4-1: Comparison Test; Absolute Convergence Theorem; Limit Comparison Test

Prakash Balachandran Department of Mathematics Duke University

February 1, 2010

• Please don’t send me short-term illness reports if you can’t make it to class. You can always make up participation by speaking up next time around! • Guest Lecturer next week [I’ll still be here]. 1. Comparison Test 2. Absolute Convergence Theorem 3. Limit Comparison Test

1 Comparison Test

P∞ • Recall that we’re trying to test when a series k=1 ak converges. This allows us to approximate the inﬁnite sum by the nth partial sum if necessary, or allows us to compute various quantities of interest in probability. ∞ • Before we state the theorem, let’s do a straight forward example. Consider the two series P √1 and k=1 k P∞ 1 k=1 k . We know by the integral test that both these series diverge. √ Let’s take another approach. We know that k ≤ k for k ≥ 1, so that 1 ≤ √1 for k ≥ 1 and k k

n n X 1 X 1 S = ≤ = S0 . n k k n k=1 k=1 Since this is true for all n, taking the limit as n → ∞ we ﬁnd that

∞ ∞ X 1 X 1 ≤ √ . k k=1 k=1 k So, if we knew that the harmonic series diverged, then this “comparison” would imply divergence of ∞ P √1 . k=1 k P∞ 1 P∞ 1 2 3 1 1 Similarly, let’s take the series k=1 k2 and k=1 k3 . Since k ≤ k for k ≥ 1, we have that k3 ≤ k2 for k ≥ 1 so that n n X 1 X 1 S = ≤ = S0 . n k3 k2 n k=1 k=1

1 Since this is true for all n, taking the limit as n → ∞ we ﬁnd that

∞ ∞ X 1 X 1 ≤ . k3 k2 k=1 k=1 P∞ 1 P∞ 1 Thus, if we knew that k=1 k2 converges, then we would know that k=1 k3 is ﬁnite (converges). These ideas essentially embody the proof of:

• Theorem 1 (The Comparison Test) Suppose 0 ≤ am ≤ bm for all m beyond a certain value. Then, P P 1. If bn converges, then an converges. P P 2. If an diverges, then bn diverges. • Examples: P∞ 1 – k=1 n3+1 , ∗ ∞ ∞ 1 1 X 1 X 1 n3 ≤ n3 + 1 ⇒ ≤ ⇒ ≤ . n3 + 1 n3 k3 + 1 k3 k=1 k=1 P∞ 1 P∞ 1 k=1 k3 is a p-series with p = 3, so it converges, and thus so does k=1 k3+1 by the comparison test. P∞ k−1 – k=1 k3+3 , 3 3 k−1 k 1 P∞ 1 ∗ k − 1 ≤ k and k ≤ k + 3 so that k3+3 ≤ k3 = k2 . Again, k=1 k2 is a p-series with p=3, so it converges, and by the comparison test,

∞ ∞ X k − 1 X 1 ≤ k3 + 3 k2 k=1 k=1 converges. P∞ 6k2+1 – k=1 2k3−1 ,

2 2 3 3 3 6k2 6k2+1 ∗ 6k ≤ 6k + 1 and 2k − 1 ≤ 2k so that k = 2k3 ≤ 2k3−1 . By the comparison test, divergence P∞ 1 P∞ 6k2+1 of k=1 k (the harmonic series) implies divergence of k=1 2k3−1 . √ P∞ 5+ j – j=1 1+j2 . √ √ 2 2 5+ j 5+ j P∞ 1 P∞ 1 ∗ j ≤ 1 + j so that 1+j2 ≤ j2 . We know that j=1 j2 and j=1 3 converges, and so the √ j 2 P∞ 5+ j comparison test implies that j=1 1+j2 converges.

2 Absolute Convergence Theorem

• Notice we have an easy consequence of the comparison test: clearly

∞ ∞ X X ak ≤ |ak| ⇒ ak ≤ |ak|. k=1 k=1 P P Thus, if |an| converges, then so does an.

2 P P • Theorem 2 (The Absolute Convergence Theorem) If |an| converges, then so does an. • Examples: P∞ P∞ k – Easy consequence: if k=1 |ak| converges, this means that k=1(−1) ak converges. Examples are P∞ (−1)k k=1 kp for p > 1. P∞ P∞ – Notice that convergence of k=1 |ak| is sufﬁcient but not necessary for convergence of k=1 ak. That P∞ P∞ P∞ is to say, while convergence of k=1 |ak| implies convergence of k=1 ak, convergence of k=1 ak P∞ P∞ need not imply that k=1 |ak| converges. The idea is that due to cancellation, k=1 ak might have a P∞ much better chance of converging than k=1 |ak|. We’ll come back to this on Friday (4-3).

3 Limit Comparison Test

• Theorem 3 (The Limit Comparison Test) Suppose an > 0 and bn > 0 for all n. If a lim n = c n→∞ bn P P where c > 0 then the two series an and bn both converge or diverge. Idea of Proof: Since an → c and c > 0 we can expect bn → 1 . So, by the comparison test, we have bn an c an an ≤ cbn for n large enough and bn ≤ c for n large enough. • Examples: P∞ n2+6 – n=0 n4−2n+3 , n2+6 1 ∗ Let an = n4−2n+3 and bn = n2 . Then

4 2 an n + 6n lim = lim 4 = 1 = c > 0 n→∞ bn k→∞ n − 2n + 3 P P so the limit comparison test implies that an and bn both converge and diverge. Since P∞ 1 P∞ n2+6 n=1 n2 converges, we ﬁnd that n=0 n4−2n+3 converges. P∞ 1 – k=1 sin k , 1 1 ∗ Let ak = sin k and bk = k . Then a sin 1 lim k = lim k = 1 = c > 0. k→∞ k→∞ 1 bk k P P Thus, the limit comparison test implies that an and bn both converge or diverge. Since P∞ 1 P∞ 1 k=1 k diverges, thus so does k=1 sin k . P∞ 1+n n – n=1 3n 1+n n 1 n ∗ Let an = 3n and bn = 3 . Then, a 1 + nn lim n = lim = e = c > 0. n→∞ bn n→∞ n

P∞ 1 n P∞ 1+n n So, n=1 3 converges implies that n=1 3n converges.