1 Thus far, we have only dealt with series with only positive terms. An alternating series is one whose terms are alternately positive and negative. The Alternating series test ∞ The alternating series ∞ ∑ ๐๐ = ∑(−1)๐−1 ๐๐ = ๐1 − ๐2 + ๐3 − ๐4 + โฏ ๐=1 ๐๐ > 0 ๐=1 converges if ๐) ๐๐) ๐๐+1 ≤ ๐๐ ๐๐๐ ๐๐๐ ๐ ∈ ๐ + (๐๐๐๐๐๐๐ ๐๐๐) ๐๐๐ ๐๐ = 0 ๐→∞ Note: The theorem also applies if the first term is negative, since we could simply consider the series without the first term. Proof: 2๐๐กโ ๐๐๐๐ก๐๐๐ ๐ ๐ข๐ ๐๐ : โ ๐2๐ = (๐1 − ๐2 ) + (๐3 − ๐4 ) + โฏ + (๐2๐−1 − ๐2๐ ) {๐๐ } ๐๐ ๐๐๐๐๐ก๐๐๐๐๐๐๐๐ฆ ๐๐๐๐๐๐๐ ๐๐๐, ๐2๐−1 − ๐2๐ ≥ 0 ๐๐๐ ๐๐๐ ๐ ⇒ ๐2๐ ๐๐ ๐๐๐๐๐ก๐๐๐๐๐๐๐๐ฆ ๐๐๐๐๐๐๐ ๐๐๐ (If the first term is negative, consider the series in the absence of the first term.) โ ๐2๐ = ๐1 − (๐2 − ๐3 ) − (๐4 − ๐5 ) − โฏ − (๐2๐−2 − ๐2๐−1 ) − ๐2๐ {๐๐ } ๐๐ ๐๐๐๐๐ก๐๐๐๐๐๐๐๐ฆ ๐๐๐๐๐๐๐ ๐๐๐, ๐2๐−2 − ๐2๐−1 ≥ 0 ๐๐๐ ๐๐๐ ๐ ⇒ ๐2๐ ≤ ๐1 ๐๐๐๐ข๐๐๐๐ {๐2๐ } ๐๐ ๐๐๐๐๐ก๐๐๐๐๐๐๐๐ฆ ๐๐๐๐๐๐๐ ๐๐๐ ๐๐๐ ๐๐๐ข๐๐๐๐. ๐โ๐ข๐ ๐กโ๐ ๐ ๐๐๐ข๐๐๐๐ {๐2๐ } โ๐๐ ๐ ๐๐๐๐๐ก ๐๐ฆ ๐กโ๐ ๐๐๐๐๐ก๐๐๐ ๐ถ๐๐๐ฃ๐๐๐๐๐๐๐ ๐โ๐๐๐๐๐. ๐ฟ๐๐ก ๐ = lim ๐2๐ ๐→∞ โ ๐๐๐ค ๐ค๐ ๐๐๐๐ ๐๐๐๐ ๐กโ๐ ๐๐๐ ๐๐๐๐ก๐๐๐ ๐ ๐ข๐๐ : ๐2๐+1 = ๐2๐ + ๐2๐+1 lim ๐2๐+1 = lim ๐2๐ + lim ๐๐ = ๐ ๐→∞ ∴ ๐→∞ ๐→∞ ๐๐ → 0 ๐๐ฆ ๐๐ข๐ โ๐ฆ๐๐๐กโ๐๐ ๐๐ . ๐โ๐ข๐ ๐กโ๐ ๐๐ฃ๐๐ ๐๐๐ ๐๐๐ ๐ ๐๐๐๐๐ ๐๐๐กโ ๐๐๐๐ฃ๐๐๐๐ ๐ก๐ ๐กโ๐ ๐ ๐๐๐ ๐๐๐๐๐ก ๐๐ ๐๐๐ ๐๐๐๐ฆ ๐๐ lim ๐๐ = 0 ๐→∞ Q.E.D. 2 ๐ธ๐ฅ๐๐๐๐๐: 1 1 1 Show that 1 − + − + โฏ 2 3 4 ∞ (−1)๐−1 =∑ converges. ๐ ๐=1 ๐๐๐๐๐ 1 1 < , ๐กโ๐ ๐ ๐๐๐๐๐ ๐ ๐๐ก๐๐ ๐๐๐๐ ๐๐+1 ≤ ๐๐ ๐๐๐ ๐๐๐ ๐ ∈ ๐ + ๐+1 ๐ ๐ด๐๐ ๐ ๐๐๐ ๐๐ = ๐๐๐ ๐→∞ ∞ ∴ ∑ ๐=1 1 (Alternating series for which ๐๐ = ) ๐ ๐→∞ 1 =0 ๐ (−1)๐−1 ๐๐๐๐ฃ๐๐๐๐๐ ๐๐ฆ ๐กโ๐ ๐ด๐๐ก๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐๐ ๐๐๐ ๐ก ๐ ∞ (๐๐ฃ๐๐ ๐กโ๐๐ข๐โ ∑ ๐=1 1 ๐๐ ๐๐๐ก ๐๐๐๐ฃ๐๐๐๐๐๐ก) ๐ The Alternating Series Remainder Estimates A partial sum Sn of any