n ) converged, it would be bounded. Thus 9 M > 0 3 n = n apple M 8 n N. Then n = log n apple log M 8 n N, contradicting the Archimedean Property.
12 3.. LIMIT THEOREMS 45 Theorem (3..3). Suppose lim(x n ) = x and lim(y n ) = y. (a) lim(x n + y n ) = lim(x n ) + lim(y n ) = x + y. Proof. [The technique.] Since lim(x n ) = x, 9 K N 3 8 n K, x n x <. Since lim(y n ) = y, 9 K N 3 8 n K, y n y <. Let K = max{k, K }. Then n K =) n K and n K =) (x n + y n ) (x + y) = (x n x) + (y n y) apple x n x + y n y < + =. (b) lim(x n y n ) = lim(x n ) lim(y n ) = x y. Proof. Similar to the above.
13 46 3. SEQUENCES AND SERIES (c) lim(x n y n ) = lim(x n ) lim(y n ) = xy. Proof. Let > 0 be given. Note x n y y xy = (by smuggling) (x n y n x n y) + (x n y xy) apple (x n y n x n y) + (x n y xy) apple x n y n y + y x n x. [We are now able to gain control over all of the variable parts.] Since lim(x n ) = x: () 9 M > 0 3 x n apple M 8 n N by Theorem 3... Let M = max M, y. () 9 K N 3 8n K, x n x < M. Since lim(y n ) = y, 9 K N 3 8n K, y n y < M. Let K = max K, K. Then, 8n K, x n y y xy apple x n y n y + y x n x apple M (d) lim(cx n ) = c lim(x n ) = cx for c R. Proof. This is a special case of (c). M + M M =.
14 3.. LIMIT THEOREMS 47 (e) If lim(z n ) = z, z n 6= 0 8 n N, and z 6= 0, then lim = z n z. Proof. Let > 0 be given. Note z n z = z h We need to find a bound for z n z n z Let = z > 0. Since lim(z n) = z: = z n z z n z. i z n in the first factor. () 9 K N 3 8n K, z n z <. Then < z n z {z} apple z n z =) Cor...4(a) Th...(c) z = z apple z n {z } z n is bounded away from 0 () 9 K N 3 8n K, z n z < z. =) z n apple z. Let K = max K, K. Then, 8n K, z n z = z n z z n z < z z =. (f) If lim(z n ) = z, z n 6= 0 8 n N, and z 6= 0, then lim xn z n x z. Proof. This follows directly from parts (c) and (e) above. = lim(x n) lim(z n ) =
15 48 3. SEQUENCES AND SERIES Example. Find lim Proof. lim 3n n + n 3n n + n. = lim lim(3) lim n lim() + lim n 3 n + = lim 3 n n lim + n = = 3. = Theorem (3..4). If lim(x n ) = x and x n 0 8 n N, then x 0. Proof. [Use contradiction by picking an appropriate.] Suppose x < 0 =) x > 0. Since lim(x n ) = x, for = x, 9 K N 3 8 n K, x n x < x or ( x) < x n x < x or x + x < x n < x + x = 0. Thus, for n = K, x K < 0, contradicting our hypotheses. Thus x 0. Theorem (3..5). If (x n ) and (y n ) are convergent sequences and if x n apple y n 8 n N, then lim(x n ) apple lim(y n ). Proof. Let z n = y n x n. Then z n 0 8 n N, so 0 apple lim(z n ) = lim(y n ) lim(x n ) =) lim(x n ) apple lim(y n ). Theorem (3..6). If (x n ) is convergent and a apple x n apple b 8 n N, then a apple lim(x n ) apple b. Proof. This follows from Theorem 3..5 by comparing (a) and (b) with (x n ).
