§3.3–Series, Part A Tom Lewis Fall Term 2006 Tom Lewis () §3.3–Series, Part A Fall Term 2006 1 / 18 Outline 1 Definition 2 Two examples 3 Linearity of convergent series 4 Geometric series 5 Harmonic series 6 Cauchy’s condition 7 nth-term test for divergence 8 Absolute convergence 9 Tails Tom Lewis () §3.3–Series, Part A Fall Term 2006 2 / 18 Let (ak , k ≥ 1) be a sequence of real numbers. For each n ≥ 1, define Sn = a1 + a2 + ··· + an The sequence (Sn) is called the sequence of partial sums. P∞ If Sn → S, then we say that the infinite series k=1 ak converges to S and we write ∞ X ak = S. k=1 P∞ If (Sn) diverges, then we say that the infinite series k=1 ak diverges. Definition Definition Tom Lewis () §3.3–Series, Part A Fall Term 2006 3 / 18 P∞ If Sn → S, then we say that the infinite series k=1 ak converges to S and we write ∞ X ak = S. k=1 P∞ If (Sn) diverges, then we say that the infinite series k=1 ak diverges. Definition Definition Let (ak , k ≥ 1) be a sequence of real numbers. For each n ≥ 1, define Sn = a1 + a2 + ··· + an The sequence (Sn) is called the sequence of partial sums. Tom Lewis () §3.3–Series, Part A Fall Term 2006 3 / 18 P∞ If (Sn) diverges, then we say that the infinite series k=1 ak diverges. Definition Definition Let (ak , k ≥ 1) be a sequence of real numbers. For each n ≥ 1, define Sn = a1 + a2 + ··· + an The sequence (Sn) is called the sequence of partial sums. P∞ If Sn → S, then we say that the infinite series k=1 ak converges to S and we write ∞ X ak = S. k=1 Tom Lewis () §3.3–Series, Part A Fall Term 2006 3 / 18 Definition Definition Let (ak , k ≥ 1) be a sequence of real numbers. For each n ≥ 1, define Sn = a1 + a2 + ··· + an The sequence (Sn) is called the sequence of partial sums. P∞ If Sn → S, then we say that the infinite series k=1 ak converges to S and we write ∞ X ak = S. k=1 P∞ If (Sn) diverges, then we say that the infinite series k=1 ak diverges. Tom Lewis () §3.3–Series, Part A Fall Term 2006 3 / 18 Definition Remark An infinite series does not need to begin at k = 1. For example, the series ∞ X 1 k2 k=4 would coincide with the sequence of partial sums 1 1 1 1 1 1 , + , + + ,... 16 16 25 16 25 36 Tom Lewis () §3.3–Series, Part A Fall Term 2006 4 / 18 1 1 1 Observe that = − . k2 + k k k + 1 Thus 1 1 1 1 1 1 1 S = − + − + ··· + − = 1 − . n 1 2 2 3 n n + 1 n + 1 Since Sn → 1, we say that the series converges to 1. Solution Two examples Problem Show that ∞ X 1 = 1. k2 + k k=1 Tom Lewis () §3.3–Series, Part A Fall Term 2006 5 / 18 1 1 1 Observe that = − . k2 + k k k + 1 Thus 1 1 1 1 1 1 1 S = − + − + ··· + − = 1 − . n 1 2 2 3 n n + 1 n + 1 Since Sn → 1, we say that the series converges to 1. Two examples Problem Show that ∞ X 1 = 1. k2 + k k=1 Solution Tom Lewis () §3.3–Series, Part A Fall Term 2006 5 / 18 Thus 1 1 1 1 1 1 1 S = − + − + ··· + − = 1 − . n 1 2 2 3 n n + 1 n + 1 Since Sn → 1, we say that the series converges to 1. Two examples Problem Show that ∞ X 1 = 1. k2 + k k=1 Solution 1 1 1 Observe that = − . k2 + k k k + 1 Tom Lewis () §3.3–Series, Part A Fall Term 2006 5 / 18 Since Sn → 1, we say that the series converges to 1. Two examples Problem Show that ∞ X 1 = 1. k2 + k k=1 Solution 1 1 1 Observe that = − . k2 + k k k + 1 Thus 1 1 1 1 1 1 1 S = − + − + ··· + − = 1 − . n 1 2 2 3 n n + 1 n + 1 Tom Lewis () §3.3–Series, Part A Fall Term 2006 5 / 18 Two examples Problem Show that ∞ X 1 = 1. k2 + k k=1 Solution 1 1 1 Observe that = − . k2 + k k k + 1 Thus 1 1 1 1 1 1 1 S = − + − + ··· + − = 1 − . n 1 2 2 3 n n + 1 n + 1 Since Sn → 1, we say that the series converges to 1. Tom Lewis () §3.3–Series, Part A Fall Term 2006 5 / 18 Clearly S1 = 1,S2 = 0,S3 = 1,S4 = 0, and, in general, ( 1 if n is odd Sn = 0 if n is even Since the sequence