convergent series can be used as an approximation to the total sum S. The error involved in using ๐ ≈ ๐๐ , for the sum of a converging alternating series is the remainder ๐ ๐ = ๐ − ๐๐ and |๐ ๐ | = |๐ − ๐๐ | ≤ ๐๐+1 (error = remainder = truncation error) Proof: ∞ ๐ ๐ = ๐ − ๐๐ = ∑ (−1)๐−1 ๐๐ = (−1)๐ ๐๐+1 + (−1)๐+1 ๐๐+2 + โฏ ๐=๐+1 โ ๐ ๐ = (−1)๐ [(๐๐+1 − ๐๐+2 ) + (๐๐+3 − ๐๐+4 ) + โฏ ] {๐๐ } ๐๐ ๐๐๐๐๐ก๐๐๐๐๐๐๐๐ฆ ๐๐๐๐๐๐๐ ๐๐๐, ๐๐+๐ − ๐๐+๐+1 ≥ 0 ๐๐๐ ๐๐๐ ๐ ⇒ |๐ ๐ | = (๐๐+1 − ๐๐+2) + (๐๐+3 − ๐๐+4 ) + โฏ ≥ 0 โ ๐ ๐ = (−1)๐ {๐๐+1 − [(๐๐+2 − ๐๐+3 ) + (๐๐+4 − ๐๐+5 ) + โฏ ]} ≥ 0 ๐๐๐ ๐๐๐ ๐ [{๐๐ } ๐๐ ๐๐๐๐๐ก๐๐๐๐๐๐๐๐ฆ ๐๐๐๐๐๐๐ ๐๐๐] ⇒ |๐ ๐ | ≤ ๐๐+1 Q.E.D. 3 ∞ EX: Find the sum of ∑ ๐=1 0< (−1)๐−1 correct to 3 decimal places. ๐! 1 1 < (๐ + 1)! ๐! ๐ด๐๐ ๐, 0< ๐๐๐๐๐ lim ๐→∞ ( ๐๐ = 1 ) ๐! ∴ ๐๐+1 ≤ ๐๐ ๐๐๐ ๐๐๐ ๐ ∈ ๐ + 1 1 < ๐! ๐ 1 =0 ⇒ ๐ lim ๐→∞ 1 =0 ๐! ๐๐ฆ ๐กโ๐ ๐๐๐ข๐๐๐ง๐ ๐โ๐๐๐๐๐ ∴ ๐กโ๐ ๐ ๐๐๐๐๐ ๐๐๐๐ฃ๐๐๐๐๐ ๐๐ฆ ๐กโ๐ ๐ด๐๐ก๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐๐ ๐๐๐ ๐ก ๐ =1− 1 1 1 1 1 1 + − + − + +โฏ 2 6 24 120 720 5040 ๐7 = 1 1 < 5040 2000 = 0.0005 ๐๐๐๐ข๐๐๐๐ฆ ๐6 = 1 − 1 1 1 1 1 + − + − = 0.631944 2 6 24 120 720 ๐ต๐ฆ ๐ธ๐ ๐ก๐๐๐๐ก๐๐๐ ๐โ๐๐๐๐๐, |๐ − ๐6 | ≤ ๐7 ∴ 0.631944 − 1 1 ≤ ๐ ≤ 0.631944 + ⇒ 0.6317456 ≤ ๐ ≤ 0.6321424 5040 5040 ๐ ≈ ๐6 = 0. 632 (3 ๐๐๐๐๐๐๐ ๐๐๐๐๐๐ ) ∞ EX: Aproximate the sum of ∑ ๐=1 (−1)๐ by its first 5 terms (5th partial sum). ๐2 ๐น๐๐๐ ๐กโ๐ ๐๐๐ก๐๐๐ฃ๐๐ ๐๐ ๐คโ๐๐โ ๐ ๐ข๐ ๐ ๐๐๐๐ . ๐5 = −1 + 1 1 1 1 − + − ≈ −0.838611 4 9 16 25 1 1 ≤ (๐ + 1)2 ๐2 ⇒ lim ๐→∞ 1 =0 ๐2 ∴ ๐๐+1 ≤ ๐๐ ๐๐๐ ๐๐๐ ๐ ∈ ๐ + ∴ ๐กโ๐ ๐ ๐๐๐๐๐ ๐๐๐๐ฃ๐๐๐๐๐ ๐๐ฆ ๐กโ๐ ๐ด๐๐ก๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐๐ ๐๐๐ ๐ก ๐ธ๐๐๐๐ ๐๐๐ข๐๐ ๐๐ ๐๐๐ฃ๐๐ ๐๐ฆ: |๐ − ๐5 | = |๐ 5 | ≤ ๐๐ ๐๐๐ก๐ข๐๐ ๐ ๐ข๐ ๐๐๐๐ ๐๐๐ก๐ค๐๐๐ 1 ≈ 0.027778 36 − 0.838611 − 0.027778 ≤ ๐ ≤ −0.838611 + 0.027778 4 