CHAPTER 2 INFINITE SUMS (SERIES) Lecture Notes PART We extend now the notion of a finite sum Σ n k= a k to an INFINITE SUM which we write as Σ n= a n as follows. For a given a sequence {a n } n N {0}, i.e the sequence a, a 2, a 3,...a n,... we consider a following (infinite) sequence S = a, S 2 = a + a 2,..., S n = Σ n k= a k, S n+ = Σ n+ k= a k,... and define the infinite sum as follows. DEFINITION If the limit of the sequence {S n = Σ n k= a k} exists we call it an INFINITE SUM of the sequence S n and write it as Σ n= a n = lim S n = lim Σn k= a k. The sequence {S n } is called its sequence of partial sums. DEFINITION 2 If the limit lim S n exists and is finite, i.e. lim S n = S, then we say that the infinite sum Σ n= a n CONVERGES to S and we write otherwise the infinite sum DIVERGES. Σ n= a n = lim Σn k= a k = S, In a case that lim S n exists and is infinite, i.e. lim S n =, then we say that the infinite sum Σ n= a n DIVERGES to and we write Σ n= a n =.
In a case that lim S n does not exist we say that the infinite sum Σ n= a n DIVERGES. OBSERVATION In a case when all elements of the sequence {a n } n N {0} are equal 0 starting from a certain k the infinite sum becomes a finite sum, hence the infinite sum is a generalization of the finite one, and this is why we keep the similar notation. EXAMPLE The infinite sum of a geometric sequence a n = x k for x 0, i.e. converges if and only if x < because Σ n= x n hence x( x n ) lim = lim x Σ n k= x k = S n = x xn+ x, and x x ( xn ) = Σ n=x k = x x. x x iff x <, EXAMPLE 2 The series Σ n= DIVERGES to as S n = Σ n k= = n and lim S n = lim n =. EXAMPLE 3 The infinite sum Σ n= ( ) n DIVERGES. EXAMPLE 4 The infinite sum Σ n=0 (k+)(k+2) CONVERGES to ; i.e. Σ n=0 (k + )(k + 2) =. Proof: first we evaluate S n = Σ n k= (k+)(k+2) as follows. S n = Σ n k=0 (k + )(k + 2) = Σn k=k 2 = x + n+ 0 = n + 2 + and lim S n = lim n + 2 + =. 2
DEFINITION 3 For any infinite sum (series) a series is called its n-th REMINDER. Σ n=a n r n = Σ m=n+ a m FACT If Σ n=a n converges, then so does its n-th REMINDER r n = Σ m=n+ a m. Proof: first, observe that if Σ n=a n converges, then for any value on n so does r n = Σ m=n+ a m because So we get r n = lim (a n+ +... + a n+k ) = lim S n+k S n = Σ m=a m S n. lim r n = Σ m=a m lim S n = Σ m=s m Σ n=a n = S S = 0. General Properties of Infinite Sums THEOREM If the infinite sum Σ n=a n converges, then Proof: observe that a n = S n S n and hence lim a n = 0. lim a n = lim S n T he lim S n = 0, as lim S n = lim S n. REMARK The reverse statement to the theorem If lim a n = 0. then Σ n=a n converges is not always true. There are infinite sums with the term converging to zero that are not convergent. 3
EXAMPLE 5 The infinite HARMONIC sum H = Σ n= n DIVERGES to, i.e. Σ n= n = but lim n = 0. DEFINITION 4 Infinite sum Σ n=a n is BOUNDED if its sequence of partial sums S n = Σ n k= a k is BOUNDED; i.e. there is a number M such that S n < M, for all n, n N. FACT 2 Every convergent infinite sum is bounded. THEOREM 2 If the infinite sums Σ n=a n, Σ n=b n CONVERGE, then the following properties hold. Σ n=(a n + b n ) = Σ n=a n + Σ n=b n, Σ n=ca n = cσ n=a n, c R. Alternating Infinite Sums. Abel Theorem DEFINITION 5 An infinite sum Σ n=( ) n+ a n, for a n 0 is called ALTERNATING infinite sum (alternating series). EXAMPLE 6 Consider If we group the terms in pairs, we get Σ n=( ) n+ = + +... ( ) + ( ) +... = 0 but if we start the pairing one step later, we get ( ) ( )... = 0 0 0... =. 4