of partial sums (Sn) diverges, the series P∞ k+1 k=1(−1) diverges. Solution Two examples Problem P∞ k+1 Show that the series k=1(−1) diverges. Tom Lewis () §3.3–Series, Part A Fall Term 2006 6 / 18 Clearly S1 = 1,S2 = 0,S3 = 1,S4 = 0, and, in general, ( 1 if n is odd Sn = 0 if n is even Since the sequence of partial sums (Sn) diverges, the series P∞ k+1 k=1(−1) diverges. Two examples Problem P∞ k+1 Show that the series k=1(−1) diverges. Solution Tom Lewis () §3.3–Series, Part A Fall Term 2006 6 / 18 Since the sequence of partial sums (Sn) diverges, the series P∞ k+1 k=1(−1) diverges. Two examples Problem P∞ k+1 Show that the series k=1(−1) diverges. Solution Clearly S1 = 1,S2 = 0,S3 = 1,S4 = 0, and, in general, ( 1 if n is odd Sn = 0 if n is even Tom Lewis () §3.3–Series, Part A Fall Term 2006 6 / 18 Two examples Problem P∞ k+1 Show that the series k=1(−1) diverges. Solution Clearly S1 = 1,S2 = 0,S3 = 1,S4 = 0, and, in general, ( 1 if n is odd Sn = 0 if n is even Since the sequence of partial sums (Sn) diverges, the series P∞ k+1 k=1(−1) diverges. Tom Lewis () §3.3–Series, Part A Fall Term 2006 6 / 18 P P Let Sn, Tn, and Un be the partial sums of ak , bk and P (αak + βbk ) respectively. Then Un = αSn + βTn. Since Sn → A and Tn → B, Un → αA + βB, as was to be shown. Proof. Linearity of convergent series Theorem P P If ak and bk converge to A and B respectively and α, β ∈ R, then P (αak + βbk ) converges to αA + βB. Tom Lewis () §3.3–Series, Part A Fall Term 2006 7 / 18 P P Let Sn, Tn, and Un be the partial sums of ak , bk and P (αak + βbk ) respectively. Then Un = αSn + βTn. Since Sn → A and Tn → B, Un → αA + βB, as was to be shown. Linearity of convergent series Theorem P P If ak and bk converge to A and B respectively and α, β ∈ R, then P (αak + βbk ) converges to αA + βB. Proof. Tom Lewis () §3.3–Series, Part A Fall Term 2006 7 / 18 Then Un = αSn + βTn. Since Sn → A and Tn → B, Un → αA + βB, as was to be shown. Linearity of convergent series Theorem P P If ak and bk converge to A and B respectively and α, β ∈ R, then P (αak + βbk ) converges to αA + βB. Proof. P P Let Sn, Tn, and Un be the partial sums of ak , bk and P (αak + βbk ) respectively. Tom Lewis () §3.3–Series, Part A Fall Term 2006 7 / 18 Since Sn → A and Tn → B, Un → αA + βB, as was to be shown. Linearity of convergent series Theorem P P If ak and bk converge to A and B respectively and α, β ∈ R, then P (αak + βbk ) converges to αA + βB. Proof. P P Let Sn, Tn, and Un be the partial sums of ak , bk and P (αak + βbk ) respectively. Then Un = αSn + βTn. Tom Lewis () §3.3–Series, Part A Fall Term 2006 7 / 18 Linearity of convergent series Theorem P P If ak and bk converge to A and B respectively and α, β ∈ R, then P (αak + βbk ) converges to αA + βB. Proof. P P Let Sn, Tn, and Un be the partial sums of ak , bk and P (αak + βbk ) respectively. Then Un = αSn + βTn. Since Sn → A and Tn → B, Un → αA + βB, as was to be shown. Tom Lewis () §3.3–Series, Part A Fall Term 2006 7 / 18 Geometric series Theorem If |ρ| < 1, then ρn → 0; if |ρ| > 1, then (ρn) diverges. In particular, if ρ > 1, then ρn → +∞. Tom Lewis () §3.3–Series, Part A Fall Term 2006 8 / 18 Let 0 < ρ < 1 and observe that 1/ρ = 1 + h for some h > 0. By the n n Binomial Theorem, 1/ρ ≥ nh for each n ≥ N; thus, 0 ≤ ρ ≤ 1/(nh) n for each n ∈ N. Since 1/(nh) → 0 as n → ∞, ρ → 0 as n → ∞. Since |ρn| = |ρ|n, it follows that ρn → 0 for |ρ| < 1. If ρ > 1, then ρ = 1 + h for h > 0. By the Binomial Theorem, ρn = (1 + h)n ≥ nh. Since nh → ∞ as n → ∞, it follows that (ρn) diverges. If |ρ| > 1, then (ρn) diverges because (|ρ|n) diverges. Geometric series Proof. Tom Lewis () §3.3–Series, Part A Fall Term 2006 9 / 18 Since |ρn| = |ρ|n, it follows that ρn → 0 for |ρ| < 1.