Administrative/Copyright Notice Transcript begins with an ANU copyright notice (Section 113P, Copyright Act 1968).
Students are reminded not to reproduce or further communicate the material and not to remove the notice.
Big-Picture Road-Map of Lecture Previous lectures:
• Sequences → Series.
• Started discussing convergence/divergence of series.
Today’s main goals: 1. Show how four “calculus objects” (functions, sequences, series, integrals) inter-relate.
Develop & practise the major convergence tests (integral test, comparison tests, etc.).
(If time) finish a few remaining sequence slides (else to be posted on Canvas).
Four Core Calculus Objects & Their Web of Connections Functions (continuous objects over an interval).
Sequences a < e m > n < / e m > n ∈ N {a<em>n}</em>{n\in\mathbb N} a < e m > n < / e m > n ∈ N N \mathbb N N
Series ∑ a n \sum a_n ∑ a n
Integrals
• Definite ∫ < e m > a b f ( x ) d x \int<em>a^b f(x)\,dx ∫ < e m > a b f ( x ) d x Improper ∫ < / e m > a ∞ f ( x ) d x \int</em>a^{\infty} f(x)\,dx ∫ < / e m > a ∞ f ( x ) d x
“Discrete vs. Continuous” theme:
• Sequence ↔ Function (discrete sampling).
• Riemann sum ↔ Definite integral (finite discrete approximation).
• Series ↔ Improper integral (infinite discrete approximation).
When Do Discrete & Continuous Agree? Good agreement requires a well-behaved function: smooth, positive, decreasing, bounded derivative, etc.
Cartoon examples:
• Flat, slowly varying f f f
• Highly oscillatory f f f
Integral Test (Key Bridge Between Worlds) Statement Let f : [ 1 , ∞ ) → R f:[1,\infty)\to\mathbb R f : [ 1 , ∞ ) → R continuous , positive , decreasing .
Set a n = f ( n ) . a_n=f(n). a n = f ( n ) .
∑ < e m > n = 1 ∞ a < / e m > n \displaystyle \sum<em>{n=1}^{\infty} a</em>n ∑ < e m > n = 1 ∞ a < / e m > n ∫ 1 ∞ f ( x ) d x \displaystyle \int_1^{\infty} f(x)\,dx ∫ 1 ∞ f ( x ) d x
They “share the same fate.”
Why Hypotheses Matter Classic Corollary: p p p Evaluate ∫ 1 ∞ 1 x p d x \displaystyle \int_1^{\infty} \frac{1}{x^{p}}\,dx ∫ 1 ∞ x p 1 d x
• Convergent if p>1, divergent if p ≤ 1 p\le 1 p ≤ 1
Hence ∑ n = 1 ∞ 1 n p \displaystyle \sum_{n=1}^{\infty}\frac1{n^{p}} n = 1 ∑ ∞ n p 1
Worked Example 1 Determine ∑ n = 2 ∞ 1 n ln n \displaystyle \sum_{n=2}^{\infty}\frac1{n\,\ln n} n = 2 ∑ ∞ n ln n 1
Let f ( x ) = 1 x ln x f(x)=\frac1{x\ln x} f ( x ) = x l n x 1 x ≥ 2 x\ge 2 x ≥ 2
Improper integral:
∫ < e m > 2 ∞ 1 x ln x d x = lim < / e m > t → ∞ ∫ 2 t 1 x ln x d x . \int<em>{2}^{\infty}\frac1{x\ln x}\,dx=\lim</em>{t\to\infty}\int_{2}^{t}\frac1{x\ln x}\,dx. ∫ < e m > 2 ∞ x l n x 1 d x = lim < / e m > t → ∞ ∫ 2 t x l n x 1 d x .
Use u = ln x ⇒ d u = 1 x d x u=\ln x\Rightarrow du=\frac{1}{x}dx u = ln x ⇒ d u = x 1 d x ln ( ln x ) \ln(\ln x) ln ( ln x )
Limit diverges to + ∞ +\infty + ∞
Therefore series diverges (by integral test).
“Challenge Problem” Mentioned Invent alternative hypotheses (other than “continuous, positive, decreasing”) that still guarantee the integral test conclusion.
• Suggested idea: f ∈ C 1 , ∣ f ′ ( x ) ∣ ≤ C f ( x ) f\in C^1,\ |f'(x)|\le C\,f(x) f ∈ C 1 , ∣ f ′ ( x ) ∣ ≤ C f ( x ) x x x
Purely conceptual; not examinable but deepens understanding.