ABSOLUTE AND CONDITIONAL CONVERGENCE (for loco series +, - ,+,+,+,-,-,-) ∞ Given ๐๐๐ series ∞ ∑ ๐๐ we can consider the corresponding series ∑|๐๐ | = |๐1 | + |๐2 | + โฏ ๐=1 ๐=1 Absolute Convergence ∞ ∞ The series ∑ ๐๐ is ๐๐๐ฌ๐จ๐ฅ๐ฎ๐ญ๐ฅ๐ฒ ๐๐จ๐ง๐ฏ๐๐ซ๐ ๐๐ง๐ญ if ∑|๐๐ | converges. ๐=1 ๐=1 If ๐๐ ≥ 0 for all n, absolute convergence is the same as convergence. Conditional Convergence ∞ ∞ ∞ The series ∑ ๐๐ is ๐๐จ๐ง๐๐ข๐ญ๐ข๐จ๐ง๐๐ฅ๐ฅ๐ฒ ๐๐จ๐ง๐ฏ๐๐ซ๐ ๐๐ง๐ญ if ∑ ๐๐ converges but ∑|๐๐ | diverges . ๐=1 ๐=1 ๐=1 ∞ (−1)๐−1 ๐ธ๐: ∑ ๐๐ ๐๐๐๐๐๐ก๐๐๐๐๐ฆ ๐๐๐๐ฃ๐๐๐๐๐๐ก ๐ ๐=1 Theorem of Absolute Convergence ∞ If ∑ ๐๐ is absolutely convergent then it is also convergent. ๐=1 Proof: by definition of absolute value: − |๐๐ | ≤ ๐๐ ≤ |๐๐ | ⇒ 0 ≤ ๐๐ + |๐๐ | ≤ 2|๐๐ | ∞ ∞ If ∑ ๐๐ is absolutely convergent then 2 ∑|๐๐ | is convergent. ๐=1 ๐=1 ∴ by the Comparison Test ∑∞ ๐=1(๐๐ + |๐๐ |) is also convergent. ∞ ๐๐๐๐๐ ∑ ๐๐ ๐๐ ๐กโ๐ ๐๐๐๐๐๐๐๐๐๐ ๐๐ ๐ก๐ค๐ ๐๐๐๐ฃ๐๐๐๐๐๐ก ๐ ๐๐๐๐๐ ๐=1 ∞ ∞ ∞ ∑ ๐๐ = ∑(๐๐ + |๐๐ |) − ∑|๐๐ | ๐=1 ๐=1 ๐=1 ∞ ∴ ∑ ๐๐ is convergent ๐=1 Q.E.D. Absolute convergence is a “stronger” type of convergence. Series that are absolutely convergent are guaranteed to be convergent. However, series that are convergent may or may not be absolutely convergent (they do converge, but the sum is loco). 5 ∞ EX: Show that ∑ ๐=1 ∞ ∑ ๐=1 cos ๐ is convergent. ๐2 cos ๐ cos 1 cos 2 = + 2 +โฏ 2 ๐ 12 2 ๐ก๐๐๐๐ ๐ค๐๐กโ ๐๐๐๐๐๐๐๐๐ก ๐ ๐๐๐๐ , ( ๐๐๐ก ๐๐๐ก๐๐๐๐๐ก๐๐๐ ๐ ๐๐๐๐๐ ). ∞ cos ๐ 1 ๐ป๐๐ค๐๐ฃ๐๐ | 2 | ≤ 2 ๐ ๐ ๐๐๐ ๐๐๐ ๐ ∈ ๐ + ๐๐๐ ∑ ๐=1 ∞ ∴ ๐๐ฆ ๐กโ๐ ๐ถ๐๐๐๐๐๐๐ ๐๐ ๐๐๐ ๐ก, ∑| ๐=1 1 ๐๐ ๐๐๐๐ฃ๐๐๐๐๐๐ก. ๐2 cos ๐ | ๐๐ ๐๐๐๐ฃ๐๐๐๐๐๐ก, ๐2 ๐๐๐ ๐๐ฆ ๐กโ๐ ๐โ๐๐๐๐๐ ๐๐ ๐ด๐๐ ๐๐๐ข๐ก๐ ๐ถ๐๐๐ฃ๐๐๐๐๐๐๐, ∞ ๐ ๐ ๐๐ ∑ ๐=1 cos ๐ ๐2 So what is important about absolute and conditional convergence? Algebra: a + b = b + a. If we have a finite