CHAPTER 10, INFINITE SERIES Infinite Sequence

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1 CHAPTER INFINITE SERIES Definition. = f(n).. Infinite Sequence a a a 3 { } n= { } { }. () { n } { n } { 3n 7} 3 {sin nπ } {3+( )n }. () { } (3) { } (4) Fibonacii sequemce F =F =F n+ = F n + F n n>(recurrently definded) (5) A = A n+ = A n (.)(recurrently definded). Limit of Sequence. Definition. A sequence { } converges to a limit L if for any ɛ> there is N(ɛ) such that n N(ɛ) implies that L <ɛinthiscasewewritelim n = L. A sequence diverges if it does not converge. ( lim n n =. {( )n } diverges. Limit Laws. Theorem. Suppose that lim n = A lim n b n = B then () lim n c = ca () lim n ( + b n )=A + B (3) lim n b n = AB a (4) lim n n bn = A B if B. Theorem. (Substitution law) If lim n = a and lim n f( )=L thenlim n f( )= L. Theorem. (Squeeze law) If b n c n and lim n = L = lim n c n then lim n b n = L. 7n () lim n 5n 3 = 7 5 cos n () lim n = n (3) If a> lim n a n = Typeset by AMS-TEX

2 CHAPTER INFINITE SERIES (4) lim n 4n n+ = (5) If r < then lim n r n =. (6) lim n ln n n = (7) lim n n n = (8) lim n 3n 3 e n =. Definition. A sequence { } is bounded sequence if there exists M such that M for all n. Theorem. Convergent sequences are bounded sequence. Definition. A sequence { } is monotonic sequence if ( )+ for all n. Bounded monotonic sequences. Theorem. Bounded monotonic sequences are convergent sequences. a = 6+ = 6+ n In general (Exploration page 73) a = q + = q + p n Exer. 56 page 73. (Exploration page 73) a = p + = p + q. a = a + = +a n converges if and only if a. Definition. A sequence {b n } is a subsequence of { } if b n = a λ(n) for n = 3 and λ is an increasing function from N to N. = n b n = n c n =. n Theorem. If { } converges to L then every subsequence also converges to L Definition. A sequence { } is a Cauchy sequence if for any ɛ> there is N(ɛ) such that m n N(ɛ) implies that a m <ɛ. Theorem. A sequence is a convergent sequence iff it is a Cauchy sequence. Proof. () Choose c d such that c d for all n andletl = d c. () Let c = c d = d and choose N such that for all m>n N we have a m < L 4. (3) Let c =max(c a N L 4 )d =min(d a N + L 4 ) then c d for all n N and d c L.ChooseN such that for all m>n N we have a m < L 8. (4) Assume we have done k step. Let c k+ =max(c a Nk ) L d k =min(d a Nk )+ L k then c k+ le d k+ for all n N k+ and d k+ c k+ L.ChooseN k+ k+ such that for all m>n N k+ we have a m < L k+.

3 CHAPTER INFINITE SERIES 3 (5) Now we have c c cdotsc n d n d d.so{c n } converges to the l.u.b. C and {d n } converges to the g.l.b.d and D C. But since d n D C c n for all n sod = C. (6) Let L = D = C thenforn N k we have L L. k Remark. In case of a recursiveely definded sequence + = f( ) you can use the solution x = f(x) to determine the limit only after you show that the sequence converges. Infinite series and their partial sum..3 Infinite series and convergence Infinite series partial sum up to m-th terms S m = Remark. Every sequence can be written as the squence of partial sums of some series. Definition. Aseries converges (or is convergent) to a sum S provided that the limit S = lim m S m exists. Otherwise the series diverges. () ( )n = ( )n diverges (3) n(n+) =. Definition. Geometry series a = a + = r. 3 n Theorem. A geometric series converges to () a =r = 3 n. 3 n = A b n = B then () ( + b n )=A + B c = ca. a r ().555 ().788 (3) Paul and Mary r =( 5 6 ) a = 6 b = m if r <. Divergesif r.

4 4 CHAPTER INFINITE SERIES Remark. A real number is a rational number if and only if its decimal form is periodic. Theorem. (n-th term test) If converges then lim n =. Theorem. ( )n n n diverges n 3n+ both diverge Theorem. If = b n for all n N then either both b n converge or both diverge. Theorem. converges if and only if for any ɛ> there is N(ɛ) such that for all N(ɛ) n<m implies m n a i <ɛ..4 Taylor polynomials and Taylor series Taylor polynomials. m f (k) (a) (x a) k k! is the Taylor polynomial of f center at a up to m-th terms. () x a = m xk () ln xa = m (3) e x a = m ( ) k (x ) k k x k k!. Theorem. (Taylor s formulae) Suppose that f is (n +)-times differentiable on an open interval I contains a thenforeachx I f(x) = n f (k) (a) k! (x a) k + f (n+) (z) (x a)(n+) (n +)! where z is between a and x. Proof. Let F (t) = (x t)n+ (n+)! thenf(x) F (a) = (x a)n+ (n+)! Let G(t) =f(x) n and F (z) = (x z)n n!. f (k) (t) k! (x t) k theng(x) G(a) =f(x) n f (k) (a) k! (x a) k and G (z) = f n+ (z) n! (x z) n. Apply the generalized M.V.T. to F G we get [F (x) F (a)]g (z) =[G(x) G(a)]F (z) to complete the proof. ln. =(.) (.) + (.)3 3 (.)4 4z.ln

