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 ∑ 𝒃 𝒐 , 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) Is ∑ 𝑏 % a known series? (geometric series or p-series) 2. 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): 2 3 4 3 is finite and > 0, then ∑ 𝑏 % and ∑ 𝑐 % behave in the same way. If lim %→*

One more test….let’s start with an observation: * 𝑏 𝑠 % For a geometric series ∑𝑏 % = ∑ , the ratio of any two %89 subsequent terms is constant 𝑏 %:; lim = 𝑠 𝑏 % %→* We need 𝑠 < 1 for convergence. The ratio test extends this idea.

The Ratio Test 2 3>? = 𝑴 < 𝟐 ⇒ the series ∑ 𝑏 % converges • if lim 2 3 %→* 2 3>? 2 3>? = ∞ ⇒ the series ∑ 𝑏 % diverges • if lim = 𝑴 > 𝟐 or lim 2 3 2 3 %→* %→* 2 3>? • if lim = 1 ⇒ the test is inconclusive. 2 3 %→* 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: ;9 C * • ∑ D8; D! % 3 * • ∑ %8; %! 𝑓 GH ( 𝑘 K + 4) * • ∑ H8; D * • ∑ D8; N C %! * • ∑ %8; (K%)!

Determine whether the following series converge or diverge: ;9 C ;9 (C>?) 2 3>? D:; ! O D! ;9 * • ∑ lim 2 3 = lim ;9 C = lim D:; = 0 < 1 converges D8; D! %→* D→* D→* % % 3 (%:;) (3>?) %:; 3 O %! %:; %→* 1 + ; * • ∑ lim % 3 = lim = lim = 𝑓 > 1 diverges %8; %:; % 3 %! %:; ! % %→* %→* P Q(R>?) ( (H:;) S :T) (H S :KH:U) 𝑓 GH ( 𝑘 K + 4) P(H S :;) = ; * • ∑ lim = lim P < 1 converges H8; P QR ( H S :T) H→* H→* N C>? O N C D D:; D:; ND = ; * • ∑ lim D = lim N < 1 converges D8; N C D→* D→* %! K(%:;) ! O K% ! %:; ! %:; K% ! ; * • ∑ lim %! = lim K% ! = lim K K%:; = 0 < 1 %8; (K%)! K%:K K%:; %→* %→* %→* converges

Warning: The ratio test may be inconclusive… Simple example: 2 3>? 1 + 1 + 1 + 1 + ⋯ we know it diverges but lim 2 3 = 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… 2 3>? lim 2 3 = 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 (%) * • ∑ %8; % S (G;) WQ? * • ∑ c8; KcG;

Examples Determine whether the following series converge absolutely ; ; * * * (−1) D:; • ∑ ∑ ∑ 𝑏 D = p-series with p>1, converges D8; D8; D8; D _/S D _ `ab (%) * • ∑ (series with both positive and negative terms) %8; % S `ab (%) converges by comparison with ∑ ; * * ∑ ∑ 𝑏 % = %8; %8; % S % S |`ab (%)| ; ≤ % S % S (G;) WQ? ; diverges by comparison with ∑ ; * * * • ∑ ∑ ∑ 𝑏 c = c8; c8; c8; KcG; KcG; K%

Special case: Alternating series G; C>? = 1 − ; K + ; N − ; T + ; * Signs strictly alternate ∑ U ⋯ D8; D 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 % = ∑ 𝑏 9 − 𝑏 ; + 𝑏 K − 𝑏 N + 𝑏 T … %89 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 ; − 𝑏 K + 𝑏 N − 𝑏 T + ⋯ (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>? * • ∑ D8; D ; * • ∑ cos(𝑜𝜌) %8; K 3 P 3 * (−1) % • ∑ %8K % j %! * (−1) % • ∑ %8; % 3

Examples G; C>? * • ∑ alternating harmonic series, converges D8; D ; * • ∑ cos(𝑜𝜌) alternating geometric series, converges %8; K 3 P 3 P 3 * (−1) % • ∑ diverges because lim % j = ∞ %8K % j %→* %! %! * (−1) % • ∑ converges because lim % 3 = 0 and %8; % 3 %→* 𝑜 + 1 ! (𝑜 + 1) % ≤ 𝑜! 𝑜! (𝑜 + 1) %:; = 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>? = 1 − ; K + ; N − ; T + ; U − ; \ + ; k − ; [ + ; * ∑ l ⋯ = ln 2 (see next week) D8; D Rearrange (1 − ; K − ; T − ; [ ⋯ ) + ( ; N − ; \ − ; ;K ⋯ ) + ( ; U − ; ;9 − ; K9 ⋯ ) % % % ; + ; ; + ; ; 1 − ∑ N 1 − ∑ U 1 − ∑ + ⋯ we get 0 = ln2 (!) K K K → 0 → 0 → 0