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[PDF] Week 6 - School of Mathematics and Statistics, University of Sydney

n=3 1 nlog n(log(log n))p ; Solution: We again use the Cauchy condensation test and observe that it is sufficient to prove the convergence properties of ∞ ∑

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so ∑n≥4 1/n log n log log n diverges by the integral test Alternatively, by the Cauchy condensation test, ∑n≥4 1/n log n log log n converges if and only

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1 n log n = 1 2 log 2 + 1 3 log 3 + ··· + 1 N log N Utiliser le résultat de la question précédente pour obtenir une minoration de cette somme : ∫ N 2

[PDF] Analysis 1

says, for n ≥ 6: pn < n log(n) + n log(log(n)) Use this to prove the divergence of the prime harmonic series: ∞ ∑ n=1 1 pn Note: Isn't this a little bit impressive?

[PDF] An O(logn/log logn)-approximation Algorithm for - MIT Mathematics

O(log n/ log log n) of the optimum with high probability 1 Introduction In the Asymmetric Traveling Salesman problem (ATSP) we are given a set V of n points  

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Traveling salesman problem is one of the most celebrated and intensively studied problems in combinato- rial optimization [30, 2, 13] It has found applications in 

[PDF] also called Cauchys condensation test - mathchalmersse

3 mai 2018 · to integral test, but in many cases the condensation test requires less work than the integral n log n diverges (c) With an = 1/n2, we have ∞ ∑ n=1 2na2n = ∞ ∑ n=1 2n 1 n=16 1 n log 2(log n log 2)(log log n log 2)p

[PDF] MATH 370: Homework 5

where M > 0 exists because the sequence log n n has a limit (= 0); hence, the original series converges (e) ∞ ∑ n=4 1 n(log n)(log log n) diverges b/c ∫ ∞

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Analysis 1

Some extra, or interesting, or challenging questions

Seehttp://www.maths.dur.ac.uk/users/steven.charlton/analysis1_1314for updates.Below are come extra, or interesting, or challenging questions related to, or extending, the

concepts and results covered in the Analysis 1 course.

Questions marked with:

?are tricky, but require nothing more than what is covered in the course are more difficult - they ma yrequire some clev erinsigh t,or tric k ?require using results from a previous question, so may exceed the scope of the course

1 Limit Computations

Q1) Calculate limn→∞xnfor the following sequences: i)xn=3⎷n

3+ 2n2+ 7n-3-3⎷n

3-3n2-6n+ 4.

Hint:Make use ofa3-b3= (a-b)(a2+ab+b2)

ii)

More generally xn=3⎷n

3+an2+bn+c-3?n

3+dn2+en+f

Q2) Use the squeezing theorem to ev aluatelimn→∞xnfor the following sequences: i)xn=n2+ cos(n/5)2n2-sin(n2) ii)xn=?3cosn+ 4sinn6 n Hint:Can you write3cosn+ 4sinnin the formRsin(n+θ)? iii)xn= (cosh(n))1/n

Hint:Squeezelogxn.

Q3) GivenStirling"s approximationlimn→∞n!⎷2πn?ne n= 1, computelimn→∞xnfor the following sequences: i)xn=n(3n)!3

3n(n!)3

ii)xn=24nn ?2n n? 2

2?-NProofs of Limit Computations

Q1) Evaluatelimn→∞xn, and then give a formal?-Nproof of this result, for the following sequences: 1 i)xn=n+ 5n 2+ 3 ii)xn=2n2+ 7n

2+ 2n+ 1

iii)xn=3n3-2n+ 7n

4+ 7n+ 1

iv)xn=3 + 2n22-n+n2 Q2)Evaluatelimn→∞xn, and then give a formal?-Nproof of this result, for the following sequences: i)xn=?2 +x1 +x ii)xn=3cos(n) + 2n22-n+n2 iii)xn=3 + 2n22-nsin(n) +n2

3 Limit Theorems

Q1)COLT:

Ifxn→Landyn→Masn→ ∞, use the?-Ndefinition of limits to prove that asn→ ∞: i)xnyn→LM Hint:Rewritexnyn-LMas(xn-L)(yn-M) +L(yn-M) +M(xn-L). ii) 1y n→1M (assumingMand allynare non-zero) Q2) If an→Lasn→ ∞, give direct?-Nproofs of the following: i)a2n→L2(don"t just appeal to calculus of limits) ii) ⎷a n→⎷L(assumingan≥0for alln) iii)logan→logL(assumingan>0for alln, andL >0) iv)exp(an)→exp(L) Q3)Cesàro Mean:Suppose that the sequencexn→Lasn→ ∞. Consider the sequence y n:=1n (x1+x2+···+xn).

