6 Series We call a normed space (X, ) a Banach space provided that every Cauchy sequence (x n ) in X converges. For example, R with the norm = is an example of Banach space. Now let (x n ) be a sequence in X. Define a new sequence (s n ) in X by s n = x k, n N. k= The sequence (s n ) is called a series in X and is written as x k or k x k. The n-th term s n of the sequence (s n) is called the nth partial sum and x k is called the kth summand of the series k x k. Definition 6.. The series k x k converges or is convergent if the sequence (s n ) of partial sums converges. The limit lim n s n is called the value of the series k x k and is written as k= x k Finally, k x k diverges or is divergent if the sequence (s n ) diverges in X. Proposition 6. (Necessary condition for convergence). If the series xk converges, then x n 0. Proof. Denoting by s n the nth partial sum of the series x k, we have s n x for some x. Since the result follows. x n = s n+ s n x x = 0, Example 6.3. The series (/k!) converges in R. Its value is /k! = e. Example 6.4. The series /k converges in R. Indeed, if s n = n k= /k, then s n = /k ( + k(k ) = + k ) = + k n <. k= k= So, the sequence (s n ) of partial sums is bounded in R and since it is also increasing, it converges. Example 6.5. The harmonic series /k diverges. Indeed, if s n = n k= /k, then s n s n = n + +... + n n n = showing that (s n ) is not a Cauchy sequence and this implies that k /k is dirvergent. 30 k=
Example 6.6. The geometric series a k, where a R satisfies a <, converges. Indeed, s n = a k = ak+ a and since a <, we have a k+ 0 showing that lim s n = n If a, then k ak diverges. a k = a. Proposition 6.7. Let a k and b k be convergent series in a Banach space X and let α R. Then The series (a k + b k ) converges and k= (a k + b k ) = k= a k + k= b k. The series k= (αa k) converges and k= (αa k) = α k= a k. Theorem 6.8 (Cauchy criterion). For a series x k in a Banach space (X, ) the following conditions are equaivalent. (a) The series x k converges. (b) For every ε > 0 there is N N such that m x k < ε for all m > n N. k=n+ Proof. Let s n = n k= x k be the nth partial sum. Then s m s n = m k=n+ x k if m > n.thus (s n ) is Cauchy in X if and only if the condition (b) holds true. By assumption, (X, ) is a Banach space and the result follows. Theorem 6.9. If a k is a series in R and a k 0 for all k N, then a k converges if and only if the sequence (s n ) of partial sums is bounded. In this case, the series has the value equal to sup n N s n. Proof. Since a k 0 for all k N, the sequence (s n ) of partial sums is increasing. Hence (s n ) converges if and only if (s n ) is bounded. 3
6. Alternating series Definition 6.0. A series a k in R is called alternating if a k and a k+ have opposite signs. Any such series can be written as ± ( ) k a k with a k 0 for all k N. Theorem 6. (Leibniz criterion). Let (a k ) be a decreasing null sequence such that a k 0 for all k N. Then the alternating series ( ) k a k converges in R. Proof. Since s n+ s n = a n+ + a n+ 0, the sequence (s n ) is decreasing. Similarly, s n+3 s n+ = a n+ a n+3 0, implying that (s n+ ) is increasing. Moreover, s n+ s n so that s n+ s and s n 0 for all n. Hence, there are s and t such that s n s and s n+ t. Also t s = lim n (s n+ s n ) = lim n a n+ = 0 implying that t = s. Take ε > 0. Then there are N,N N such that s n s < ε, n N and s n+ s < ε, n + N. Thus, s n s < ε for all n > max{n,n }. Example 6.. The assumption that (a n ) is decreasing is necessary. For example, set a n = n and a n = (n ). Then a n = (n ) n so that 0 a n n showing that a n 0. Since a n < a n for n, the sequence (a n ) is not decreasing. However, the series ( ) n a n diverges. Arguing by contradiction we assume that the sequence (s n ) of partial sums converges. We have s n = k= = ( ) k a k = k= = A n + B n, ( )a k + k= (k ) + k 3 k= k= a k