It shows that grouping terms in a case of infinite sum can lead to inconsistencies (contrary to the finite case). Look also example on page 59. We need to develop some strict criteria for manipulations and convergence/divergence of alternating series. THEOREM 3 The alternating infinite sum Σ n=( ) n+ a n, (a n 0) such that a a 2 a 3... and lim a n = 0 always CONVERGES. Moreover, its partial sums S n = Σ n k= ( )n+ a n fulfil the condition S 2n Σ n=( ) n+ a n S 2n+. Proof: observe that the sequence of S 2n is increasing as S 2(n+) = S 2n+2 = S 2n +(a 2n+ a 2n+2, and a 2n+ a 2n+2 0, i.e.s 2n+2 S 2n. The sequence of S 2n is also bounded as S 2n = a ((a 2 a 3 ) + (a 4 a 5 ) +...a 2n ) a. We know that any bounded and increasing sequence is is convergent, so we proved that S 2n converges. Let denote lim S 2n = g. To prove that Σ n=( ) n+ a n = lim S n converges we have to show now that lim S 2n+ = g. Observe that S 2n+ = S 2n + a 2n+2 and we get lim S 2n+ = lim S 2n + lim a 2n+2 = g as we assumed that lim a n = 0. We proved that the sequence {S 2n } is increasing, in a similar way we prove that the sequence {S 2n+ } is decreasing. Hence S 2n lim S 2n = g = Σ n=( ) n+ a n and S 2n+ lim S 2n+ = g = Σ n=( ) n+ a n, i.e S 2n Σ n=( ) n+ a n S 2n+. EXAMPLE 7 Consider the INHARMONIC series (infinite sum) AH = Σ n= ( ) n+ n = 2 + 3 4... Observe that a n = n, and n n+ i.e. a n a n+ for all n, so the assumptions of the theorem 3 are fulfilled for AH and hence AH converges. In fact, it is proved (by analytical methods, not ours) that AH = Σ n=( ) n+ = ln 2. n 5
EXAMPLE 8 A series (infinite sum) Σ n=0( ) n 2n + = 3 + 5 7 + 9... converges, by Theorem 3 (proof similar to the one in the example 7). It also is proved that Σ n=0( ) n 2n + = π 4. THEOREM 4 (ABEL Theorem) IF a sequence {a n } fulfils the assumptions of the theorem 3 i.e. a a 2 a 3... and lim a n = 0 and an infinite sum (converging or diverging) Σ n=b n is bounded, THEN the infinite sum Σ n=a n b n always converges. Observe that Theorem 3 is a special case of theorem 4 when b n = ( ) n+. Convergence of Infinite Sums with Positive Terms We consider now infinite sums with all its terms being positive real numbers, i.e. S = Σ n=a n, for a n 0, a n R. Observe that if all a n 0, then the sequence {S n } of partial sums S n = Σ n k= a k is increasing; i.e. S S 2... S n... and hence the lim S n exists and is finite or is. This proves the following theorem. THEOREM 5 The infinite sum always converges, or diverges to. S = Σ n=a n, for a n 0, a n R 6
THEOREM 6 (Comparing the series) Let Σ n=a n be an infinite sum and {b n } be a sequence such that for all n, 0 b n a n. If the infinite sum Σ n=a n converges, then Σ n=b n also converges and Σ n=b n Σ n=a n. Proof: Denote S n = Σ n k=a k, T n = Σ n k=b k. As 0 b n a n we get that also S n T n. But S n lim S n = Σ n=a n so also T n Σ n=a n = S. The inequality T n Σ n=a n = S means that the sequence {T n } is a bounded sequence with positive terms, hence by theorem 5, it converges. By the assumption that Σ n=a n we get that Σ n=a n = lim Σn k=a k = lim S n = S. We just proved that T n = Σ n k= b k converges, i.e. lim T n = T = Σ n=b n. But also we proved that S n T n, hence lim S n lim T n what means that Σ n=b n Σ n=a n. EXAMPLE 9 Let s use Theorem 5 to prove that the series Σ n= (n + ) 2 converges. We prove by analytical methods that it converges to π2 6, here we prove only that it does converge. First observe that the series below converges to, i.e. Σ n= n(n + ) =. 7