Prove thatlimn→∞yn=L, as well.

Hint: Sincexn→L, findN0such thatn≥N0implies|xn-L|< ?/2. Split up the sum in ynat the fixed numberN0. Q4) Use the result in the previous question to evaluatelimn→∞ynfor the following sequences: 2 i)yn=1n log?11/121/231/3···n1/n?

Can you evaluate any of these more directly?

Q5)Supposeanis a sequence such thatlimn→∞? ???a n+1a n? ???=L. Prove thatlimn→∞n?|an|=L, as well. Does the converse hold? Note: This says that whenever the ratio test succeeds (reaches a conclusion on convergence or divergence) then the root test (see Section 5, Q1) below) will also succeed (and reach the same conclusion, of course). So the root test isstrongerthan the ratio test. Q6) Use the result in the previous question to evaluatelimn→∞xnfor the following sequences: i)xn=n?1 n! ii)xn=nn ⎷n! iii)xn=n1n

Can you evaluate any these more directly?

Q7)Tail of a sequence:

It"s been mentioned several times that changing the first few terms of a sequence does not change its convergence, or the limit. Let"s make this precise: Suppose xn→Lasn→ ∞. Sety1=A1,y2=A2,...,yM-1=AM-1andyn=xnforn≥M.

Give an?-Nproof thatyn→Ltoo.

Q8) It"s also been mentioned that 'shifting" a sequence does not change its convergence, or the limit, either. Let"s make this precise: Supposexn→Lasn→ ∞. Setyn=xn+k. Give an ?-Nproof thatyn→Ltoo.

4 Completeness

Q1) Recall the completeness ofR: every non-empty subsetS?Rwhich is bounded above has a supremumsupS?R. Show that this does not hold ifRis replaced byQby looking at the following example:

S={x?Q|x2<2}

Specifically, show that noq?Qcan be the least upper bound ofS. So we can concludeQ is not complete.

Q2)Cauchy Criterion:

A sequencexnis calledCauchyif: given any? >0, there existsN such that forn,m≥N, we have|xn-xm|< ?. (Eventually all terms are within?of each other.) Working inR, prove the following are equivalent: a)

The sequence xnconverges (toL?R).

b)

The sequence xnis Cauchy.

3 Hint:For a)=?b), writexn-xmas(xn-L) + (L-xm), and use the?-Ndefinition of xn→L. (With what choice of??) For b)=?a), first show{xn}is bounded. Then use Bolzano-Weierstrass to find a convergent subsequencexni→L. And finally showxn→L, itself. Note: a)=?b) always holds, but showing b)=?a) requires using the completeness ofR. That every Cauchy sequence converges can be taken as adefinitionof completeness. You might meet the concept of a Cauchy sequence again Complex Analysis 2.

5 Infinite Series Convergence Tests

Q1)Root Test:

Let?∞

n=1anbe a series, and suppose thatL=limn→∞n?|an|exists. Prove that: •ifL <1then the series converges absolutely, •ifL >1then the series diverges. Can you say anything ifL= 1? What if you also know thatn?|an| →1from above, i.e. eventually alln?|an| ≥1?

Hint:Imitate the proof of the ratio test.

Q2) Use the ro ottes tab oveto dete rminewhethe ror not the follo wingseries con verge: i) n=21log(n)n ii) n=2? log?3n2+nn

2-2n+ 1??

n iii) n=3? log?2n3-5n2n

3+ 4n-3??

n iv) n=1?

3n6sin3?2n+ 73n3+ 5n2-6n-1??

n

Q3)Generalised Comparison:

for alln?Z>0. Prove that: •If the series?∞ n=1anand?∞ n=1cnboth converge, then?∞ n=1bnalso converges. •If the series?∞ n=1andiverges to+∞, then?∞ n=1bnalso diverges to+∞. •If the series?∞ n=1cndiverges to-∞, then?∞ n=1bnalso diverges to-∞. Hint: test applicable? For the second and third, partial sums? Note: Since you haven"t seen this in lectures, you maynotuse this when solving homework or exam questionsunlessyou prove it first. If youreallyfeel the need to use this generalisation, it may be worth looking at?∞ n=1|xn|instead. 4 Q4)Recall the Alternating Sign Test from lectures: It says that if the sequenceynis positive, decreasing, andlimn→∞yn= 0, then the alternating series?∞ n=1(-1)nynconverges. i) Firstly, show that the condition thatynis positive can be removed: Ifynis a decreasing sequence, andlimn→∞yn= 0, then the alternating series?∞ n=1(-1)nynconverges. Hint: Useynis decreasing, andlimn→∞yn= 0to showyn≥0for alln. What happens ifyN= 0for someN, and otherwise? ii) Now show that the other two hypotheses are necessary, even if you restrict toyn positive. That is, find a sequenceanwhich is positive, decreasing, but doesn"t have limit 0, such that the alternating series?∞ n=1(-1)nandoesn"t converge. And find another sequencebnwhich is positive, has limit 0, but is not decreasing, such that the alternating series?∞ n=1(-1)nbnalso doesn"t converge. Note: This is a nice example of how mathematicians think. When told of a theorem requiring many hypotheses, a mathematician will naturally wonder whether all the hypotheses are necessary. Where are these conditions used in the proof? Does theorem fail if I weaken or remove any of the hypotheses? Can I generalise the theorem in any way?