Direct Comparison Test (Series Version) Statement Let 0 ≤ a < e m > n ≤ b < / e m > n 0\le a<em>n\le b</em>n 0 ≤ a < e m > n ≤ b < / e m > n n n n
Practical Tips Real-world sequences rarely satisfy a < e m > n ≤ b < / e m > n a<em>n\le b</em>n a < e m > n ≤ b < / e m > n every n n n n ≥ N n\ge N n ≥ N
Finite truncations never affect convergence because adding/removing finitely many terms just shifts partial sums by a constant.
Worked Example 2 Test ∑ k = 1 ∞ ln ( k + 10 ) k \displaystyle \sum_{k=1}^{\infty}\frac{\ln(k+10)}{k} k = 1 ∑ ∞ k ln ( k + 10 )
For k ≥ 1 k\ge1 k ≥ 1 ln ( k + 10 ) ≥ 1 \ln(k+10)\ge1 ln ( k + 10 ) ≥ 1 ln ( k + 10 ) k ≥ 1 k . \frac{\ln(k+10)}{k}\ge\frac1k. k l n ( k + 10 ) ≥ k 1 .
∑ 1 k \sum\frac1k ∑ k 1 diverges .
Note: shift “+ 10 +10 + 10
Concept of Finite Truncation (Formal Justification) Given b < e m > n = a < / e m > n ( n ≥ N ) , b < e m > n = 0 ( n k = 1 n a < / e m > k = T < e m > n + ∑ < / e m > k = 1 N − 1 a k b<em>n=a</em>n\,(n\ge N),\ b<em>n=0\,(n{k=1}^{n}a</em>k=T<em>n+\sum</em>{k=1}^{N-1}ak b < e m > n = a < / e m > n ( n ≥ N ) , b < e m > n = 0 ( n k = 1 n a < / e m > k = T < e m > n + ∑ < / e m > k = 1 N − 1 ak S < e m > n S<em>n S < e m > n T < / e m > n T</em>n T < / e m > n
*\n
Statement For positive sequences a < e m > n , b < / e m > n a<em>n,b</em>n a < e m > n , b < / e m > n
c = lim < e m > n → ∞ a < / e m > n b < e m > n , 0 n \displaystyle c=\lim<em>{n\to\infty}\frac{a</em>n}{b<em>n},\qquad 0n c = lim < e m > n → ∞ b < e m > n a < / e m > n , 0 n ∑ b ∗ n \sum b*n ∑ b ∗ n
*\n
Worked Example 3 Test ∑ n = 1 ∞ n + sin n n 2 + 1 . \displaystyle \sum_{n=1}^{\infty}\frac{n+\sin n}{n^{2}+1}. n = 1 ∑ ∞ n 2 + 1 n + sin n .
Dominant terms: numerator ∼ n \sim n ∼ n ∼ n 2 \sim n^{2} ∼ n 2 b n = n n 2 = 1 n . b_n=\frac{n}{n^{2}}=\frac1n. b n = n 2 n = n 1 .
Compute limit:
a n b n = n + sin n n 2 + 1 ⋅ n = 1 + sin n n 1 + 1 n 2 → n → ∞ 1. \frac{an}{bn}=\frac{n+\sin n}{n^{2}+1}\cdot n=\frac{1+\frac{\sin n}{n}}{1+\frac1{n^{2}}} \xrightarrow{n\to\infty}1. bn an = n 2 + 1 n + s i n n ⋅ n = 1 + n 2 1 1 + n s i n n n → ∞ 1.