where A n = n k= and B (k ) n = n k= k = n k= k. Since the series k converges, the sequence (A (k ) n ) converges. Consequently, the sequence B n = s n +A n converges. But this contradicts the fact that the series k k diverges. 6. Absolute convergence Definition 6.3. The series x k in a Banach space (X, ) converges absolutely or is absolutely convergent if the series x k is convergent in R. The series x k is conditionally convergent if x k converges but x k diverges. Example 6.4. The alternating harmonic series ( ) k /k converges in view of Leibniz criterion, however, the series of absolute values /k diverges. Hence the series ( ) k /k converges conditionally. Proposition 6.5. Every absolutely convergent series converges. Proof. The proposition is a consequence of Cauchy criterion and the triangle inequality. Indeed, m m x k x k. k=n+ k=n+ 6.3 Comparison, Root, and Ratio Tests From now on we only consider real sequences, however, you are invited to figure out which of the results can be stated in the full generality of Banach spaces. Proposition 6.6 (Comparison test). Let a k and b n be two sequences of nonnegative terms and assume that Then the following holds true, a n b n for all n. (a) If b k converges, then so is a k. (b) If a k diverges, then so is b k. 33
Example 6.7. Let m. Then the series /k m converges absolutely. Recall from Example 6.4 that the series converges. Since for m, k we have k m k for all k, it follows from Proposition 6.6 that the series /k m is indeed convergent. Proposition 6.8 (Limit comparison test). Let a k and b n be two sequences of positive terms and assume that a n b n L (Clearly, L [0, ) or L = ). Then the following holds true, (a) If L (0, ), then the convergence of b k implies the convergence of ak and the divergence of a k implies the divergence of b k. (b) Assume that L = 0. Then a k converges if b k converges. (c) Assume that L =. Then a k diverges, if b k diverges. Proof. We only prove (a). If L (0, ), then there is N so that L/ = L L/ < a n /b n < L+L/ = 3L/ for n N. So, (L/)b n < a n < (3L/)b n for n N and the result follows from Proposition 6.6. Theorem 6.9 (Root test). Let a k be a series of nonnegative real numbers and let α = lim sup n a n. Then a k converges absolutely if α <. a k diverges if α >. If α =, then both convergence and divergence of a k are possible. Theorem 6.0 (Ratio test). Let a k be a series of real numbers such that a k 0 for all k N. Then If lim a k+ a <, k k then a k converges absolutely. 34
If then a k diverges. lim a k+ a >, k k a If lim k+ k a k =, then both convergence and divergence of ak are possible. Proposition 6. (Cauchy s condensation test). Let (a k ) be a sequence of nonnegative and decreasing terms. Then a k converges if and only if the series k a k converges. Proof. Denote by s n = n k= and t n = n k= k a k. Then, if n < m, we have s n a + (a + a 3 ) +... + (a m +... + a m+ ) a + a +... + m a m = t m. If k a k converges, then the sequence (t m ) is bounded and above estimate shows that (s n ) is bounded. Hence a k converges. For the converse, note that if n > m, then s n a + a + (a 3 + a 4 ) +... + (a m + +... + a m) a + a + a 4 +... + m a m = t m. This shows that if a k converges, then the sequence (t m ) is bounded so that k a k converges. Example 