Q5)Cauchy Condensation Test:

Supposeanis a positive, decreasing sequence. By showing: n=1a n=12 n=1a n, conclude that: n=1a nconverges??∞? n=12 na2nconverges

Hint:Write out the partial sums.

Note: This generalises the proof from lectures that the harmonic series diverges. Taking an=1n, one gets?∞ n=11nconverges if and only if?∞ n=12n12 n=?∞ n=11converges, and the latter series obviously diverges. You willnotneed to use this in any homework, tutorial of exam questions. Just consider it a glimpse of the vast zoo of more specialised convergence tests. Q6) In lectures you used the integral test to show the following series diverges. This time, use

Cauchy"s Condensation Test to show it diverges:

n=21nlog(n) 5

6 Infinite Series Convergence

Q1)

T rueor false?

i) n=11n 1+1n converges since1 +1n >1. ii) n=1? 1-1n 2? n2 converges. Q2)

Do the fol lowingseries con verge?

i) n=1(n-1)!n n-1? 197
n-1

Hint:You may assume that197

< e. ii) n=11log(n!)

Hint:How donnandn!compare?

iii) n=1log?n+ 1n

Hint:What are the partial sums?

Q3) Use the in tegraltest to determine whether the follo wingseries con verge: i) n=31nlog(n) log(log(n)) ii) n=21nlog(n)2 iii) n=21n ?log(n) iv) n=31nlog(n) log(log(n))1/3 v) n=151nlog(n) log(log(n)) log(log(log(n))) Considering the above examples, can you guess a more general result? Can you prove it? Note:Series like the above, and things involving nested logarithms appear a lot in analytic number theory. This leads to the appalling joke: What sound does a drowning analytic number theorist make? Log, log, log, ... 6 Q4)Linking to analytic number theory... Ifpnis then-th prime number, Dusart"s inequality says, forn≥6: p n< nlog(n) +nlog(log(n)). Use this to prove the divergence of theprime harmonic series: n=11p n Note: Isn"t this a little bit impressive? Even after getting rid of all the composite numbers from the harmonic series, the sumstilldiverges. So in some sense you might say there are more primes than squares, since?∞ n=11n

2converges.

Q5) Find all v aluesof pfor which the following series converges: n=21nlog(n)p Q6) You can determine whether the following series converge or diverge using other means, but for this question use the integral test to determine whether they converge or diverge. Make sure the function satisfies all the hypotheses of the integral test! i) n=1n

2exp(-n)

ii) n=1log(n)n 2 iii) n=111 +n2 iv) n=112 ⎷x

7 Infinite Series Theorems

Q1)Tail of a series:

It"s been mentioned that only the tail of an infinite series is important when discussing convergence. Make this precise: By looking at the partial sums, show that?∞ n=1xnconverges if and only?∞ n=Mxnconverges.

Q2)Riemann Series Theorem:Let?∞

n=1anbe a conditionally convergent series. i) Show that a conditionally convergent series must have an infinite number of positive terms and an infinite number of negative terms. ii)

T ake:

a +n:=? a nifan≥0

0otherwise

a -n:=?

