Science One Math March 21, 2018 Announcements Webwork MATH 2.9 due - - PowerPoint PPT Presentation

Science One Math

March 21, 2018

Announcements

• Webwork MATH 2.9 due on Saturday

☛ short, on sequences and geometric series

• WebWork MATH 2.10 due a week on Saturday

☛ on convergence tests, start working on it NOW

• Today’s evening Math workshop is CANCELLED

☛ next workshop is on Monday

What we know so far….

Main question: Given a series ∑ 𝒃𝒐, does it converge? (harder: if ∑ 𝑏% converges, what does it converge to?) Things you can check : 1. Do the terms go to 0? If lim

%→*𝑏% ≠ 0 then the series diverges. (test for divergence)

2. Is ∑ 𝑏% a known series? (geometric series or p-series) 3. Otherwise test for convergence (if positive terms, 𝑏% > 0) i. Integral test ii. Comparison tests:

• Compare with a known series:

If 𝑏% ≤ 𝑐% and ∑ 𝑐% converges, then ∑ 𝑏% converges. If 𝑏% ≥ 𝑐% and ∑ 𝑐% diverges, then ∑ 𝑏% diverges.

• Compare the limit of the n-th terms (limit comparison test):

If lim

%→* 23 43 is finite and > 0, then ∑ 𝑏% and ∑ 𝑐% behave in the same way.

One more test….let’s start with an observation:

For a geometric series ∑𝑏% = ∑ 𝑏 𝑠%

* %89

, the ratio of any two subsequent terms is constant lim

%→*

𝑏%:; 𝑏% = 𝑠 We need 𝑠 < 1 for convergence. The ratio test extends this idea.

The Ratio Test

• if lim

%→* 23>? 23

= 𝑴 < 𝟐 ⇒ the series ∑ 𝑏% converges

• if lim

%→* 23>? 23

= 𝑴 > 𝟐 or lim

%→* 23>? 23

= ∞ ⇒ the series ∑ 𝑏% diverges

• if lim

%→* 23>? 23

= 1 ⇒ the test is inconclusive. Rationale: If the limit above exists, then the tail of the series behaves like a geometric series Note: The ratio test works well when 𝑏% involves exponentials and factorials.

Determine whether the following series converge or diverge:

• ∑

;9C D! * D8;

• ∑

%3 %! * %8;

• ∑

𝑓GH( 𝑘K + 4)

* H8;

• ∑

D NC * D8;

• ∑

%! (K%)! * %8;

Determine whether the following series converge or diverge:

• ∑

;9C D! * D8;

lim

%→* 23>? 23 = lim D→* ;9(C>?) D:; ! O D! ;9C = lim D→* ;9 D:; = 0 < 1

converges

• ∑

%3 %! * %8;

lim

%→* (%:;)(3>?) %:; !

O %!

%3 = lim %→* %:; %:; 3 %:; %3

= lim

%→* 1 + ; % %

= 𝑓 > 1 diverges

• ∑

𝑓GH( 𝑘K + 4)

* H8;

lim

H→* PQ(R>?)( (H:;)S:T) PQR( HS:T)

= lim

H→* (HS:KH:U) P(HS:;) = ; P < 1

converges

• ∑

D NC * D8;

lim

D→* D:; NC>? O NC D = lim D→* D:; ND = ; N < 1

converges

• ∑

%! (K%)! * %8;

lim

%→* %:; ! K(%:;) ! O K% ! %! = lim %→* %:; K% ! K%:K K%:; K% ! = lim %→* ; K K%:; = 0 < 1

converges

Warning: The ratio test may be inconclusive…

Simple example: 1 + 1 + 1 + 1 + ⋯ we know it diverges but lim

%→* 23>? 23 = 1

the test cannot detect the divergence! Another example: ∑ ;

%W p-series, we know it converges for 𝑞 > 1 and diverges for 𝑞 ≤ 1, but…

lim

%→* 23>? 23 = 1 for any 𝑞, the ratio test does not feel the difference between

𝑞 = 2 (convergence) and 𝑞 = 1 (divergence). The integral test is sharper! (but computing integrals may be hard!)

Series with negative terms

1 −

; K − ; T + ; [ + ; ;\ − ; NK − ; \T … does it converge?