16 3.. LIMIT THEOREMS 49 Theorem (3..7 Squeeze Theorem). Suppose x n apple y n apple z n 8n N and lim(x n ) = lim(z n ). Then (y n ) converges and lim(x n ) = lim(y n ) = lim(z n ). Proof. Let w = lim(x n ) = lim(z n ). Given > 0. 9 K N 3 8 n K, < x n w <, and also 9 K N 3 8 n K, < z n w <. Let K = max K, K. Then for n K, {z} < x n w apple y n w apple z n w {z} < =) y n w <. n K n K Thus lim(y n ) = w. Note. The hypotheses of Theorem 3..4 thru Theorem 3..7 can be weakened to apply to tails of the sequences rather than to the sequences themselves. Example. cos n () Find lim. n Solution. apple cos n apple =) cos n apple n n apple n. Since lim = lim = 0, n n cos n lim = 0 by the Squeeze Theorem. n
17 50 3. SEQUENCES AND SERIES () Find lim n /n. Solution. [This one is tricky.] For n >, n /n >, so x n = n /n = + t n, where t n = n /n from the Binomial theorem, so Thus Since lim() = lim n = ( + t n ) n = + nt n + n(n ) t n + positive terms, n(n ) t n < n =) t n < n =) t n < < x n = + t n < + + p p n =, p p n =) < x n < + p p n. > 0. Then, p p n. lim(x n ) = lim n /n = by the Squeeze Theorem. np (3) Find lim n. Solution. lim np n = lim np n np n = lim np n lim np n = =. Theorem (3..9). Suppose lim(x n ) = x. Then lim( x n ) = x. Proof. We know x n x apple x n x. Thus, given > 0, if 9 K N 3 8 n K, x n x <, we also get x n x <.
18 3.. LIMIT THEOREMS 5 Theorem (3..0). Suppose lim(x n ) = x and x n lim( p x n ) = p x. Proof. [Using the conjugate]. First, x 0 by Theorem Let > 0 be given. Case x = 0 9 K N 3 8 n K, x n 0 < () 0 apple x n < () 0 apple p x n < () p x n 0 <. 0 8 n N. Then Case x > 0 Then p x > 0. 9 K N 3 8 n K, x n x < p x =) p x n p x = ( p x n p x) ( p xn + p x) p xn + p x = x n x p xn + p x apple x n x p < x p x p x =. Homework Pages #d, 5b, 6bd (both find and prove) xn Extra: If lim = x 6= 0, then (x n ) is not bounded. n xn bounded =) lim = 0. n Hint: Prove (x n )
19 5 3. SEQUENCES AND SERIES 3.3. Monotone Sequences Definition (3.3.). Let X = (x n ) be a sequence. X is increasing if x apple x apple apple x n apple x n+ apple. X is decreasing if x x x n x n+. X is monotone if it is either increasing or decreasing. Example. () (,,, 3, 5, 8,... ) is increasing. () For 0 < b <, (b, b, b 3,... ) is decreasing. (3) (, 0,, 0,... ) is not monotone. (4) (4,,, 3, 3, 5, 5, 7, 7,... ) is ultimately increasing. Theorem (3.3. Monotone Convergence Theorem (MCT)). A monotone sequence converges () it is bounded. Further: (a) if X = (x n ) is a bounded, increasing sequence, then lim(x n ) = sup{x n }. (b) if X = (x n ) is a bounded, decreasing sequence, then lim(x n ) = inf{x n }. Proof. (=)) Follows from Theorem 3... ((=) (a) Since (x n ) is bounded, 9 M > 0 3 x n apple M 8 n N =) x n apple M 8 n N. Thus, by Completeness, x? = sup{x n } exists. Let > 0 be given. By Property S, 9 K N 3 x? < x K (x K = s ). Since (x n ) is increasing, x K apple x n 8 n K. Thus, 8 n K, {z } < x K apple {z} x n apple x? < x? {z + } =) x n x? <. Thus lim(x n ) = x?. x? (b) Similar to (a), except uses Property I.
20 3.3. MONOTONE SEQUENCES 53 Example. Determine whether lim(x n ) exists and, if so, its value where x = and x n+ = p + x n for n. Solution. x = p + = p, x 3 = (a) [Show monotone increasing.] q + p, x 4 = x < x since < p. Assume x n apple x n+. Then x n+ = p + x n apple p + x n+ = x n+, so by induction x n apple x n+ 8 n N. Thus (x n ) is increasing. (b) [Show (x n ) is bounded above by using induction.] x = <. Suppose x n apple. Then r + x n+ = p + x n apple p + = p 3 < p 4 =. Thus, by induction, x n apple 8 n N, and so is an upper bound of (x n ). (c) Thus lim(x n ) = x for some x R by the MCT. Since (x n+ ) is a tail of (x n ), lim(x n+ ) = x also. Then q + p,... x = lim(x n+ ) = lim( p + x n ) = p lim( + x n ) = p lim() + lim(xn ) = p + x =) Since p 5 x = + x =) x x = 0 =) x = ± p 5. < 0, we conclude x = lim(x n ) = + p 5.
