Series

∑ κ=1 6 1 = 1 +1 +1 +1 +1 +1 ⏟6 ∑ κ=1 6 κ = 1+2+⋯+6 ∑ κ=1 1000 3κ = 31 +32 +⋯ +31000
∑ κ=1 ∞ κ = 1+ 2+ 3+ ⋯+ n+ ⋯ s1 = 1 = ∑ κ=1 1 κ s2 = 3 = 1+ 2 = ∑ κ=1 2 κ s3 = 6 = 1+ 2+ 3 = ∑ κ=1 3 κ ⋮ sn = n⁢(n+1) 2 = 1+ 2+ 3+ ⋯+ n = ∑ κ=1 n κ

Terminology

Infinite series

An infinite series is the sum of an infinite sequence

∑ κ=1 ∞ 𝒶κ

The value 𝒶κ is the κ'th term of the series.

Partial sums

The sum sn of the first n-terms of a series ∑ κ=1 ∞ 𝒶κ is the nth partial sum. The sequence { sn }

s1 = 𝒶1 s2 = 𝒶1 +𝒶2 = ∑ κ=1 2 𝒶κ ⋮ sn = 𝒶1 +𝒶2 +𝒶3 +⋯ +𝒶n = ∑ κ=1 n 𝒶κ ⋮

is the sequence of partial sums.

Convergence

For a series ∑ κ=1 ∞ 𝒶κ , if { sn } →𝒮 then the series converges and

∑ κ=1 ∞ 𝒶κ =𝒮

If the series is not convergent then it is the divergent.

Convergence properties

If ∑ κ=1 ∞ 𝒶κ =𝒜 , ∑ κ=1 ∞ 𝒷κ =ℬ and α,β ∈ℝ then

∑ κ=1 ∞ ( α⁢𝒶κ + β⁢𝒷κ ) = α⁢𝒜 + β⁢ℬ

Examples

Part I

Examples

∑ κ=1 ∞ 12κ ∑ κ=1 ∞ 1 κ⁢ (κ+1) ∑ κ=1 ∞ κ κ+1 ∑ κ=1 ∞ (-1) κ

Geometric series

∑ κ=1 ∞ α 𝓇κ

Harmonic series

∑ κ=1 ∞ 1κ

Properties

∑ κ=1 ∞ ( 1 2κ-3 - 3 κ⁢ (κ+1) )

Integral test

∑ κ=1 ∞ 1κ2 ∑ κ=1 ∞ 1 κ2 3 ∑ κ=1 ∞ 1 2⁢κ-1 ∑ κ=1 ∞ 1 κ3

Remainder estimate

Approximate ∑ κ=1 ∞ 1κ2 to within an error bound of 0.005.

Divergence test

Integral tests

Necessary convergence condition

If ∑ κ=1 ∞ 𝒶κ converges then { 𝒶n } →0.

Divergence test

If { 𝒶n } ↛0 then ∑ κ=1 ∞ 𝒶κ is divergent.

Area comparison

Integral

Integral test

Let { 𝒶κ } be a sequence with positive terms and f(x) be continuous, positive and decreasing real valued function. Suppose also there is N0∈ℕ such that

∀κ∈ℕ N0≤κ ⇒ f(κ) = 𝒶κ

then

∑ κ=N0 ∞ 𝒶κ and ∫ N0 ∞ f(x) ⅆx

either both diverge or both converge.

p -series test

The p-series ∑ κ=1 ∞ 1 κp is convergent for p>1 and divergent otherwise.

nth–remainder

The nth–remainder ℛn of a series ∑ κ=1 ∞ 𝒶κ is

ℛn = ∑ κ= n+1 ∞ 𝒶κ

nth–remainder estimate

If ∑ κ=1 ∞ 𝒶κ is convergent then

∫ n+1 ∞ f(x) ⅆx ≤ ℛn ≤ ∫ n ∞ f(x) ⅆx

Comparison Tests

Area comparison

Comparison test

Let ∑ κ=1 ∞ 𝒶κ and ∑ κ=1 ∞ 𝒷κ be series with non-negative terms. Suppose also there is N0∈ℕ such that

∀κ∈ℕ N0≤κ ⇒ 𝒶κ ≤ 𝒷κ

then

  • if ∑ κ=1 ∞ 𝒶κ diverges, then ∑ κ=1 ∞ 𝒷κ diverges;
  • if ∑ κ=1 ∞ 𝒷κ converges, then ∑ κ=1 ∞ 𝒶κ converges.