6.. Consider the series n with p > 0. Then the condensation test implies that p n converges if and only if n p = ( ) n np p converges. We already know that the geometric series ( ) n converges p if and only if <, i.e., p >. p The next theorem is a generalization of the condensation test. Theorem 6.3 (Schlömlich theorem). Let (g k ) be a strictly increasing sequence of positive integers such that g k+ g k C(g k g k ) for some C > 0 and all k N. If (a k ) is a decreasing sequence of nonnegative numbers, then ak converges if and only if (g k+ g k )a gk converges. 35
Proof. The proof is similar to the proof of the condensation test. Let (S n ) be the sequence of the partial sums of the series a k and (T n ) the sequence of the partial sums of the series (g k+ g k )a gk. Now, if n < g k, then S n S gk (a +... + a g ) + (a g +... + a g ) +... + (a gk +... + a gk+ ) (a +... + a g ) + (g g )a g +... + (g k+ g k )a gk (a +... + a g ) + T k. Consequently, if the sequence (T k ) converges and hence is bounded, then the sequence (S n ) is bounded so that a k converges. Conversely, assume that a k converges. If n > g k, then CS n CS gk C(a g + +... + a g ) +... + C(a gk + +... + a gk C(g g )a g +... + C(g k g k )a gk (g 3 g )a g +... + (g k+ g k )a gk = T k (g g )a g. Consequently, the sequence (T k ) is bounded and the series (g k+ g k )a gk converges. Example 6.4. Take g k = 3 k. Then the sequence (g k ) satisfies the assumptions of the above theorem and g k+ g k = 3 k+ 3 k = 3 k. So, the following holds true. Assume that (a k ) is a decreasing sequence of nonnegative numbers, then ak converges if and only if 3 k a 3 k converges. Now consider the series. Since the sequence ( ) is decreasing 3 ln n 3 ln n and converges to 0, converges if and only if 3 n 3 ln n 3 = 3 n = ( ) ln3n 3 n ln3 n 3 converges. Since ln 3 >, it follows that 3 ln 3 > and ln3 < so that the series ( ) n 3 ln3 3 converges. Hence converges. ln3 3 ln n Consider g k = k, then the sequence (g k ) satisfies the assumptions of the theorem. Then the following is true. If (a k ) is a decreasing sequence of nonnegative numbers, then ak converges if and only if ka k converges. Indeed, since (k + ) k ) = k +, it follows from Theorem?? that a k converges if and only if (k +)a k converges. Since ka k (k +)a k 36
3ka k, the comparison test shows that (k +)a k converges if and only if kak converges. showing our claim. Now consider the series. Since n the sequence / n ) is decreasing and converges to 0, the series n converges if and only if the series n n = n converges. Using the n ratio test, we see that n converges and so n converges. n 6.4 The Dirichlet and Abel Test We start with the following lemma. Lemma 6.5 (Abels lemma). Let (x n ) and (y n ) be two sequences of real numbers. Let (s n ) be a sequence of partial sums of (y n ) and s 0 = 0. Then m k=n+ x k y k = (x m s m x n+ s n ) + Proof. Note that y k = s k s k. So, m k=n+ x k y k = m k=n+ x k (s k s k ) m k=n+ (x k x k+ )s k. = x n+ (s n+ s n ) + x n+ (s n+ s n+ ) +... + x m (s m s m ) = x m (x m x n+ s n ) + s n+ (x n+ x n+ ) +... + s m (x m x m ) = x m (x m x n+ s n ) + m k=n+ x k (s k s k ) We use this lemma to obtain a test for convergence of the series x k y k. Theorem 6.6 (Dirichlet s test). Suppose that (x n ) is a decreasing sequence such that x n 0 and the sequence (s n ) of the partial sums of (y n ) is bounded. Then the series x k y k converges. Proof. Since (s n ) is bounded, s n C for all n. Since (x k ) is decreasing, x k x k+ 0. Thus, by the above lemma, m k=n+ x k y k x m s m + x n+ s n + m k=n+ (x k x k+ ) s k ((x n+ + x m ) + (x n+ x m ))C = x n+ C. 37