Consider S n = 2 + 2 3...+ n(n + ) = ( 2 )+( 2 3 )+...( n n + ) = n + so we get Σ n= n(n + ) = lim S n = lim ( n + ) =. Now we observe (easy to prove) that 2 2 2, 3 2 3,... (n + ) 2 n(n + ),... i.e. we proved that all assumptions of Theorem 5 hold, hence Σ n= converges and moreover Σ n= (n + ) 2 Σ n= n(n + ). (n+) 2 THEOREM 7 (D Alambert s Criterium) Any series with all its terms being positive real numbers, i.e. Σ n=a n, for a n 0, a n R converges if the following condition holds: a n+ lim <. a n a Proof: let h be any number such that lim n+ a n < h <. It means that there is k such that for any n k we have a n+ a n < h, i.e. a n+ < ha n and a k+ < a k h, a k+2 = a k+ h < a k h 2, a k+3 = a k+2 h < a k h 3,... i.e. all terms a n of Σ n=k a n are smaller then the terms of a converging (as 0 < h < ) geometric series Σ n=0a k h n = a k + a k h + a k h 2 +... By Theorem 5 the series Σ n=a n must converge. THEOREM 8 (Cauchy s Criterium) Any series with all its terms being positive real numbers, i.e. Σ n=a n, for a n 0, a n R converges if the following condition holds: lim n an <. 8
Proof: we carry the proof in a similar way as the proof of theorem 6. Let h be any number such that lim n a n < h <. So it means that there is k such that for any n k we have n a n < h, i.e. a n < h n. This means that all terms a n of Σ n=k a n are smaller then the terms of a converging (as 0 < h < ) geometric series Σ n=k hn = h k + h k+ + h k+2 +... By Theorem 5 the series Σ n=a n must converge. THEOREM 9 (Divergence Criteria Any series with all its terms being positive real numbers, i.e. Σ n=a n, for a n 0, a n R diverges if a n+ lim > or lim n an > a n a Proof: observe that if lim n+ a n >, then for sufficiently large n we have that a n+ >, and hence a n+ > a n. a n This means that the limit of the sequence {a n } can t be 0. By theorem we get that Σ n=a n diverges. Similarly, if lim n a n >, then then for sufficiently large n we have that n an > and hence a n >, what means that the limit of the sequence {a n } can t be 0. By theorem we get that Σ n=a n diverges. REMARK 2 It can happen that for a certain infinite sum Σ n=a n ) a n+ lim = = lim n an. a n In this case our Divergence Criteria do not decide whether the infinite sum converges or diverges. In this case we say that the infinite sum DOES NOT REACT on the criteria. EXAMPLE 0 The Harmonic series H = Σ n= n does not react on D Alambert s Criterium (Theorem 7) because lim n n + = lim ( + n ) =. 9
EXAMPLE The series from example 9 Σ n= (n + ) 2 does not react on D Alambert s Criterium (Theorem 7) because (n + ) 2 lim (n + 2) 2 = lim n 2 + 2n + n 2 + 4n + = lim + 2 n + n 2 + 4 n + 4 n 2 =. REMARK 3 Both series Σ n= n and Σ n= (n+) do not react on D Alambert s, 2 but first in divergent and the second is convergent. There are more criteria for convergence, most known are Kumer s criterium and Raabe criterium. EXAMPLE 2 The series converges for c > 0. Σ c n n= n! Proof: Use D Alambert Criterium. a n+ a n = cn+ c n n! (n + )! = c n + and EXAMPLE 3 The series converges. a n+ c lim = lim a n n + = 0 <. n! Σ n= n n Proof: Use D Alambert Criterium. a n+ a n = n!(n + ) nn (n + ) n+ n! = (n+) n n (n + ) n+ = a n+ = a n (n + )n n (n + ) n (n + ) = ( n n + )n = ( + n )n 0