7.1 Alternating Series

Ratio and Root Tests, Alternating Series, and Conditional & Absolute Convergence Week 7

P n n+1 Deﬁnition A series an is called alternating if we may write an = (−1) bn or an = (−1) bn for some positive sequence bn.

Example The Alternating Harmonic series is the series

∞ X 1 (−1)n+1 n n=1 1 n+1 This is an alternating series since we may say bn = n and then an = (−1) bn. Does this converge or diverge? If all terms were positive, then this would just be the harmonic series and would diverge. However, the alternating sign means we will have some cancellation and it is conceivable that the sum would remain ﬁnite with that in mind.

P n n+1 Alternating Series Test Let the sum an be such that it can be written an = (−1) bn or an = (−1) bn for a positive sequence bn. Suppose that both

1. lim bn = 0 n→∞

2. {bn} is (eventually) decreasing. P In that case the series an converges.

Examples

∞ X 1 (i) (−1)n+1 n n=1

1 This is an alternating series with bn = n so we may apply the above test. 1 1. lim bn = lim = 0 n→∞ n→∞ n

bn+1 n 2. bn decreases since = n+1 < 1 bn Therefore the series converges by the Alternating Series Test.

∞ X 1 (ii) cos(nπ) n2 n=1

n 1 Since cos(nπ) = (−1) , we may consider this an alternating series with bn = n2 and apply the test. 1 1. lim bn = lim = 0 n→∞ n→∞ n2 1 2. bn decreases since for f(x) = x2 , −2 f 0(x) = < 0 when x > 0 x3 Therefore the series converges by the Alternating Series Test.

∞ ( 1 X n+1 2n if n is even (iii) (−1) bn where bn = 1 if n is odd n=0 3n This is an alternating series so we may apply the above test.

1 Ratio and Root Tests, Alternating Series, and Conditional & Absolute Convergence Week 7

 1  lim = 0 n even n→∞ 2n 1. lim bn = =⇒ lim bn = 0 n→∞ n→∞  1  lim = 0 n odd = 0 n→∞ 3n 2. bn is not decreasing however. Since (where inequalities denote eventual inequalities)

 n+1 n bn+1 2 2 if n is even b = = 2 · 3 < 1  n 3n

 3n+1 n if n is odd bn+1 = = 3 · 3 > 1 bn 2n 2 Therefore the Alternating Series Test DOES NOT APPLY. We cannot conclude that the series diverges just because the test is not applicable. In fact this series converges,

∞ ∞ ∞ X X 1 X 1 1 1 4 9 (−1)n+1b = − + − = − + = − + n 22n 32n 1 − 1 1 − 1 3 8 n=0 n=0 n=0 4 9

7.2 Conditional & Absolute Convergence P P P Deﬁnition A series an is said to converge conditionally if an converges, but an diverges. P P Deﬁnition A series an is said to converge absolutely if an converges.

Examples

∞ X 1 (i) (−1)n+1 n n=1 This series converges conditionally. We have seen that it converges by the Alternating Series Test, but ∞ ∞ X 1 X 1 (−1)n+1 = is the harmonic series and diverges. n n n=1 n=1

∞ X 1 (ii) cos(nπ) n2 n=1 ∞ ∞ X 1 X 1 This converges absolutely. As we saw it converges by the AST, but also (−1)n+1 = n2 n2 n=1 n=1 converges since it is a p-series with p > 1. In some sense we want our series to converge absolutely and this will be the ﬁrst thing we check for often times. That is because of the following theorems.

∞ ∞ X X Theorem If |an| converges , then an converges.

∞ ∞ X X Proof Note that 0 ≤ |an| + an ≤ 2|an|. Since |an| converges, so too does 2|an|. Therefore by the X comparison test |an| + an converges. Given this we may write,

∞ ∞ X X X an = |an| + an − |an|

2 Ratio and Root Tests, Alternating Series, and Conditional & Absolute Convergence Week 7

∞ X and therefore an converges as a sum of convergent series.

∞ ∞ X X Riemann Rearrangement Theorem Let an be a series. If an converges absolutely to S, then any ∞ X 0 rearrangement of the series also converges to S. If instead an converges conditionally, and S is any real ∞ X 0 number (or ±∞), then it is possible to reorder the sum so that an converges to S .

”Proof” of the Second Bit Keep adding positive terms of the sequence together until your partial sum > S0. Then keep adding negative terms of the sequence until the partial sum < S0. Rinse and repeat. You can show the sum will now converge to S0.

7.3 Ratio Test P Theorem Given the series an, consider the quantity a L = lim n+1 n→∞ an 1. If L < 1 then the series converges absolutely. 2. If L > 1 or L = ∞ then the series diverges. 3. If L = 1 or the limit diverges, then the series may be divergent, conditionally convergent, or absolutely convergent.

