5 Series

Question 1

Definition 5.1:
Infinite series

Answer 1

If (an)n∈N is a real sequence, then ∑∞n=1 an is called an infinite series (or simply series).

Question 2

Definition 5.1:
Partial sum

Answer 2

For any N ∈ N, the number SN = ∑Nn=1 an is called a partial sum of the series.

Question 3

Definition 5.2:
Convergent series

Answer 3

The series ∑∞n=1 an converges if the sequence of partial sums (SN)N∈N converges. i.e. ∑∞n=1(an) = lim.N→∞(SN).

Question 4

Proposition 5.1:
Convergent series and its relation to an

Answer 4

If ∑∞n=1(an) is a convergent series, then an→0 as n→∞.

Question 5

Theorem 5.1:
Algebra of limits for series

Answer 5

Suppose that ∑∞n=1(an) and ∑∞n=1(bn) are convergent series. Then
1. ∑∞n=1(an + bn) = ∑∞n=1(an) + ∑∞n=1(bn),
2. ∑∞n=1(an − bn) = ∑∞n=1(an) − ∑∞n=1(bn),
3. ∑∞n=1(Can) = C ∑∞n=1(an) for any constant C ∈ R.

Question 6

Theorem 5.2:
Converegent series if an≤bn for all n in N

Answer 6

Suppose that ∑∞n=1(an) and ∑∞n=1(bn) are convergent series. If an ≤ bn for all n ∈ N, then ∑∞n=1(an) ≤ ∑∞n=1(bn).

Question 7

Proposition 5.2:
Relation between convergent series and sequence of partial sums

Answer 7

Suppose that (an)n∈N is a sequence with an ≥ 0 for all n ∈ N.
Then ∑∞n=1(an) convergent ↔ (SN)n∈N bounded

Question 8

Proposition 5.3:
Limit of (sum of an from N to infinity)

Answer 8

Suppose that (an)n∈N is a sequence with an ≥ 0 for all n ∈ N. If ∑∞n=1(an) convergent, then lim.N→∞[ ∑∞n=1(an) ] = 0.

lim.N→∞[ ∑∞n=1(an) ] = lim.N→∞[ aN + a(N+1) + a(N+2) + … ]

Question 9

Definition 5.3:
Absolute convergence

Answer 9

A series ∑∞n=1(an) converges absolutely if
∑∞n=1(|an|) < ∞.

Question 10

Theorem 5.3:
Relation between |∑an| and ∑|an| (0 to infinity)

Answer 10

If ∑∞n=1(an) converges absolutely, then it converges and
|∑∞n=1(an)| ≤ ∑∞n=1(|an|).

Question 11

Definition 5.4:
Conditionally convergent series

Answer 11

A series that converges but does not converge absolutely.

Question 12

Theorem 5.4:
Rearrangement of series

Answer 12

Suppose that 􏰎∞n=1 an converges absolutely. Then for any
bijection σ : N → N, then the convergence is absolute and
∑∞n=1[ aσ(n) ] = ∑∞n=1(an).

Proof non-examinable

Question 13

Theorem 5.5:
Comparison test

Answer 13

Suppose that (an)n∈N and (bn)n∈N are real sequences and |an|≤ bn for all n ∈ N. If
∑∞n=1(bn) convergent ⇒ ∑∞n=1(an) absolutely convergent.

Question 14

Corollary 5.1:
Contrapositive of comparison test

Answer 14

Let (an)n∈N and (bn)n∈N are real sequences and 0 ≤ an ≤ bn for all n ∈ N.
Then ∑∞n=1(an) divergent ⇒ ∑∞n=1(bn) divergent.

Question 15

Theorem 5.6:
D’Alembert’s ratio test

Answer 15

Suppose that (an)n∈N is a real sequence such that an≠0 for all n∈N and r = lim.n→∞[ |a(n+1)| / |an| ].
1. If r < 1, then ∑∞n=1(an) converges absolutely.
2. If r > 1, then ∑∞n=1(an) diverges.
3. If r = 1, then the test is inconclusive.

(r = ∞ is allowed)

Question 16

Theorem 5.7:
Cauchy condensation test

Answer 16

• Let (an)n∈N be a real decreasing [ a(n+1) ≤ an for all n∈N ] sequence with an≥0.
• Define (bk)k∈N with bk = 2ᵏa₂^ₖ.
• Then ∑∞n=1(an) convergent ↔ ∑∞k=1(bk) convergent.

b1 = 2a₂, b2 = 4a₄, b3 = 8a₈, …

Question 17

Theorem 5.8:
Leibniz’s alternating series test

Answer 17

The alternating series ∑∞n=1(-1)ⁿ(an) converges if:
* an ≥ 0 for all n∈N
* lim.n→∞(an) = 0
* (an)n∈N decreasing

Does not guarantee absolute convergence

Question 18

Theorem 5.9:
Cauchy theorem for product of series

Answer 18

If A = ∑∞n=0(an) and B = ∑∞n=0(bn) are absolutely convergent, then
* A * B = ∑∞n=0(cn)

where
* cn = ∑ⁿi=0 [ ai * b(i-1) ]

Question 19

Definition 5.6:
Power series

Answer 19

Has the form ∑∞n=0(aₙxⁿ).

* aₙ∈R are the coefficients of the series.
* x∈R.

Question 20

Theorem 5.10 / Definition 5.6:
Radius of convergence (R)

Answer 20

For any power series ∑∞n=0(aₙxⁿ), there exists R ∈ [0, ∞) ∪ {∞} such that
1. If |x| < R, then ∑∞n=0(aₙxⁿ) converges absolutely, and
2. If |x| > R, then ∑∞n=0(aₙxⁿ) diverges
3. If |x| = R, then inconclusive.

Question 21

Theorem 5.11:
Radius of convergence (R) calculation

Answer 21

For any power series ∑∞n=0(aₙxⁿ),
* if lim.n→∞( |a(n+1)| / |an| ) = L exists,
* then R = 1/L.