If we replace all negative signs with +, we have a convergent geometric series. Changing from 𝑏% to |𝑏%| increases the sum (replace negative numbers with positive numbers). The smaller series ∑ 𝑏% will converge if the larger series ∑|𝑏%| converges. ⇒ checking the convergence of ∑ |𝑏%| gives us another test for convergence of ∑ 𝑏%. Let’s see…

Absolute and Conditional Convergence

Definition of Absolute Convergence

• ∑𝑏% is said to converge absolutely if the series ∑ |𝑏%| converges.

Definition of Conditional Convergence

• ∑ 𝑏% is said to converge conditionally if ∑ 𝑏% converges but ∑ |𝑏%| diverges.

Test for Absolute Convergence:

• If ∑|𝑏%| converges, then ∑𝑏% converges (absolutely).

Note: if ∑ |𝑏%| diverges, ∑𝑏% may or may not converge.

Examples

Determine whether the following series converge absolutely

• ∑

(−1)D:;

* D8; ; D_

• ∑

`ab (%) %S * %8;

• ∑

(G;)WQ? KcG; * c8;

Examples

Determine whether the following series converge absolutely

• ∑

(−1)D:;

* D8; ; D_

∑ 𝑏D =

* D8;

∑

; D_/S * D8;

p-series with p>1, converges

• ∑

`ab (%) %S * %8;

(series with both positive and negative terms) ∑ 𝑏% =

* %8;

∑

`ab (%) %S * %8;

converges by comparison with ∑ ;

%S |`ab (%)| %S

≤

; %S

• ∑

(G;)WQ? KcG; * c8;

∑ 𝑏c =

* c8;

∑

; KcG; * c8;

diverges by comparison with ∑ ;

K%

Special case: Alternating series

Signs strictly alternate ∑

G; C>? D

= 1 − ;

K + ; N − ; T + ; U ⋯ * D8;

This is an alternating harmonic series. We know it doesn’t converge

• absolutely. Does it converge conditionally?

Look at the behaviour of the partial sums! ∑ (−1)%𝑏% =

* %89

𝑏9 − 𝑏; + 𝑏K − 𝑏N + 𝑏T … where 𝑏%:; ≤ 𝑏% Intuitively: if the terms are alternating, decreasing, and go to zero, then the partial sums approaches a finite number ⇒ series converges

Alternating Series Test

If ∑(−1)%:;𝑏% = 𝑏; − 𝑏K + 𝑏N − 𝑏T + ⋯ (with 𝑏% > 0) is such that

• 𝑏%:; ≤ 𝑏% (monotone decreasing)
• lim

%→* 𝑏% = 0

then the series ∑(−1)%:;𝑏% converges.

Examples

Determine if the following series converge

• ∑

G; C>? D * D8;

• ∑

cos(𝑜𝜌)

; K3 * %8;

• ∑

(−1)%

P3 %j * %8K

• ∑

(−1)%

%! %3 * %8;

Examples

• ∑

G; C>? D * D8;

alternating harmonic series, converges

• ∑

cos(𝑜𝜌)

; K3 * %8;

alternating geometric series, converges

• ∑

(−1)%

P3 %j * %8K

diverges because lim

%→* P3 %j = ∞

• ∑

(−1)%

%! %3 * %8;

converges because lim

%→* %! %3 = 0 and

𝑜 + 1 ! (𝑜 + 1)%:; = 𝑜! (𝑜 + 1)% ≤ 𝑜! 𝑜%

The algebra of convergent series

Can a convergent series be manipulated as a finite sum? Yes, if it converges absolutely, otherwise no!

The delicacy of conditionally convergent series

If a series converges only conditionally, the order of the terms is important. ∑

G; C>? D * D8;

= 1 − ;

K + ; N − ; T + ; U − ; \ + ; k − ; [ + ; l ⋯ = ln 2

(see next week) Rearrange (1 − ;

K − ; T − ; [ ⋯ ) + (; N − ; \ − ; ;K ⋯ ) + (; U − ; ;9 − ; K9 ⋯)

1 − ∑

; K %

+ ;

N 1 − ∑ ; K %

+ ;

U 1 − ∑ ; K %

+ ⋯ we get 0 = ln2 (!) → 0 → 0 → 0