Limit comparison test

Positive limit

Let 𝒶κ>0 and 𝒷κ>0 for all κ≥N0∈ℕ, let

limκ→∞ 𝒶κ 𝒷κ =ℒ

then

if ℒ>0
then ∑ κ=1 ∞ 𝒶κ and ∑ κ=1 ∞ 𝒷κ both converge or both diverge;

Limit comparison test

Zero or infinite limit

if ℒ=0 and ∑ κ=1 ∞ 𝒷κ converges
then ∑ κ=1 ∞ 𝒶κ converges;
if ℒ=∞ and ∑ κ=1 ∞ 𝒷κ diverges
then ∑ κ=1 ∞ 𝒶κ diverges.

Examples

Part II

Comparison tests

∑ κ=1 ∞ 1 2κ+3 ∑ κ=1 ∞ 1 κ3 +3⁢κ +1 ∑ κ=1 ∞ 2 lnκ ∑ κ=1 ∞ 1 κ2-1

Limit tests

∑ κ=1 ∞ 1 κ2 -1 ∑ κ=1 ∞ 2κ 3κ-1 ∑ κ=2 ∞ lnκ κ2 ∑ κ=2 ∞ 1 κ-1

Alternating series

Definition

A series whose terms alternate between positive and negative is called alternating series, i.e.,

∑ κ=1 ∞ (-1) κ+1 𝒷κ = 𝒷1 -𝒷2 +𝒷3 -𝒷4 +⋯+ (-1) κ+1 𝒷κ +⋯ or ∑ κ=1 ∞ (-1) κ 𝒷κ = -𝒷1 +𝒷2 -𝒷3 +𝒷4 -⋯+ (-1) κ 𝒷κ +⋯

where 𝒷κ>0 for all κ.

Convergence

The series

∑ κ=1 ∞ (-1) κ+1 𝒷κ

converges if

  • 𝒷κ>0
  • { 𝒷κ } is monotone decreasing i.e., 𝒷κ ≥ 𝒷κ+1
  • { 𝒷κ } →0

Remainder

If 𝒮= ∑ κ=1 ∞ (-1) κ+1 𝒷κ and

  • 𝒷κ>0
  • { 𝒷κ } is monotone decreasing i.e., 𝒷κ ≥ 𝒷κ+1
  • { 𝒷κ } →0

then

| ℛn | = | 𝒮- sn | ≤ 𝒷n

Absolute convergence

A series

∑ κ=1 ∞ 𝒶κ

converges absolutely or is absolutely convergent if

∑ κ=1 ∞ | 𝒶κ |

converges.

Absolute implies conditional

If ∑ κ=1 ∞ | 𝒶κ | converges then ∑ κ=1 ∞ 𝒶κ converges.

Examples

Part III

Alternating series

∑ κ=1 ∞ (-1) κ ∑ κ=1 ∞ (-1) κ κ ∑ κ=1 ∞ (-1) κ κ2-1 ∑ κ=1 ∞ (-1) κ ⁢κ κ+1

Remainder estimate

Approximate ∑ κ=1 ∞ (-1) κ κ! correct up to three decimal points.

Conditional vs absolute

∑ κ=1 ∞ (-1) κ+1 3⁢κ+1 ∑ κ=1 ∞ sinκ κ2

Ratio and root tests

∑ κ=1 ∞ 1κ ∑ κ=1 ∞ 1κ2 ∑ κ=1 ∞ 2κ +5 3κ ∑ κ=1 ∞ (-1) κ ⁢κ! ⁢κ! ( 2⁢κ ) ! ∑ κ=1 ∞ κ κ κ! ∑ κ=1 ∞ 1 κ κ

Ratio and root tests

∑ κ=1 ∞ 𝒶κ where 𝒶κ = { κ 2κ κ= 2⁢m+1 1 2κ κ= 2⁢m ∑ κ=1 ∞ 2κ κ3

Ratio and Root test

Ratio test

Let ∑ κ=1 ∞ 𝒶κ be a series with positive terms and let

limκ→∞ 𝒶κ+1 𝒶κ =ρ

then

  • if ρ<1 then ∑ κ=1 ∞ 𝒶κ converges;
  • if ρ>1 or ρ=∞ then ∑ κ=1 ∞ 𝒶κ diverges;
  • if ρ=1 test is inconclusive.

Root test

Let ∑ κ=1 ∞ 𝒶κ be a series with positive terms and let

limκ→∞ 𝒶κ κ =ρ

then

  • if ρ<1 then ∑ κ=1 ∞ 𝒶κ converges;
  • if ρ>1 or ρ=∞ then ∑ κ=1 ∞ 𝒶κ diverges;
  • if ρ=1 test is inconclusive.