21 54 3. SEQUENCES AND SERIES Note. () An increasing sequence is bounded below by its first term. Thus if x? = sup{x n : n N}, M = max x, x? is a bound for the sequence. () A decreasing sequence is bounded above by its first term. Homework Page 77 #, Hint for # : (a) Show x n x n+ 0 8 n N. Thus (x n ) is decreasing. (b) Show (x n ) is bounded below. Then (x n ) is bounded by M = max x, l.b.. (c) Find and solve an equation to get x = lim(x n ). Example. () Establish convergence or divergence of (x n ) where x n = +! +! + + n!. Solution. x n+ = x n + (n + )! > x n, so (x n ) is increasing. Noting that (n + )! < n, we have x n < = 3 n + ( )n = + so (x n ) is bounded above and so (x n ) converges by the MCT. n < 3, Although we now know the limit exists, we do not have a technique for finding the exact limit.
22 (3) 3.3. MONOTONE SEQUENCES 55 Let A n = the sum of the semicircular areas. Let L n = the sum of the semicircumferences. It appears lim A n = 0 and lim L n =. Well, n! n! A n = n = n 8! 0 as n!, n but L n = n = n 8 n N, so lim L n = n!.
23 56 3. SEQUENCES AND SERIES Problem (Page 77 # 0). Establish convergence or divergence of (y n ) where y n = + n + n n N. n {z } largest term {z} smallest term Solution. It might seem obvious that lim(y n ) = 0, but incorrect. Note that y n n + + = n {z n} n terms and y n apple n = n {z n + } n terms so (y n ) is bounded by. Now so y n+ y n = n = n + = n n + <, y n+ = n + + n n + n + + n +, n + + n + n + = n + + n + n + so (y n ) is increasing. Thus (y n ) converges by the MCT and apple lim(y n) apple. Can we find lim(y n )? n + = n + = (n + )(n + ) > 0,
24 Note y n = [What is this latter sum?] 3.3. MONOTONE SEQUENCES 57 nx k= n + k = nx k= + k n n. y n is a righthand Riemann sum for f(x) = + x Thus lim(y n ) = Z 0 for 0 apple x apple. + x dx = ln + x 0 = ln ln = ln.
25 58 3. SEQUENCES AND SERIES 3.4. Subsequences and the BolzanoWeierstrass Theorem Definition (3.4.). Let X = (x n ) be a sequence and let n < n < < n k < be a strictly increasing sequence of natural numbers. Then the sequence X 0 = (x nk ) = (x n, x n,..., x nk,... ) is a subsequence of X. Example. Let X = = n, 4, 6,..., n,.... Some subsequences: () (x nk ) = = 4k 4, 8,,..., 4k,.... () (x nk ) = = 4k, 6, 0,..., 4k,.... (3) (x nk ) = = 4k 4, 6, 36,..., 4k,.... (4) (x nk ) = = (k)!!, 4!, 6!,..., (k)!,... =, 4, 70,..., (k)!,.... (5) X itself (6) Any tail of X In general, to form a subsequence of X, just pick out any infinite selection of terms of X going from left to right.
26 3.4. SUBSEQUENCES AND THE BOLZANOWEIERSTRASS THEOREM 59 Theorem (3.4.). If X = (x n ) converges to x, so does any subsequence (x nk ). Proof. Let > 0 be given. Since (x n ) converges to x, 9 K N 3 8 n K, x n x <. Since n < n < < n k < is an increasing sequence in N, n k k 8 k N. Let K 0 = n K. Then, 8 n k K 0 = n K, n k K =) x nk x <. Thus (x nk ) converges to x. Example. For c >, find lim(c n) if it exists. Solution. (a) x n = c n > 8 n N, so (x n ) is bounded below. (b) x n x n+ = c n c n+ = c n+ c n(n+) > 0 8 n N, so (x n ) is decreasing. (c) Thus lim(x n ) = x exists. [Using a subsequence to find x.] Now x n = c n = c n = x n, so x = lim(x n ) = lim x n = x =) x = x =) x x = 0 =) x(x ) = 0 =) x = 0 or x =. Since x n > 8 n N, lim(x n ) =.