5 Definition. Taylor series of f center at a is CHAPTER INFINITE SERIES 5. f (k) (a) (x a) k k! () e x = x k () cos x = (3) e t = (4) sin t = k! ( ) n x n n! sinx = ( ) n t n n! ( ) n n+ t n+ (n+)!. ( ) n x n+ (n+)! From ()() we have the Euler formulae e iθ =cosθ + i sin θ. The following example is to find an approximation π. +x = x + x x 3 + +( ) n x n + ( x)n+ +x +t = t + t 4 t 6 + +( ) n t n + ( t ) n+ +t tan t = t3 3 + t5 5 ( )n t n+ t ( t ) n+ + n + +t dt π 4 = 3 + ( )n + 5 n + + R n with R n n+3..5 Integral Test Theorem. Let be a series with for all n. Then {S m } is an increasing sequence so it converges if and only if it is bounded. Theorem. (Integral test) Suppose that f is a positive decreasing continuous function on [ ) and = f(n) for all n. Then the series and the improper integral f(x)dx are either both convergent or both divergent. Proof. Since m m m+ a + f(x)dx f(x)dx so they are either both bounded or both unbounded. () n n. p

6 6 CHAPTER INFINITE SERIES Theorem. (Remainder estimate) f(x)dx R n n+ n f(x)dx. π n with n+ Rn n..6 Comparison Test and b n are two series with positive terms such that b n for all n then () If b n convegres then also converges. () If divegres then b n also diverges. and b n are two series with positive terms such that b n for all n N for some N and b n convegres then also converges. () (3) n(n+(n+) n n!. Limit comparison test. and b a n are two series with positive terms such that lim n n bn = L for some <L<. Then either both and b n are both convergent or and b n are both divergent. Proof. There exists N such that for all n N L and b n L for n N. bn 3L. That means 3L b n Remark. Suppose that converges and b n C for all n N then converges. () (3) 3n +n n 4 + n n+ln n n + n R n n. Rearrangement and Grouping. Definition. Aseries b n is a rearrangement of a series if there is a one to one onto function σ : N N such that b n = a σ(n) for all n. b n

7 CHAPTER INFINITE SERIES 7 is a convergent series with positive terms then any rearrangement of it will converge to the same limit. Proof. Suppose that b n = a σ(n) be a rearrangement of.letλ(n) =max n {σ(j) it is clear that B n A λ(n) for all n. Hence B n for all n which implies b n. Conversely is also a rearrangement of b n.wealsohave b n. os a convergent series then any grouping of terms will also converge to the same limit. ( )n..7 Alternting series and Absolute convergence Definition. Suppose that then ( )n is an alternting series. Theorem. (Alternating series theorem) Suppose that () + for all n and () lim n =then ( )n converges. Proof. () Since S k+ = S k (a k a k+ )then {S k+ } is decreasing. () Since S k+ = S k +(a k+ a k+ ) then {S k } is increasing. (3) Since S k+ = S k + a k+ thens k+ S k. (4) For l<k S l+ S k+ S k and for l>k S l+ S l S k. (5) Every S k is a lower bound of {S n+ }so{s n+ } converges to S o which is the g.l.b. of {S n+ }. (6) Every S k+ is an upper bound of {S n }so{s n } converges to S e mwhichisthe l.u.b of {S n }. (7) S o S n for all n and S e S n+ for all n and S o S e. (8) From S o S e S k+ S k = a k+ letk we get S o = S e. () ( ) n+ n ( ) n+ n n.

8 8 CHAPTER INFINITE SERIES Theorem. (Remainder estimate) R n +. e! + ( )n 3! n! R n < (n+)!. Definition. If converges we say converges absolutely. If converges but diverges we say converges conditionally. converges then converges. Proof. Let a + n = + a n = (a + n =max( ) (a n =min( )). Then = a + n + a n = a + n a n and a + n a n. If converges then both a+ n and a n converge. Which implies that converges. Remark. If converges conditionally then both a+ n and a n diverge. () ( 3 )n = 3 4 ( 3 )n = 3 (3) cos n n. converges conditionally then for any given real number L thereis a rearrangement b n of such that b n = L. Ratio Test. Theorem. If lim n + = ρ then () ρ< implies converges () ρ> implies diverges (3) ρ =no conclusion. Theorem. If + ρ<for all n N for some N then converges. () (3) Root Test. ( ) n n! n n 3 n n. Theorem. If lim n n = ρ then () ρ< implies converges () ρ> implies diverges (3) ρ =no conclusion. Theorem. If n ρ<for all n N for some N then converges. n+( )n.