Example 6.7. Consider ( ) n x n where x n is a decreasing sequence of nonnegative numbers converging to 0. Then the series converges in view of alternating series test. However, this also follows from Dirichlet s test. Indeed, let y n = ( ) n. Then s n = n k= b k = n k= ( )k is equal to if n is odd and is equal to 0 if n is even. Hence the sequence (s n ) is bounded, and Dirichlet s test implies that the series ( ) n a n converges. Example 6.8. One can show, using for example the principle of mathematical induction, that sin ϑ (cos ϑ + cos ϑ +... + cos nϑ) = cos (n + )ϑ Hence if ϑ πk for all k Z, then Similarly, cos ϑ + cos ϑ +... + cos nϑ = sin ϑ which implies that cos (n+)ϑ sin nϑ sin ϑ (sin ϑ + sin ϑ +... + sin nϑ) = sin (n + )ϑ sin ϑ + sinϑ +... + sin nϑ sin ϑ sin nϑ. sin ϑ. sin nϑ if ϑ πk for all k Z. Hence, in view of Dirichlet s test, one has the following result. Corollary 6.9. If (x n ) is a decreasing sequence converging to 0 and ϑ πk for all k Z, then the series x n cos nϑ and x n sinnϑ converge. Theorem 6.30 (Abel s test). If (x n ) is a monotone convergent sequence and the series y k converges, then x k y k converges. Proof. Without loss of generality, we may assume that (x n ) is decreasing. If x = lim x n, set u n = x n x is decreasing and converging to 0. Thus, x n = x + u n and x n y n = xy n + u n y n. Now note that the series xy k converges since y k converges and the series u n y n converges in view of the Dirichlet s test. Hence, the series x n y n converges. 38
6.5 Cauchy Product Definition 6.3. The Cauchy product of two series n 0 a n and n 0 b n is the series n 0 c n where c n = a k b n k. n 0. Example 6.3. Let n 0 c n be the Cauchy product of the alternating series ( ) n n+ with itself. That is, c n = ( ) k k + ( ) n k = ( ) n (n k) + (k + )(n k). Note that, in view of the alternating series test, the series ( ) n n+ converges conditionally. However, the Cauchy product diverges. To see this we show that the sequence (c n ) does not converge to 0. Indeed, we have ( n ) ( n ) (n k + )(k + ) = + k and hence (n k + )(k + ) = (n/ + ) (n/ k) n/ + ) = n +. Using this we find that c n = (k + )(n k) n/ + = (n + ) n + showing that the sequence (c n ) does not converge to 0. Theorem 6.33 (Mertens theorem). Assume that n 0 converges absolutely, n=0 a n = A, n=0 b n = B, and n 0 is the Cauchy product of n 0 a n and n 0 b n. Then n 0 c n converges and n=0 c n = AB. Proof. Define A n = a n, B n = b n, C n = c n, β n = B n B. 39
Then C n = c 0 + c +... + c n = a 0 b 0 + (a 0 b + a b 0 ) +... + (a 0 b n + a b n +... + a n b 0 ) = a 0 (b 0 + b +... + b n ) + a (b 0 + b +... + b n ) +... + a n b 0 = a 0 B n + a B n +... + a n B 0 = a 0 (β n + B) + a (β n + B) +... + a n (β 0 + B) = (a 0 +... + a n )B + (a 0 β n + a β n +... + a n β n ) = A n B + γ n where we have abbreviated γ n = a 0 β n + a β n +... + a n β n. Since A n B AB, it suffices to show that γ n 0. In order to show that γ n 0, pick ε > 0. Then, since β n 0, there is N such that Let  = β n < ε for all n N. a n and β n M. n=0 (The series n 0 a n converges and (β n ) is bounded since it converges.) Then γ n = a 0 β n + a β n +... + a n β 0 a 0 β n + a β n +... + a n N β N + a n N+ β N + a n N+ β N +... + a n β 0 ε  + M ( a n N+ + a n N+ +... + a n ) Keeping N fixed and letting n, we get lim sup γ n εâ since a n 0. Since ε > 0 was arbitrary, we conclude that lim sup γ n = 0 which implies that lim γ n = 0 as claimed Theorem 6.34 (Abel s theorem). If n=0 a n = A, n=0 b n = B, and n=0 c n = C, where c n = n a kb n k for all n 0, then C = AB. 40