Since c=1\in(0,\infty)a n d and an d d i v e r g e s , o r i g i n a l s e r i e s < / e m > < s t r o n g > < e m > d i v e r g e s < / e m > < / s t r o n g > < e m > b y L C T . < / e m > < / p > < / l i > < / o l > < h 5 i d = " f 15931 c 9 − 3826 − 4338 − 8 c 14 − 59 d 29 d d b d d 0 c " d a t a − t o c − i d = " f 15931 c 9 − 3826 − 4338 − 8 c 14 − 59 d 29 d d b d d 0 c " c o l l a p s e d = " f a l s e " s e o l e v e l m i g r a t e d = " t r u e " > < e m > S t r a t e g y f o r S o l v i n g C o n v e r g e n c e P r o b l e m s < / e m > < / h 5 > < o l > < l i > < p > < s t r o n g > < e m > D i v e r g e n c e T e s t f i r s t < / e m > < / s t r o n g > < e m > ( d o e s diverges, original series </em><strong><em>diverges</em></strong><em> by LCT.</em></p></li></ol><h5 id="f15931c9-3826-4338-8c14-59d29ddbdd0c" data-toc-id="f15931c9-3826-4338-8c14-59d29ddbdd0c" collapsed="false" seolevelmigrated="true"><em>Strategy for Solving Convergence Problems</em></h5><ol><li><p><strong><em>Divergence Test first</em></strong><em> (does d i v er g es , or i g ina l ser i es < / e m >< s t ro n g >< e m > d i v er g es < / e m >< / s t ro n g >< e m > b y L CT . < / e m >< / p >< / l i >< / o l >< h 5 i d = " f 15931 c 9 − 3826 − 4338 − 8 c 14 − 59 d 29 dd b dd 0 c " d a t a − t oc − i d = " f 15931 c 9 − 3826 − 4338 − 8 c 14 − 59 d 29 dd b dd 0 c " co ll a p se d = " f a l se " seo l e v e l mi g r a t e d = " t r u e " >< e m > St r a t e g y f or S o l v in g C o n v er g e n ce P ro b l e m s < / e m >< / h 5 >< o l >< l i >< p >< s t ro n g >< e m > D i v er g e n ce T es t f i rs t < / e m >< / s t ro n g >< e m > ( d oes ? ) . < / e m > < / p > < / l i > < l i > < p > < e m > I d e n t i f y d o m i n a n t b e h a v i o u r : a l g e b r a i c ( ?).</em></p></li><li><p><em>Identify dominant behaviour: algebraic ( ?) . < / e m >< / p >< / l i >< l i >< p >< e m > I d e n t i f y d o minan t b e ha v i o u r : a l g e b r ai c ( ) , l o g a r i t h m i c , e x p o n e n t i a l , t r i g . < / e m > < / p > < / l i > < l i > < p > < e m > C h o o s e b e t w e e n : < b r > < / e m > < / p > < p > < e m > • I n t e g r a l t e s t ( n i c e a n t i d e r i v a t i v e ) . < b r > < / e m > < / p > < p > < e m > • D i r e c t c o m p a r i s o n ( i n e q u a l i t y o b v i o u s ) . < b r > < / e m > < / p > < p > < e m > • L i m i t c o m p a r i s o n ( r a t i o s e a s i e s t ) . < b r > < / e m > < / p > < p > < e m > • O t h e r t e s t s ( r a t i o , r o o t , a l t e r n a t i n g ) t o b e s t u d i e d l a t e r . < / e m > < / p > < / l i > < l i > < p > < e m > D o n ’ t f o r g e t f i n i t e t r u n c a t i o n s a n d c o m b i n i n g m u l t i p l e t e s t s p i e c e − w i s e . < / e m > < / p > < / l i > < / o l > < h 5 i d = " 04 c 30 f 28 − d e 7 d − 42 a f − 9 b 37 − 1 f 1 a 007 d 9 f b d " d a t a − t o c − i d = " 04 c 30 f 28 − d e 7 d − 42 a f − 9 b 37 − 1 f 1 a 007 d 9 f b d " c o l l a p s e d = " f a l s e " s e o l e v e l m i g r a t e d = " t r u e " > < e m > S e r i e s o f F u n c t i o n s ( P r e v i e w / M o t i v a t i o n ) < / e m > < / h 5 > < u l > < l i > < p > < e m > O f t e n n e e d t o a p p r o x i m a t e c o m p l i c a t e d ), logarithmic, exponential, trig.</em></p></li><li><p><em>Choose between: <br></em></p><p><em>• Integral test (nice antiderivative). <br></em></p><p><em>• Direct comparison (inequality obvious). <br></em></p><p><em>• Limit comparison (ratios easiest). <br></em></p><p><em>• Other tests (ratio, root, alternating) to be studied later.</em></p></li><li><p><em>Don’t forget finite truncations and combining multiple tests piece-wise.