9 CHAPTER INFINITE SERIES 9.8 Power series x n. Domain of convergence. The set {x : x n converges}. Theorem. Suppose that x n converges for some x then x n converges absolutely for all {x : x < x }. Proof. Since x n converges x n asn. In particalar there exist N such ( x x )n. So that for n N x n <. Now for fix x with x < x x n = x n N x n N ( x x )n which converges due to x < x. The proof is complete. Remark.The domain of convergence must be one of the following () {} n!xn () R x n n! (3) [ R R] (4) [ R R) (5) ( R R] (6) ( R R) x n R n n x n R n n ( x) n R n n xn R is called the radius of convergence. Theorem. Either lim n + = ρ or lim n n = ρ thenr = ρ. () x n n3 n ( x) n (3) nn x n (4) (n)! ( ) n x n n!. Power series center at c. ( ) n (x 3) n n4 n. (x c) n. Power series representation of function center at a. f(x) = f (k) (a) (x a) k. k!

10 CHAPTER INFINITE SERIES Taylor series if a = is called Maclaurin series. () e x cosh x sinh x e x () Bessel function J (x) = ( ) n x n n (n!) (3) Binomial series ( + x) α =+ α(α )(α ) (α n+) n! x n. Continuity integration and differentiation of power series. Theorem. Power series define a continuous function on the interior of the interval of conveggence. Since the partial sums of a power series are polynomials hence are continuous functions. We will use the following theorem to get the continuity of the power series from its partial sums. We define the distance between two functions fg by f g = l.u.b. x ( RR) f(x) g(x). Definition. A sequence of functions {f k } converges uniformly to a limit function f if for any given ɛ> there is a N(ɛ) such that n N(ɛ) implies that f k f <ɛ.inthiscase we will write lim k f k = f. Theorem. Suppose that sequence of continuous functions {f k } on ( R R) converges uniformly to a limit function f. Thenf is a continuous function on ( R R). Proof. For any a ( R R) since f(x) f(a) f(x) f k (x) + f k (x) f k (a) + f k (a) f(a). For given ɛ> we can fix some large k such that f k f < ɛ 3. And since f k is continuous function there is a δ>such that x a <δimplies f k (x) f k (a) < ɛ 3. Put them together we have f(x) f(a) < ɛ 3 + ɛ 3 + ɛ 3 = ɛ. Proof of the continuity of the power series. For any x ( R R) fix r<rsuch that x [ r r]. Then x n m x n x n r n. m+ m+ As we know from r<r r n converges so the m+ r n goes to zero as m goes to. So x n is continuous on [ r r] and this is true for all <rr. So x n is continuous on ( R R).

11 CHAPTER INFINITE SERIES Theorem. The power series x n its term by trem differentiation power series nx n and its term by term integration power series n+ xn+ have same radius of convergence. Proof. Letr their radius convergence be R R d R i respectively. Since the series x n and x n have same radius R. Theseries n+ xn+ n+ xn+ andx n+ xn have the same radius of convergence R i. Hence R i R and also R R d since x n is the term by term integration series of nx n. In order to show that R = R d we will show that for any x ( R R) nx n converges absolutely. Fix some r such that x <r<rand apply the limit comarison theorem sine n x n = n r n r ( x r )n as n which prove that R = R d and similarly R = R i. Next we will show that integration of power series can be done by term by term integration. Theorem. x t n dt = Proof. For given x ( R R) x m t n dt n+ xn+. n + xn+ = x t n dt m+ x x n dt and last integral is x m+ x n which will goes to zero as m due to the absolute convergence of x n In order to show that differentiation of power series can be done by term by term differentiation we need the following theorem (we will skip the proof). Theorem. Suppose that a sequence of differentiable function {f k } converges uniformly to a limit function f and{f k } also converges to a limit function g. Thenf is differentiable and f = g. d Theorem. dx ( x n )= nx n. Proof. Inorder to apply the theorem above we have to make sure that the derivetives of the partial sums converges uniformly on [ r r] for all r<r. But since R d R thisis true so the proof is complete. Remark. A function defined by a power series is call a real analytic function. Suppose that f(x) = x n thenf (k) () = k!a k. So real analytic function can only be deinned by unique powerseries. () ( x) =( ( x) ) = nxn dt () ln( + x) = x (3) tan x = x (4) sin x = x +t = ( )n xn n dt +t = xn+ ( )n n+ dt. t m+

12 CHAPTER INFINITE SERIES.9 Power series computation () 5 =.5 () A = π sin x π x dx. Algebra of power series. Theorem. Suppose that x n b nx n have radius of convergence R> then () (x n + b n x n )=( x n )+( b nx n ) c nx n =( x n )( b nx n ) where c n = n a jb n j have radius of convergence R. Proof. () is clear we only have to prove (). Let () A = x n B = b nx n d n = n Since a jb n j () à = x n à m = m x n B = b nx n B m = m (3) C = c nx n C m = m c nx n D = d nx n D m = m C m D m Ãm Bm à B that proves c nx n converges absolutely. Also b nx n d nx n. AB m c n x n à B D [ m ] that proves (). tan x cos x =sinx Indeterminate forms. sin x tan () lim x x x ln(+x) ln x () lim x x. Series solution of difrferential equation () y +y = () (x 3)y +y = (3) x y = y x (4) y + y = (5) y xy = (6) xy + y + xy = Bessel equation.