27 60 3. SEQUENCES AND SERIES Theorem (3.4.7 Monotone Subsequence Theorem). If X = (x n ) is a sequence in R, then there is a subsequence of X that is monotone. Proof. We will call x m a peak if n the right of x m is greater than x m ). Case : X has infinitely many peaks. Order the peaks by increasing subscripts. Then x m x m x mk, so (x m, x m,..., x mk,... ) is a decreasing subsequence. Case : X has finitely many (maybe 0) peaks. Let x m, x m,..., x mr denote these peaks. m =) x n apple x m (i.e, if no term to Let s = m r + (the first index past the last peak) or s = if there are no peaks. Since x s is not a peak, 9 s > s 3 x s < x s. Since x s is not a peak, 9 s 3 > s 3 x s < x s3. Continuing, we get an increasing subsequence. Theorem (3.4.8 BolzaonoWeierstrass Theorem). A bounded sequence of real numbers has a convergent subsequence. Proof. If X = (x n ) is bounded, by the Monotone Subsequence Theorem it has a monotone subsequence X 0 which is also bounded. Then X 0 is convergent by the MCT.
28 3.4. SUBSEQUENCES AND THE BOLZANOWEIERSTRASS THEOREM 6 Theorem (3.4.4). Let X = (x n ) be a sequence. The following are equivalent: (a) (x n ) does not converge to x R. (b) 9 0 > k N, 9 n k N 3 n k k and x nk x 0. (c) 9 0 > 0 and a subsequence X 0 = (x nk ) of X 3 x nk x 0 8 k N. Proof. [(a) =) (b)] This is the negative of the definition of convergence. [(b) =) (c)] Take the 0 from (b). Let n N 3 x n x 0. Let n N 3 n > n and x n x 0. Let n 3 N 3 n 3 > n and x n3 x 0. Continuing, we generate the subsequence. [(c) =) (a)] Suppose X = (x n ) has a subsequence X 0 = (x nk ) satisfying (c). If x n! x, so would (x nk )! x. Then 9 K N 3 8k K, x nk x < 0. But this contradicts (c).
29 6 3. SEQUENCES AND SERIES Example. cos n p does not converge to 4. Proof. cos n p p p p = 4, 0,,,, 0,,,.... p Let 0 = 4. 8 k N, let n k = 8k + 3. (8k + 3) p Then (x nk ) = cos =. 4 p p p Then 8 k N, x nk = = p p 4 = 0. Thus cos n p does not converge to 4. Theorem (3.4.5 Divergence Criterion). If a sequence X = (x n ) has either of the following properties, then X is divergent. (a) X has two convergent subsequences X 0 = (x nk ) and X 00 = (x rk ) whose limits are not equal. (b) X is unbounded. Homework Pages # 4b, 9 (Hint: Use Theorem 3.4.4), (Hint: What is the only possible limit?), 4 (extra credit)
30 3.4. SUBSEQUENCES AND THE BOLZANOWEIERSTRASS THEOREM 63 Theorem (3.4.9). Let X = (x n ) be a bounded sequence such that every convergent subsequence converges to x. Then lim(x n ) = x. Proof. Let M be a bound for X. Suppose x n 6! x. By Theorem 3.4.4, 9 0 > 0 and a subsequence X 0 = (x nk ) 3 x nk x 0 8 k N. Now M is also a bound for X 0 = (x nk ), so it has a convergent subsequence X 00 = (x nk r ) with lim(x n k r ) = x. Then 9 K N 3 8 r K, x nk r x < 0, a contradiction. Example. We cannot drop the bounded hypothesis:,, 3, 4, 5, 6,....