</em></p></li></ol><h5 id="04c30f28-de7d-42af-9b37-1f1a007d9fbd" data-toc-id="04c30f28-de7d-42af-9b37-1f1a007d9fbd" collapsed="false" seolevelmigrated="true"><em>Series of Functions (Preview / Motivation)</em></h5><ul><li><p><em>Often need to approximate complicated ) , l o g a r i t hmi c , e x p o n e n t ia l , t r i g . < / e m >< / p >< / l i >< l i >< p >< e m > C h oose b e tw ee n :< b r >< / e m >< / p >< p >< e m > • I n t e g r a lt es t ( ni ce an t i d er i v a t i v e ) . < b r >< / e m >< / p >< p >< e m > • D i rec t co m p a r i so n ( in e q u a l i t yo b v i o u s ) . < b r >< / e m >< / p >< p >< e m > • L imi t co m p a r i so n ( r a t i ose a s i es t ) . < b r >< / e m >< / p >< p >< e m > • Ot h er t es t s ( r a t i o , roo t , a lt er na t in g ) t o b es t u d i e d l a t er . < / e m >< / p >< / l i >< l i >< p >< e m > Do n ’ t f or g e t f ini t e t r u n c a t i o n s an d co mbinin g m u lt i pl e t es t s p i ece − w i se . < / e m >< / p >< / l i >< / o l >< h 5 i d = "04 c 30 f 28 − d e 7 d − 42 a f − 9 b 37 − 1 f 1 a 007 d 9 f b d " d a t a − t oc − i d = "04 c 30 f 28 − d e 7 d − 42 a f − 9 b 37 − 1 f 1 a 007 d 9 f b d " co ll a p se d = " f a l se " seo l e v e l mi g r a t e d = " t r u e " >< e m > S er i eso f F u n c t i o n s ( P re v i e w / M o t i v a t i o n ) < / e m >< / h 5 >< u l >< l i >< p >< e m > O f t e nn ee d t o a pp ro x ima t eco m pl i c a t e d b y < / e m > < s t r o n g > < e m > i n f i n i t e s u m s o f s i m p l e r f u n c t i o n s < / e m > < / s t r o n g > < e m > . E x a m p l e s : < b r > < / e m > < / p > < p > < e m > • < / e m > < s t r o n g > < e m > T a y l o r s e r i e s < / e m > < / s t r o n g > < e m > : by </em><strong><em>infinite sums of simpler functions</em></strong><em>. Examples: <br></em></p><p><em>• </em><strong><em>Taylor series</em></strong><em>: b y < / e m >< s t ro n g >< e m > in f ini t es u m so f s im pl er f u n c t i o n s < / e m >< / s t ro n g >< e m > . E x am pl es :< b r >< / e m >< / p >< p >< e m > • < / e m >< s t ro n g >< e m > T a y l orser i es < / e m >< / s t ro n g >< e m >: ( p o l y n o m i a l a p p r o x i m a t i o n ) . < b r > < / e m > < / p > < p > < e m > • < / e m > < s t r o n g > < e m > F o u r i e r s e r i e s < / e m > < / s t r o n g > < e m > : s u m s o f s i n e s / c o s i n e s f o r s i g n a l a n a l y s i s . < / e m > < / p > < / l i > < l i > < p > < e m > B e n e f i t s : p o l y n o m i a l s / t r i g f u n c t i o n s a r e e a s y t o d i f f e r e n t i a t e , i n t e g r a t e , a n d a n a l y s e ; p a r t i a l s u m d e g r e e (polynomial approximation). <br></em></p><p><em>• </em><strong><em>Fourier series</em></strong><em>: sums of sines/cosines for signal analysis.</em></p></li><li><p><em>Benefits: polynomials/trig functions are easy to differentiate, integrate, and analyse; partial sum degree ( p o l y n o mia l a pp ro x ima t i o n ) . < b r >< / e m >< / p >< p >< e m > • < / e m >< s t ro n g >< e m > F o u r i erser i es < / e m >< / s t ro n g >< e m >: s u m so f s in es / cos in es f ors i g na l ana l ys i s . < / e m >< / p >< / l i >< l i >< p >< e m > B e n e f i t s : p o l y n o mia l s / t r i g f u n c t i o n s a ree a sy t o d i ff ere n t ia t e , in t e g r a t e , an d ana l yse ; p a r t ia l s u m d e g ree
Miscellaneous Remarks & Practical Advice Integral test fails for oscillatory or sign-changing functions; ensure positivity and monotonicity.
Repeated practice is essential; convergence testing is “an art.”
When stuck, analyse long-term form, strip lower-order terms, and search for a familiar