§3.3--Series, Part A
§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
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. 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. 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 → ∞.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 9 / 18 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. 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.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 9 / 18 If |ρ| > 1, then (ρn) diverges because (|ρ|n) diverges.
Geometric series
Proof. 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.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 9 / 18 Geometric series
Proof. 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.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 9 / 18 Geometric series
Theorem (Geometric Series) Let a 6= 0 and r be real numbers. The series
∞ a X = if |r| < 1 ar k−1 = a + ar + ar 2 + ··· 1 − r k=1 diverges if |r| ≥ 1.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 10 / 18 Geometric series
Proof.
If r = 1, then Sn = a + a + ··· + a = na, which diverges. If r 6= 1, then 1 − r n S = a . n 1 − r n r = −1, then (r ) diverges and thus (Sn) diverges. n If |r| > 1, then (r ) diverges and thus (Sn) diverges. n If |r| < 1, then r → 0 and thus Sn → a/(1 − r).
Tom Lewis () §3.3–Series, Part A Fall Term 2006 11 / 18 Harmonic series
Theorem The harmonic series 1 1 1 1 + + + + ··· 2 3 4 diverges.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 12 / 18 Let Sn = 1 + (1/2) + ··· + (1/n). Since (Sn) is monotone increasing, it is enough to show that (Sn) is not bounded above. We will use a “dyadic-blocking argument” to obtain a lower bound for the partial sums.
Consider the partial sums of the form S2n . Notice that, for example, 1 1 1 1 1 S = 1 + + + + + ··· + 8 2 3 4 5 8 1 1 1 3 ≥ 1 + + 2 + 4 = 1 + , 2 4 8 2 where we have replaced each element in a dyadic block by the smallest element in the block. n In general, S2n ≥ 1 + 2 , which is unbounded as n → ∞. Since the sequence of partial sums is not bounded, the series diverges, as was to be shown.
Harmonic series Proof.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 13 / 18 We will use a “dyadic-blocking argument” to obtain a lower bound for the partial sums.
Consider the partial sums of the form S2n . Notice that, for example, 1 1 1 1 1 S = 1 + + + + + ··· + 8 2 3 4 5 8 1 1 1 3 ≥ 1 + + 2 + 4 = 1 + , 2 4 8 2 where we have replaced each element in a dyadic block by the smallest element in the block. n In general, S2n ≥ 1 + 2 , which is unbounded as n → ∞. Since the sequence of partial sums is not bounded, the series diverges, as was to be shown.
Harmonic series Proof.
Let Sn = 1 + (1/2) + ··· + (1/n). Since (Sn) is monotone increasing, it is enough to show that (Sn) is not bounded above.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 13 / 18 Consider the partial sums of the form S2n . Notice that, for example, 1 1 1 1 1 S = 1 + + + + + ··· + 8 2 3 4 5 8 1 1 1 3 ≥ 1 + + 2 + 4 = 1 + , 2 4 8 2 where we have replaced each element in a dyadic block by the smallest element in the block. n In general, S2n ≥ 1 + 2 , which is unbounded as n → ∞. Since the sequence of partial sums is not bounded, the series diverges, as was to be shown.
Harmonic series Proof.
Let Sn = 1 + (1/2) + ··· + (1/n). Since (Sn) is monotone increasing, it is enough to show that (Sn) is not bounded above. We will use a “dyadic-blocking argument” to obtain a lower bound for the partial sums.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 13 / 18 n In general, S2n ≥ 1 + 2 , which is unbounded as n → ∞. Since the sequence of partial sums is not bounded, the series diverges, as was to be shown.
Harmonic series Proof.
Let Sn = 1 + (1/2) + ··· + (1/n). Since (Sn) is monotone increasing, it is enough to show that (Sn) is not bounded above. We will use a “dyadic-blocking argument” to obtain a lower bound for the partial sums.
Consider the partial sums of the form S2n . Notice that, for example, 1 1 1 1 1 S = 1 + + + + + ··· + 8 2 3 4 5 8 1 1 1 3 ≥ 1 + + 2 + 4 = 1 + , 2 4 8 2 where we have replaced each element in a dyadic block by the smallest element in the block.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 13 / 18 Since the sequence of partial sums is not bounded, the series diverges, as was to be shown.
Harmonic series Proof.
Let Sn = 1 + (1/2) + ··· + (1/n). Since (Sn) is monotone increasing, it is enough to show that (Sn) is not bounded above. We will use a “dyadic-blocking argument” to obtain a lower bound for the partial sums.
Consider the partial sums of the form S2n . Notice that, for example, 1 1 1 1 1 S = 1 + + + + + ··· + 8 2 3 4 5 8 1 1 1 3 ≥ 1 + + 2 + 4 = 1 + , 2 4 8 2 where we have replaced each element in a dyadic block by the smallest element in the block. n In general, S2n ≥ 1 + 2 , which is unbounded as n → ∞.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 13 / 18 Harmonic series Proof.