Proof Intuitively, if L < 1 then we can bound the series by a convergent geometric series.

If L < 1 then for n ≥ N for some N and a > 0,

an+1 < 1 − =⇒ |an+1| < |an| · (1 − ) an m Therefore iterating gives that |aN+m| < |aN |(1 − )

∞ N−1 ∞ N−1 ∞ X X X X X m =⇒ an = an + an ≤ an + aN (1 − ) n=i n=i n=N n=i m=0 The rightmost inﬁnite series is geometric with r = (1 − ) < 1 and therefore converges. By comparison the original series diverges. If L > 1 instead, then the proof is essentially the the same using the identity m |aN+m| > |aN |(1 + ) .

Examples ∞ X 12n (i) (−1)n n! n=0 a 12n+1 n! 12 n+1 n+1 L = lim = lim (−1) · n n = lim = 0 n→∞ an n→∞ (n + 1)! (−1) 12 n→∞ n Since L < 1, this series converges absolutely by the ratio test. ∞ X nn (ii) n! n=0 a (n + 1)n+1 n! n + 1n 1 n n+1 1 L = lim = lim · n = lim = lim 1 + = e n→∞ an n→∞ (n + 1)! n n→∞ n n→∞ n Since L > 1, this series diverges by the ratio test.

3 Ratio and Root Tests, Alternating Series, and Conditional & Absolute Convergence Week 7

∞ X n + 1 (iii) 2n − 3 n=0 a (n + 2) 2n − 3 2n2 + n − 6 2 n+1 L = lim = lim · = lim 2 = = 1 n→∞ an n→∞ 2n − 1 n + 1 n→∞ 2n + n − 1 2 Since L = 1, the ratio test is inconclusive. We can say that the series diverges by the divergence test, n + 1 1 lim an = lim = 6= 0 n→∞ n→∞ 2n − 3 2

∞ X 1 (iv) (−1)n+1 n2 n=1 a 1 n2 n2 1 n+1 n+2 L = lim = lim (−1) 2 · n+1 = lim 2 = = 1 n→∞ an n→∞ (n + 1) (−1) n→∞ n + 2n + 1 1 Since L = 1, the ratio test is inconclusive. We can say that the series converges absolute though by a p-series test as we saw two sections ago.

∞ X 1 (v) (−1)n+1 n n=1 I leave it to you to show that the Ratio test gives L = 1 and is inconclusive, but the series is in fact conditionally convergent.

7.4 Root Test P Theorem Given the series an, consider the quantity

pn L = lim |an| n→∞ 1. If L < 1 then the series converges absolutely. 2. If L > 1 or L = ∞ then the series diverges. 3. If L = 1 or the limit diverges, then the series may be divergent, conditionally convergent, or absolutely convergent.

Proof Intuitively, if L < 1 then we can bound the series by a convergent geometric series. The proof is essentially the same as in the ratio test so you can ﬁgure it out with some sweat and tears.

Note: The Root Test is more powerful than the Ratio Test. Speciﬁcally, if the Ratio Test is inconclusive, then the Root Test may still provide a conclusion. However, if the Root Test is inconclusive, then the Ratio Test is necessarily inconclusive.

4 Ratio and Root Tests, Alternating Series, and Conditional & Absolute Convergence Week 7

Examples

∞ X 100000 2+n (i) nn n=1 r 2 r 2+n n +1 n n 100000 100000 L = lim an = lim = lim = 0 n→∞ n→∞ nn n→∞ n Since L < 1, this series converges absolutely by the Root Test.

∞ X 7n2 − 3n + 1n (ii) − 1 − 4n2 n=1 s r 2 n 2 n n 7n − 3n + 1 7n − 3n + 1 7 L = lim an = lim − = lim = n→∞ n→∞ 1 − 4n2 n→∞ 1 − 4n2 4

Since L > 1, this series diverges by the Root Test.

∞ X 1 (iii) (ln n)n n=1 s r n n 1 1 L = lim an = lim = lim = 0 n→∞ n→∞ (ln n)n n→∞ ln n

Since L < 1, this series converges absolutely by the Root Test.

∞ 2 X 1 n (iv) (−1)n 1 − n n=1 r r r n2 n2 n n n n 1 n 1 1 −1 L = lim an = lim (−1) 1 − = lim 1 − = lim 1 − = e n→∞ n→∞ n n→∞ n n→∞ n Since L < 1, this series converges absolutely by the Root Test.

X 24−3n (v) 37−2n n=0 r r 4−3n 4 −3 −3 n 2 2 n 2 9 L = lim n a = lim = lim = = n 7−2n 7 −2 −2 n→∞ n→∞ 3 n→∞ 3 n 3 8 Since L > 1, this series diverges by the Root Test.

X n (vi) 3−(n+(−1) ) n=0 Verify that the Root test shows that this series converges, but the Ratio test gives an inconclusive result.