31 64 3. SEQUENCES AND SERIES 3.5. The Cauchy Criterion Example. Suppose lim(x m NO! x n = p n is a counterexample. x n ) = 0? Does lim(x n ) necessarily exist?. Definition (3.5.). A sequence X = (x n ) is a Cauchy sequence if 8 > 0 9 H( ) N 3 8 n, m H( ) with n, m N, x n x m < Lemma (3.5.3). If X = (x n ) converges, then X is Cauchy. Proof. [Another argument.] Suppose lim(x n ) = x. Given > 0, 9 K N 3 8 n K, x n x <. Let H = K. Then, for m, n H = K, x n x m = (x n x) + (x x m ) apple x n x + x m x < + = Thus (x n ) is Cauchy. Note. (x n ) is not Cauchy if 9 0 > 0 3 8H N, 9 n, m H 3 x n x m 0. Example. (x n ) = p n is not Cauchy. Proof. Let 0 = and H N be given. Let m = H, so p x m = p m = p H. Since p n is unbounded, 9 n N 3 p p n H where n H. Thus p p x n xm.
32 3.5. THE CAUCHY CRITERION 65 Lemma (3.5.4). Cauchy sequences are bounded. Proof. Let X = (x n ) be Cauchy and = 9 H N 3 8 n H, x n x H <. Then x n x H apple x n x H < =) < x n x H < =) x n < x H +. Let M = max x, x,..., x H, x H +. Then x n apple M 8n N. Theorem (3.5.5 Cauchy Convergence Criterion). A sequence is convergent () it is Cauchy. Proof. [Yet another argument.] (=)) Lemma ((=) Let X = (x n ) be Cauchy. Then X is bounded, so by BW, X has a convergent subsequence, say X 0 = (x nk )! x?. [To show lim(x n ) = x?.] Let > 0 be given. Since (x n ) is Cauchy, 9 H N 3 8 n, m H, x n x m <. Since lim(x nk ) = x?, 9 K N 3 K H and x K x? <. But x n x K < also. Then, for n H, x n x? = (x n x K ) + (x K x? ) apple (x n x K ) + (x K x? ) < + =. Thus lim(x n ) = x?.
33 66 3. SEQUENCES AND SERIES Example. (x n ) = Proof. n is Cauchy. Given > 0. WLOG (without loss of generality), suppose n m. Then Take H = n h +. Then i m = m n < m < {z} (= if n m H =) n m > =) n m >. m <.
34 3.5. THE CAUCHY CRITERION 67 Problem (Page 9 # b). Show +! + 3! + + is Cauchy. n! Proof. h In this proof we use the facts that n apple n! (Example..4(e)) and that + r + r + + r n = i rn+ r. Given > 0. WLOG, suppose n m. x n x m = +! + 3! + + n! (m + )! + (m + )! + + n! + m + n m +! + 3! + + m! apple m + m+ + + n apple = m n m = n m n apple n m = n < (= m < m (= log < m (= log < m. n h i o Choose H = max, log +. Then = n m H =) m > log =) x n x m <. Definition (3.5.7). A sequence X = (x n ) is contractive if 9 0 < C < 3 x n+ x n+ apple C x n+ x n 8 n N. C is the constant of the contractive sequence.