Let Sn = 1 + (1/2) + ··· + (1/n). Since (Sn) is monotone increasing, it is enough to show that (Sn) is not bounded above. We will use a “dyadic-blocking argument” to obtain a lower bound for the partial sums.
Consider the partial sums of the form S2n . Notice that, for example, 1 1 1 1 1 S = 1 + + + + + ··· + 8 2 3 4 5 8 1 1 1 3 ≥ 1 + + 2 + 4 = 1 + , 2 4 8 2 where we have replaced each element in a dyadic block by the smallest element in the block. n In general, S2n ≥ 1 + 2 , which is unbounded as n → ∞. Since the sequence of partial sums is not bounded, the series diverges, as was to be shown.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 13 / 18 For n ≥ m,
n X |Sn − Sm−1| = ak . (2) k=m
But the series converges if and only if (Sn) converges, and (Sn) converges if and only if it satisfies Cauchy’s condition.
However, by (2), (Sn) satisfies Cauchy’s condition if and only if the series satisfies (1).
Proof.
Cauchy’s condition
Theorem P∞ The series k=1 ak converges if and only if for each ε > 0 there exists a N ∈ N such that
n X n ≥ m ≥ N implies ak < ε. (1) k=m
Tom Lewis () §3.3–Series, Part A Fall Term 2006 14 / 18 For n ≥ m,
n X |Sn − Sm−1| = ak . (2) k=m
But the series converges if and only if (Sn) converges, and (Sn) converges if and only if it satisfies Cauchy’s condition.
However, by (2), (Sn) satisfies Cauchy’s condition if and only if the series satisfies (1).
Cauchy’s condition
Theorem P∞ The series k=1 ak converges if and only if for each ε > 0 there exists a N ∈ N such that
n X n ≥ m ≥ N implies ak < ε. (1) k=m
Proof.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 14 / 18 But the series converges if and only if (Sn) converges, and (Sn) converges if and only if it satisfies Cauchy’s condition.
However, by (2), (Sn) satisfies Cauchy’s condition if and only if the series satisfies (1).
Cauchy’s condition
Theorem P∞ The series k=1 ak converges if and only if for each ε > 0 there exists a N ∈ N such that
n X n ≥ m ≥ N implies ak < ε. (1) k=m
Proof. For n ≥ m,
n X |Sn − Sm−1| = ak . (2) k=m
Tom Lewis () §3.3–Series, Part A Fall Term 2006 14 / 18 However, by (2), (Sn) satisfies Cauchy’s condition if and only if the series satisfies (1).
Cauchy’s condition
Theorem P∞ The series k=1 ak converges if and only if for each ε > 0 there exists a N ∈ N such that
n X n ≥ m ≥ N implies ak < ε. (1) k=m
Proof. For n ≥ m,
n X |Sn − Sm−1| = ak . (2) k=m
But the series converges if and only if (Sn) converges, and (Sn) converges if and only if it satisfies Cauchy’s condition.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 14 / 18 Cauchy’s condition
Theorem P∞ The series k=1 ak converges if and only if for each ε > 0 there exists a N ∈ N such that
n X n ≥ m ≥ N implies ak < ε. (1) k=m
Proof. For n ≥ m,
n X |Sn − Sm−1| = ak . (2) k=m
But the series converges if and only if (Sn) converges, and (Sn) converges if and only if it satisfies Cauchy’s condition.
However, by (2), (Sn) satisfies Cauchy’s condition if and only if the series satisfies (1).
Tom Lewis () §3.3–Series, Part A Fall Term 2006 14 / 18 We will prove the contrapositive. P Suppose that an converges and let ε > 0 be given. Choose N in accord with Cauchy’s condition. Then for n = m ≥ N we have |an| < ε,
which shows that an → 0.
Proof.
nth-term test for divergence
Theorem P If an 6→ 0, then an diverges.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 15 / 18 We will prove the contrapositive. P Suppose that an converges and let ε > 0 be given. Choose N in accord with Cauchy’s condition. Then for n = m ≥ N we have |an| < ε,
which shows that an → 0.
nth-term test for divergence
Theorem P If an 6→ 0, then an diverges.
Proof.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 15 / 18 P Suppose that an converges and let ε > 0 be given. Choose N in accord with Cauchy’s condition. Then for n = m ≥ N we have |an| < ε,
which shows that an → 0.
nth-term test for divergence
Theorem P If an 6→ 0, then an diverges.
Proof. We will prove the contrapositive.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 15 / 18 Choose N in accord with Cauchy’s condition. Then for n = m ≥ N we have |an| < ε,
which shows that an → 0.
nth-term test for divergence
Theorem P If an 6→ 0, then an diverges.
Proof. We will prove the contrapositive. P Suppose that an converges and let ε > 0 be given.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 15 / 18 nth-term test for divergence
Theorem P If an 6→ 0, then an diverges.
