Similar to sequence, note that a series converges if and only if its tail converges, that is, r 1 r ( 1 < r < 1), ( 1) k k. r k =

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1 Infinite Series We say an infinite series a k converges to s if its sequence of initial sums converges to s, that is, lim( n a k : n N) = s. Similar to sequence, note that a series converges if and only if its tail converges, that is, n= a n converges if and only if k=n a k converges for some number N. Some examples: k 2 = π2 6, converge. While diverge. k! = e, r k = k (the Harmonic series), r r ( < r < ), sin k, k ln(k + ), ( ) k k = ln 2 Cauchy Criterion Recall that a sequence converges if and only if it is a Cauchy sequence, similarly we know a series converges if and only if it satisfies the Cauchy criterion, that is given any ε > 0, there exists N N such that n-th term test a n+ + + a m < ε for all m > n N. If n= a n converges, then lim(a n ) = 0. However, its converse is not true. For example, lim(/n) = 0 but n= n diverges. Comparison test for nonnegative series If there exists N N such that 0 a n b n for all n N, then n= b n < implies n= a n <. Hence (/n p ) converges for < p < and diverges for 0 < p.

2 2 Limit comparison test for positive series If a n, b n > 0 for all n and lim(a n /b n ) = L > 0, then n= a n < if and only if n= b n <. However, if L = 0, then n= a n < if n= b n <. Question: do we need a n > 0? Alternating series test If (a n ) is a monotone decreasing sequence that converges to 0, then n= ( ) n a n converges. Moreover, the following is easy to verify. Suppose both series n= a n and n= b n converge, then n= (a n + b n ) = n= a n + n= b n. Moreover, n= ca n = c n= a n. However, in general, n= (a n b n ) n= a n n= b n. Indeed, the series n= (a n b n ) may not converge. Section 9. A series n= a n is said to be absolute convergent if n= a n <. Note that a series converges whenever it converges absolutely. Theorem 9..3 If a series converges, then any series obtained from it by grouping the terms is also convergent to the same value. The theorem follows immediately from the definition and the fact that any subsequence of a convergent sequence converges to the same limit. However, the converse of the theorem is not true. Rearrangement of series A series y k is said to be a rearrangement of x k if there is a bijection f : N N such that y k = x f(k) for all k N.

3 Theorem 9..5 If x k converges absolutely, then any rearrangement of it converges to the same value. Proof Let f : N N be any bijection and let s = x k. Then given any ε > 0, 3 there exists N N such that N x k s < ε/2 and x k < ε/2 for all m > n N. k=n+ Let N 0 = max{f (),, f (N)}. Then n N N n x f(k) s x k s + x k x f(k). But if n > N 0, then N n x k x f(k) x k with m = max{n, f(),, f(n)}. N+ Thus, if n > N 0, then n x f(k) s < ε. 9.2 Tests for Absolute Convergence Limit Comparison test root test and ratio test (this is different from your textbook) If lim( a n /n ) <, then n= a n converges absolutely. If lim( a n /n ) >, then n= a n diverges. If lim( a n+ /a n ) <, then n= a n converges absolutely. If lim( a n+ /a n ) >, then n= a n diverges. A useful fact lim( a n+ /a n ) lim( a n /n ) lim( a n /n ) lim( a n+ /a n ). In particular, if lim( a n+ /a n ) exists, then it is equal to lim( a n /n ). Example: even though we have learned that lim(n /n ) =, however, its computation is quite complicated. On the other hand, it is quite easy to compute lim((n + )/n).

4 4 Thus, even though root test is more powerful, limits of ratio are usually easier to compute. Note that lim n (/n 2 ) /n = lim n (/n) /n =, while (/n 2 ) < and (/n) =. Exercise 3.7: -6, 8-5. (4th ed: 6-5 become 9-8) 9.: 2,5-, :,2,4 a,b,f,7,4. 9.3:, 3-8, 3-5. (Note that 4th ed and 3rd ed are the same for 9. to 9.3. Homework6 to be collected 9.:,3, 2, 9.3: 2,9-2. Dirichlet s test and Abel s test Abel s Lemma Let (a n ) and (b n ) be sequences and let S n = n b k be the sequence of partial sums with S 0 = 0. Then k=n+ for m > n, m, n N. Proof LHS = a k b k = a m S m a n+ S n + = k=n+ k=n+ a k S k a k S k k=n+ m k=n = a m S m a n+ S n + m k=n+ a k S k a k+ S k m k=n+ (a k a k+ )S k (a k a k+ )S k Dirichlet s Test Let (a n ) be a decreasing sequence of real numbers that converges to 0 and there exists an M > 0 such that N b k M for N N. Then the series a k b k converges.

5 Abel s Test Let (a n ) be a convergent monotone sequence and let b k converges. Then the series a k b k converges. 5 Examples () (2) (3) Alternating series test. (4) a n 0, a n sin nx. (5) (cos(πk/2 ) /k (6) ( ) k+ log k/k (7) 0 if k is odd ( ) k+ a k /k, a k = if k is even (8) ( + k ( ) k k (9) (0) sin(kπ/2) k () sin(k + k )π (2) 0 sin( + π k ) ( )k log k

6 6 double series, Fubini s theorem and Tonelli s theorem We will call a(i, j) a double series. i,j= It can be used to represent (A) lim n i,j n a(i, j). However, to avoid some unpleasant surprises, we will often assume a stronger limit exists: n a(i, j) = lim a(i, j) = S, n,m i,j= i= j= that is, given any ε > 0, there exists N N such that n a(i, j) S < ε for n, m N. i= j= Of course, when a(i, j) 0, then the above is equivalent to the preceding definition (A). Indeed, one could interpret the first one as a subsequence of the second one. Alternatively, the 2nd one allow various grouping of series, while the st one is a specific grouping of series. Tonelli s theorem If a(i, j) 0 for all i, j, then a(i, j) = a(i, j) = a(i, j) i,j= i= j= j= i= ( ). Note that it is possible that they are all equal to. Next, we also have Fubini s theorem If a(i, j) <, then ( ) holds. counterexamples: i,j= Let a(, ) = a(2, 2) =, a(, 2) = a(2, ) =, a(3, 3) = a(4, 4) =, a(3, 4) = a(4, 3) =, a(5, 5) = a(6, 6) =, a(5, 6) = a(6, 5) =. Continue the pattern. Moreover, we will also let a(i, j) = 0 otherwise. Note that we have chosen such that j= a(i, j) = 0 for all i and i= a(i, j) = 0 for all j. It is then clear that

7 7 a(i, j) = a(i, j) = 0. However, the double series does not converge, i.e., i= j= i= j= the limit lim a(i, j) does not exist. n,m i n, j m Indeed, the above two theorems are special cases of integration theory when the measures involved are just counting measure. Infinite sum over uncountable index set We would also like to define what it means by α I a(α). In general, we do not know how to define it. However, there is a natural definition when a(α) 0. We can define a(α) = sup{ a(α) : A is a finite subset of I}. α I α A Clearly, this definition coincide with the usual definition when I is countable. Again, we can definite what we mean by absolute convergence. We say the infinite sum converges absolutely if a(α) <. α I First, we have a quick fact: the set A 0 = {α : a(α) 0} is countable if the sum converges absolutely. Proof: it suffices to show that for each n N, the set A n = {α : a(α) > /n} is finite and hence countable. However, it is clear. For simplicity, let us choose a countably infinite subset S of A that contains A 0. Thus, in the case where the infinite sum converges absolutely, we can define α A a(α) as i= a(f(i)) where f : N S is any bijection. This is well defined as we know the sum of a absolutely convergent series is independent of rearrangement.