Mathematics 116 Term Test Or Test for Divergence to Apply the Term Test

Mathematics 116 Term Test Or Test for Divergence to Apply the Term Test

Mathematics 116 Term Test or Test for Divergence To apply the Term Test (for Divergence): ∞ Testing an for convergence or divergence. X Investigate lim an. n →∞ ∞ If lim an = 0, conclude that an diverges. n →∞ 6 X∞ If lim an = 0, conclude that an has a chance to converge but further testing is needed to n →∞ decide whether it does or not. TheX series an might converge or it might diverge. In sum, the Term Test give no information in this case.X A formal look at why the Term Test is legitimate. Theorem. If lim Sn = L for some number L, then lim an = 0. In other words, if the partial-sum n n sequence converges,→∞ then the term sequence must converge→∞ to 0. Proof. The idea behind the proof is to find a way to write an by using the partial-sum sequence. Notice that an = Sn Sn 1. Now suppose that lim Sn = L (where L is a number, not or − n − →∞ ∞ ). Then lim Sn 1 = L, as well. Use the appropriate limit theorem for sequences we can write n − −∞ →∞ lim an = lim (Sn Sn 1) = lim Sn lim Sn 1 = L L = 0. n n − n n − →∞ →∞ − →∞ − →∞ − By what’s often called contraposition, we can reformulate this theorem in a different way. Corollary. If lim an = 0 (in other words if this limit either does not exist at all or is not zero), n →∞ 6 ∞ then an diverges. X Proof. Contraposing the previous theorem tells us that if lim an = 0, then lim Sn cannot be a n n →∞ 6 →∞ ∞ number, and the definition of convergence for series then tells us that an diverges. X Page 1 of 2 A.Sontag November 19, 2001 Term Test continued An informal look at why the Term Test is legitimate Suppose lim an = 3, say. This tells us that when n is huge, an 3. So n →∞ ≈ a = a + a + a + some number Q+ 3+ 3+ 3 + = n 1 2 3 · · · ≈ ≈ ≈ ≈ ··· ∞ XThis should convince us that if a 3, then S + and a diverges. n −→ n −→ ∞ n X Practice Problems What information, if any, does the Term Test give about the convergence or divergence of the following series? ∞ 1 1. 2n + 3 n=1 X ∞ 3n 2. 2n + 4 n=1 X n ∞ 1 3. 1 + n n=1 X ∞ n2 4. en n=1 X ∞ 2n2 + 4n 5. 5n2 + 6n n=1 X ∞ 1 6. n ln n n=2 X ∞ n 7. ln n + 1 n=1 X Answers. 1. No information. 2. Series diverges. 3. Series diverges. 4. No information. 5. Series diverges. 6. No information. 7. No information. Page 2 of 2 A.Sontag November 19, 2001.

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