Proof. We will prove the contrapositive. P Suppose that an converges and let ε > 0 be given. Choose N in accord with Cauchy’s condition. Then for n = m ≥ N we have |an| < ε,
which shows that an → 0.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 15 / 18 Let ε > 0 be given and choose N large enough that n ≥ m ≥ N implies n X |ak | < ε. k=m Then for n ≥ m ≥ N, by the triangle inequality,
n n X X ak ≤ |ak | < ε, k=m k=m P∞ which shows that k=1 ak converges.
Proof.
Absolute convergence
Theorem P∞ P∞ If k=1 |ak | converges, then k=1 ak converges.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 16 / 18 Let ε > 0 be given and choose N large enough that n ≥ m ≥ N implies n X |ak | < ε. k=m Then for n ≥ m ≥ N, by the triangle inequality,
n n X X ak ≤ |ak | < ε, k=m k=m P∞ which shows that k=1 ak converges.
Absolute convergence
Theorem P∞ P∞ If k=1 |ak | converges, then k=1 ak converges.
Proof.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 16 / 18 Then for n ≥ m ≥ N, by the triangle inequality,
n n X X ak ≤ |ak | < ε, k=m k=m P∞ which shows that k=1 ak converges.
Absolute convergence
Theorem P∞ P∞ If k=1 |ak | converges, then k=1 ak converges.
Proof. Let ε > 0 be given and choose N large enough that n ≥ m ≥ N implies n X |ak | < ε. k=m
Tom Lewis () §3.3–Series, Part A Fall Term 2006 16 / 18 Absolute convergence
Theorem P∞ P∞ If k=1 |ak | converges, then k=1 ak converges.
Proof. Let ε > 0 be given and choose N large enough that n ≥ m ≥ N implies n X |ak | < ε. k=m Then for n ≥ m ≥ N, by the triangle inequality,
n n X X ak ≤ |ak | < ε, k=m k=m P∞ which shows that k=1 ak converges.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 16 / 18 + − Let ak = (|ak | + ak )/2 and ak = (|ak | − ak )/2. These are the positive and negative parts of ak respectively. + − + − Observe that 0 ≤ ak ≤ |ak | and 0 ≤ ak ≤ |ak | and ak − ak = ak . P + P − By the Simple Comparison Test, ak and ak both converge and therefore the series ∞ ∞ X X + − ak = ak − ak k=1 k=1 converges by the linearity property of convergent series.
Absolute convergence
Alternative Proof.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 17 / 18 + − + − Observe that 0 ≤ ak ≤ |ak | and 0 ≤ ak ≤ |ak | and ak − ak = ak . P + P − By the Simple Comparison Test, ak and ak both converge and therefore the series ∞ ∞ X X + − ak = ak − ak k=1 k=1 converges by the linearity property of convergent series.
Absolute convergence
Alternative Proof. + − Let ak = (|ak | + ak )/2 and ak = (|ak | − ak )/2. These are the positive and negative parts of ak respectively.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 17 / 18 P + P − By the Simple Comparison Test, ak and ak both converge and therefore the series ∞ ∞ X X + − ak = ak − ak k=1 k=1 converges by the linearity property of convergent series.
Absolute convergence
Alternative Proof. + − Let ak = (|ak | + ak )/2 and ak = (|ak | − ak )/2. These are the positive and negative parts of ak respectively. + − + − Observe that 0 ≤ ak ≤ |ak | and 0 ≤ ak ≤ |ak | and ak − ak = ak .
Tom Lewis () §3.3–Series, Part A Fall Term 2006 17 / 18 Absolute convergence
Alternative Proof. + − Let ak = (|ak | + ak )/2 and ak = (|ak | − ak )/2. These are the positive and negative parts of ak respectively. + − + − Observe that 0 ≤ ak ≤ |ak | and 0 ≤ ak ≤ |ak | and ak − ak = ak . P + P − By the Simple Comparison Test, ak and ak both converge and therefore the series ∞ ∞ X X + − ak = ak − ak k=1 k=1 converges by the linearity property of convergent series.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 17 / 18 Theorem P∞ The series k=1 ak converges if and only if each tail of the series converges.
Proof. Exercise. Use the Cauchy criterion for series.
Tails
Definition P∞ P∞ A tail of the series k=1 ak is any series of the form k=m ak for some m ≥ 1.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 18 / 18 Proof. Exercise. Use the Cauchy criterion for series.
Tails
Definition P∞ P∞ A tail of the series k=1 ak is any series of the form k=m ak for some m ≥ 1.
Theorem P∞ The series k=1 ak converges if and only if each tail of the series converges.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 18 / 18 Tails
Definition P∞ P∞ A tail of the series k=1 ak is any series of the form k=m ak for some m ≥ 1.
Theorem P∞ The series k=1 ak converges if and only if each tail of the series converges.
Proof. Exercise. Use the Cauchy criterion for series.
Tom Lewis () §3.3–Series, Part A Fall Term 2006 18 / 18