35 68 3. SEQUENCES AND SERIES Theorem (3.5.8). Every contractive sequence is convergent. Proof. [We prove the sequence to be Cauchy, and thus convergent.] Let X = (x n ) be a contractive sequence. 8 n N, x n+ x n+ apple C x n+ x n apple C x n x n apple apple C n x x. Then, WLOG for m > n, x m x n apple x m x m + x m x m + + x n+ x n {z } smuggling + triangle inequality C m + C m C n x x = C n C m n + C m n + + x x = C C n m n x x apple C C n x x! 0 as n! C since lim(c n ) = 0. Thus (x n ) is Cauchy, and so convergent. apple
36 3.5. THE CAUCHY CRITERION 69 Example. x =, x =, x n = (x n + x n ) for n 3. (x n ) =,, 3, 7 4, 3 8, 7 6,.... (a) (x n ) is contractive. Thus (x n ) converges. Proof. x n+ x n+ = (x n + x n+ ) x n+ = x n (b) Note that x n+ = x n+ x n. x n+ x n = n x x = and x n n+ x n > 0 (by induction). (c) [To find lim(x n+ ) = lim(x n ).] x n+ x n = (x n + x n ) x n = x n x n = x n x n = = n n. Thus + x n+ = x n + = x n n n 3 = = h n = + n h n i = n 6 Thus lim(x n ) = lim(x n+ ) = 5 3. n i = 4 h n + n i = n 4 =! + 3 = 5 3 as n!. =
37 70 3. SEQUENCES AND SERIES 3.6. Properly Divergent Sequences Definition. Let (x n ) be a sequence. (a) We say (x n ) tends to + and write lim(x n ) = + if 8 R 9 K( ) N 3 8 n K( ), x n >. (b) We say (x n ) tends to and write lim(x n ) = if 8 R 9 K( ) N 3 8 n K( ), x n <. We say (x n ) is properly divergent in either case. Example. For C >, lim(c n ) = + Proof. Let R be given. [How to express C >.] C = + b where b > 0. By the Archimedean Property, 9 K( ) N 3 K( ) >. Then 8 n K( ), b Thus lim(c n ) = +. Homework Page 9 # a, 3b, 7, 9 C n = ( + b) n {z} + nb > + >. Bernoulli
38 3.7. INTRODUCTION TO INFINITE SERIES Introduction to Infinite Series Definition (3.7.). If X = (x n ) is a sequence in R, then the infinite series (or just series) generated by X is the sequence S = (s k ) defined by s = x s = s + x (= x + x ). s k = s k + x k (= x + x + + x k ). The x n are the terms of the series and the s k are the partial sums of the series. If lim S exists, we say the series is convergent and call this limit the sum or value of the series. If this limit does not exist, we say this series S is divergent. Notation. We can also use X (xn ) X n=0 or x n X xn If the first term of the series is x N, then the first partial sum is s N. or X n=5 or x n X n= x n
39 46 3. SEQUENCES AND SERIES Example. X () r n = + r + r + + r n + (geometric series) n=0 For r 6=, s n = + r + r + + r n rs n = r + r + + r n + r n+ s n ( r) = s n rs n = r n+ s n = Then X r n r n+ = lim n! r n=0 X and r n diverges if r. () Since n=0 X n= rn+ r = r n(n + ) = k(k + ) = k k +, s n = + X n= s n = n(n + ) = lim n! if r <, n n + =) =. n + n + =)
40 (3) X ( ) n = + +. n=0 S = (s n ) = (, 0,, 0,... ) diverges =) 3.7. INTRODUCTION TO INFINITE SERIES 47 X ( ) n diverges. Theorem (3.7.3 nth Term Test). If P x n converges, lim(x n ) = 0. Proof. P x n converges =) s = lim(s n ) exists =) s = lim(s n ) =) lim(x n ) = lim(s n s n ) = lim(s n ) lim(s n ) = s s = 0. Example. (4) Geometric series with r diverges since (r n ) diverges. n=0 (5) For But X n= p = + p + p + + p +, n 3 n lim(x n ) = lim s n = + p + p p n p n = 0. Thus lim(s n ) p + p + p + + p = n n n n n {z } n terms lim( p X n)!, so p diverges. n x= p n = p n Note. This implies the converse of Theorem is not true.
41 48 3. SEQUENCES AND SERIES (6) Consider X cos n. n= Assume lim(cos n) = 0 =) lim(cos n) = 0 =) lim(sin n) = lim( cos n) = lim() lim(cos n) = 0 =. Then lim(sin n) = as a subsequence. Now sin n = sin n cos n =) sin n = 4 sin n cos n =) lim(sin n) = 4 lim(sin n) lim(cos n) = 4 0 = 0 6=, a contradiction. Thus lim(cos n) 6= 0 =) X cos n diverges. n= Theorem (3.7.4 Cauchy Criterion for Series). P xn converges () 8 > 0 9 M N 3 if m > n M, then s m s n = x n+ + x n+ + + x m <. Theorem (3.7.5). Let (x n ) be a sequence of nonnegative numbers. Then P xn converges () S = (s k ) is bounded. In this case, nx x n = lim(s k ) = sup{s k : k N}. i= Proof. Since x n 0 8 n N, (s k ) is monotone increasing. By the MCT, S = (s k ) converges () it is bounded, in which case lim(s k ) = sup{s k }.