0ifan≥0

a notherwise, 7 so thata+nkeeps the positive terms and replaces the negative terms with 0s, while

a-nkeeps the negative terms and replaces the positive terms with 0s. Observe that:?an=?a+n+?a-n. Use this to conclude that?∞

n=1a+nand?∞ n=1a-ndiverge to +∞and-∞respectively. iii) Pick anyM?R. Use the previous part to show that you can rearrange the terms of the sequenceanto get a new sequencebnwhose sum?∞ n=1bn=M. Hint: For anyr >0?R, since?a+ndiverges to+∞, the partial sums are eventually greaterr. Once you have added enough positive terms to exceedM, make use of the negative terms. (Convince yourself that you"ve used every termanonce and only once.) iv)

Can you do the same but get that?∞

n=1bndiverges to+∞, or diverges to-∞, or diverges by oscillation1 Note: This question shows how dangerously, badly wrong some naïve ways of manipulating infinite series can go: you can"t rearrange a convergent series and be sure you"ve still got convergent series, let alone one with the same sum, at least for general infinite series. However, it can be shown that when you rearrange an absolutely convergent series, the rearranged series always converges and it converges to thesamesum as the original series. (Absolutely convergent series are very well behaved.)

8 Power Series

Q1)Dilogarithm and polylogarithms:

Recall the Taylor series for-log(1-x)is given by

-log(1-x) =∞? n=1x nn and this converges for|x|<1. By replacingnwithn2we get the definition of thedilogarithm function Li

2(x):=∞?

n=1x nn 2. More generally thep-th polylogarithm(forp= 1,2,3,...), which is defined by Li p(x):=∞? n=1x nn p. i) What is the radius of con vergenceof the p owerser iesdefinin gLip(x)? ii) Using the fact that you can differentiate a power series term-by-term, findddxLip(x) (in terms ofLip-1). iii)

Chec kthe fol lowingiden tityfor the dilogarithm

Li

2(x2) = 2(Li2(x) + Li2(-x))

(Try adding the power series together.)1

That is, the series diverges but the partial sums don"t go to±∞, the partial sums remain bounded.

8 iv)A similar identity holds for every polylogarithm. What should the coefficientλbe to make the following identity true? Li p(x2) =λ(Lip(x) + Lip(-x)) Note: The dilogarithm and higher polylogarithm functions are of considerable interest to both number theorists and particle physicists. One of the main focuses is finding 'functional equations", identities like the two examples above. For example, the fundamental property of the (usual) logarithm is that it turns multiplication into addition log(xy) = log(x) + log(y). (You can write this in terms ofLi1(x) =-log(1-x)if you want.) Similarly, the dilogarithm satisfies the so-called '5-term" equation Li

2(x) + Li2(y) + Li2?

1-x1-xy?

+ Li

2(1-xy) + Li2?1-y1-xy?

26
-log(x)log(1-x)-log(y)log(1-y) + log?

1-x1-xy?

log?1-y1-xy? This equation is the fundamental property of the dilogarithm. (Where doesπ26come from?

Try looking up the Basel Problem.)

Physicists are interested in these functional equations because of how drastically they can simplify answers to certain computations. An answer which used to span 14 pages can be condensed to just 4 lines with good knowledge of the properties of polylogarithms!

Polylogarithmfor much more information.

9 Riemann Integrals

Q1) LetPn([a,b])be the partition of the interval[a,b]intonequal parts. Write down the upper and lower Riemann sums for the given function, on the given partition of the given interval, and numerically evaluate the results: i)f(x) =xforP5([0,3]) ii)f(x) =xforP10([0,3]) iii)f(x) =x2forP4([0,1]) iv)f(x) =x2forP5([0,1]) v)f(x) =x2forP6([0,1]) vi)f(x) =⎷xforP4([0,4]) vii)f(x) =⎷xforP5([0,4]) viii)f(x) =⎷xforP6([0,4]) ix)f(x) =exforP5([1,2]) 9 x)f(x) =exforP6([1,2]) xi)f(x) = logxforP5([1,2]) xii)f(x) = logxforP5([1,2]) xiii)f(x) = sinxforP5([0,π]) xiv)f(x) = sinxforP5([0,π]) Q2)Write down the upper and lower Riemann sums for the given function on the partition Pn([a,b])of the interval[a,b]intonequal parts. Use this to show that the function is

Riemann integrable on the given interval

i)f(x) =xon[0,1] ii)f(x) =xon[0,2] iii)f(x) =xon[0,a iv)f(x) =x2on[0,1] v)f(x) =x2on[0,2] vi)f(x) =x2on[0,a] vii)f(x) = sin(x)on[0,π](For ease, do this forP2n+1([0,π]).) Q3) By viewing the following as a Riemann sum for a particular function on some interval, express the following limits as a definite integral, and evaluate the integral to find the limit: i)limn→∞?n i=11n ?i n ii)limn→∞?n i=11n ?i-1n iii)limn→∞?n i=11n ?1 +in 2 iv)limn→∞?n i=11n

1 +?in

2? v)limn→∞?n i=11n

1 +?1 +in

3? vi)limn→∞?n i=1nnquotesdbs_dbs20.pdfusesText_26