6.6 Double Sums Let (X, ) be a normed space, let (a ij ) (i,j) N N be a double sequence in X, and the double series a ij. We have a function a : N N N, (j,k) a jk. Since N N is countable, there is a bijection σ : N N N. If σ is such bijection, the series n a σ(n) is called an ordering of the double series a jk. If we fix j N (or k N ), then the series a jk is called the jth row series (or the kth column series). If every row series (or column series) converges, then we may consider the series of row sums j ( a jk) (or the series of column sums k ( j=0 a jk)). Finally, the double series a jk is summable if sup n a jk <. Theorem 6.35. Let a jk be a summable double series. Then j,k (a) Every ordering n a σ(n) of a jk is converges absolutely to a value s X which is independent of σ. (b) The series of row sums j ( a jk) and the series of column sums k ( j=0 a jk) converge absolutely and ( j=0 a jk ) = k ( a jk ). j=0 To prove (a) we will need the following result. Let a n be a series and let σ : N N be a bijection. Define b n = a σ(n). Then the series b n is called a rearrangement of of a n. Proposition 6.36. Assume that a n converges absolutely. Then any rearrangement of a n converges absolutely. (They all converge to the same value). Proof. Assume that a n converges absolutely. Take a bijection σ : N N and consider b n with b n = a σ(n). Denote by (A n ) and (B n ) the sequences of partial sums of a n and b n, respectively. Since a n converges absolutely, given ε > 0, there is N such that m a k < ε () k=n 4
for all m > n N. Let K be such that {,...,N} {σ(),...,σ(k)}. Take n > K and consider the difference A n B n Note that the terms a,...,a N appear in both partial sums A n and B n. So, by () A n B n < ε. Hence lim A n = lim B n, proving that b n converges to the same value as an. The same argument in which A n and B n are partial sums of a n and b n, respectively, shows that b n converges absolutely. Proof of Theorem 6.35. (a) Abbreviate M = n n j, a jk. Let σ : N N N be a bijection and let N N. Then there is some K N such that {σ(0),...,σ(n)} {(0,0),(,0),...,(K,0),...,(K,K)}. () Together with the summability of a jk this imples N aσ(n) n=0 K j, a jk M. This implies that n a σ(n) converges absolutely. Let δ : N N N be another bijection. Then α = σ δ : N N is a bijection. Set y m = x σ(m). Then b α(n) = a σ(α(n)) = a δ(n), n N. Thus n a δ(n) is a rearrangement of n a σ(n). Since n a σ(n) converges absolutely, the series n a δ(n) also converges absolutely. (b) The row series a jk, j N, and the column series j=0 a jk, k N, converge absolutely. This follows from the summability of a jk. So, the series of row series j ( a jk) and the series of column series k ( j=0 a jk). To see that these series converge absolutely, consider the following inequalities, l m l m m a jk a jk a jk M, l m. j=0 j=0 Taking the limit m, we get l j=0 a jk M, l N. This proves the absolute convergence of the series of row series j ( a jk). A similar argument proves the absolute convergence of the series of column series k ( j=0 a jk). j,k 4
Now let σ : N N N be a bijection and let s = n a σ(n). Then for every ε > 0, there is N N such that n=n+ aσ(n) <. Also there is some K N such that () is satisfied. Then l m N a jk a σ(n) a σ(n) < ε/, l,m K. j=0 n=0 n=n+ Taking the limit m and l, we get N a jk a σ(n) j=0 n=0 Applying the triangle inequality to N s a σ(n) n=0 n=n+ n=n+ aσ(n) ε/. aσ(n) < ε/, we get a jk s ε. j=0 A similar argument shows that the value of k ( j=0 a jk) is also s. 43