42 Example. (7) The harmonic series 5. (8) The pseries X n= 3.7. INTRODUCTION TO INFINITE SERIES 49 X n= n diverges. The proof is similar to that of Example converges for p >. np Proof. Since (s k ) is monotone, we need only to show (s k ) is bounded. But it su ces to show that some subsequence is bounded. Let k = =, so s k =. Let k = = 3. Then, since p < 3 p, s k = p + p + 3 p < + p = + p. Let k 3 = 3 = 7 =) s k3 = s k p 5 + p 6 + p 7 p < s k < + p + p 4 p. Continuing inductively, if k j = j, Thus (s kj ) is bounded and so 0 < s kj = + p + 4 p + + ( j ) p = + + p ( p ) + + ( p ) j < p. X n= converges for p >. n p
43 50 3. SEQUENCES AND SERIES (9) The pseries For p apple 0, X n= diverges for p apple. np n p = n p with p 0 =) lim n p 6= 0. For 0 < p apple, n p < n =) n apple n p. Since the partial sumes of the harmonic series are not bounded, neither are the X partial sums here. Thus diverges for p apple. np n= Theorem (9.3. Alternating Series Test). Let Z = (z n ) be a decreasing sequence of strictly positive numbers with lim(z n ) = 0. Then the alternating X series ( ) n+ z n converges. n= Proof. s n = (z z ) + (z 3 z 4 ) + + (z n z n ), and since z k z k+ > 0, (s n ) is increasing. Since s n = z (z z 3 ) (z 4 z 5 ) (z n z n ) z n, s n < z 8 n N =) (s n ) converges by the MCT. Suppose lim(s n ) = s. Then, given > 0, 9 K N 3 for n K, Then for n K, s n s < and z n+ <. s n+ s = s n + z n+ s apple s n s + z n+ < + =. Thus, for n large enough, each partial sum is within of s, so lim(s n ) = s and X ( ) n+ z n = s. n=
44 Example. (0) The alternating harmonic series For 3.7. INTRODUCTION TO INFINITE SERIES 5 X ( ) n+ n converges. n= X Note. If ( ) n+ z n = s, s s n apple z n+. n= X ( ) n+ 0X n, s ( ) n+ n = s <. n= n= Theorem (3.7.7 Comparison Test). Let (x n ) and (y n ) be sequences where 0 apple x n apple y n for n K N. Then (a) P y n converges =) P x n converges. (b) P x n diverges =) P y n diverges. Proof. (a) Suppose P y n converges and > 0 is given. 9 M N 3 m > n M =) y n+ + + y m <. Let M 0 = max{k, M}. Then m > M 0 =) so P x n converges. (b) Contrapositive of (a). 0 apple x n+ + + x m apple y n+ + + y m <,
45 5 3. SEQUENCES AND SERIES Example. X () converges since n + n () n= X n= n + n < n 8 n N and n n= X n= n +. Seems to be like X n. n converges (p = ). But n n + > 8n N. n We can show 0 < n n + apple 8n N, but this is not obvious. n Theorem (3.7.8 Limit Comparison Test). Suppose (x n ) and (y n ) are xn strictly positive sequences and suppose r = lim exists. y n (a) If r 6= 0, P x n converges () P y n converges. (b) If r = 0 and P y n converges, then P x n converges. Proof. xn (a) Since r = lim 6= 0, 9 K N 3 y n r apple x n apple r for n K =) y n r y n apple x n apple (r)y n for n K. The result follows by applying comparison twice. (b) If r = 0, 9 K N 3 0 < x n apple y n for n K. The result follows by comparison.
46 Example. ( continued) Since X n= (3) n r = lim X n= X n= n n+ n 3.7. INTRODUCTION TO INFINITE SERIES 53 n n +. n = lim = lim n n + n + n n + converges by limit comparison since X n converges. p 3n +. We compare with r = lim Thus X n= p 3n+ p n X n= p n = X n= n= =, n/, which diverges as a pseries with p apple. n / h n i / = lim = lim = (3n + ) / 3n + apple n / / lim = = p3 6= 0. 3n + 3 p 3n + diverges by limit comparison.
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