sum ∑ ๐๐ , we can also reorder the terms without affecting the sum. Infinite series which are absolute convergent behave like finite series, so for these we can again reorder the terms of the series without affecting the sum. However, the same is not true for conditionally convergent series! ๐ธ๐: 1 1 1 1 1 ๐=1− + − + − +โฏ 2 3 4 5 6 ⇒ 1 1 1 1 1 ๐ = − + − +โฏ 2 2 4 6 8 ๐๐๐ ๐กโ๐๐ ๐ก๐๐๐๐กโ๐๐ 3 1 1 1 2 1 1 1 1 1 ๐ = 1 + 0 + − + + 0 โฏ ⇒ ๐ = (1 − + − + − + โฏ ) 2 3 2 5 3 2 3 4 5 6 Thus we get a rearrangement of the original series with a different sum! โ If ∑ ๐๐ is absolutely convergent and its value is s then any rearrangement of ∑ ๐๐ will also have a value of s. โ If ∑ ๐๐ is conditionally convergent and r is any real number then there is a rearrangement of ∑ ๐๐ whose value will be r. NO KIDDING !!!!!! But the series is convergent !!!!!!!!!! Yet, we don’t know the sum. (proof shown by Riemann) -๏ (−1)๐−1 โช We found that alternating harmonic series ∑ is conditionally convergent, so no matter what value ๐ we choose there is some rearrangement of terms that will give that value. Unbelievable, right. But true. Note that above fact does not tell us what that rearrangement must be, ONLY that it does exist. (−1)๐ โช We found that alternating series ∑ 2 is absolutely convergent and so no matter how we rearrange the ๐ terms of this series we’ll always get the same value. -๏ . In fact it can be shown that the value of that series is ∞ ∑ ๐=1 (−1)๐ ๐2 = − ๐2 12 6 ALTERNATING and LOCO SERIES PROBLEMS 1. Determine if each of the following series are absolute convergent, conditionally convergent or divergent. ∞ (−1)๐−1 (๐) ∑ ๐ ๐=1 ∞ (−1)๐ ( ๐) ∑ ๐2 ๐=1 ∞ sin ๐ (๐) ∑ 3 ๐ ๐=1 ∞ (−1)๐ ๐2 (๐) ∑ 2 ๐ +5 ๐=1 ∞ (−1)๐−3 √๐ ๐) ∑ ๐+4 ๐=1 ∞ ๐) ∑ ๐=2 cos(๐๐) √๐