[PDF] 6.1 Sigma Notation & Convergence / Divergence

View - Download 6.1 Sigma Notation & Convergence / Divergence

Previous PDF

Next PDF

Math 123 - Shields Innite Series Week 6

Zeno of Elea posed some philosophical problems in the 400s BC. His actual reasons for doing so are somewhat

murky, but they are troubling regardless. In his Dichotomy paradox he argues that any objects in motion

must arrive at the halfway point before they arrive at their goal. However, we can iterate the reasoning, and

an object must also arrive at the halfway point to the halfway point rst, and so on:::.

If the goal ismmeters away and you travel at a speed ofv, then the time needed to traverse the distance is

t=x2v+x4v+x8v+x16v+::: xv 12 +14 +18 +116

The parenthetical expression is an innite sum. Zeno said that the time for the motion then must be innite

since if we add innitely many positive numbers together then this must result in a positive number. As

beings of logic we are forced to conclude that motion is an illusion. We have all been duped.

On a completely unrelated note recall from Calc I that for small angleswe can use a local linear approxi-

mation toe. That is e 1 +

This approximation is pretty good, but it's not perfect because the approximation is a line andebends.

Well in a similar way we can make our estimate better by taking a quadratic approximation and say e

1 ++12

2

This is better, but still ultimately an approximation with some built in error. However, we can keep making

this more precise by adding on a cubed term, then a quadratic term, etc. What if we go o our rockers and

decide to add innitely many terms. Then maybe our approximation wouldn't be an approximation at all, but it would become an identity? e = 1 ++12 2+16

3+:::+1n!n+:::

Maybe that could make senes. But right now it doesn't because we haven't dened what an innite series even means. The same thing holds for Mr. Zeno. He may have a point, but we need to be sure we have things dened before we start talking about properties that they may or may not have.

6.1 Sigma Notation & Convergence / Divergence

First we start by discussing nite sums. Consider the sum

1 + 2 + 3 + 4:

It's all well and good to write this out when the sum is short, but it would be frustrating if the sum were

instead

S= 1 + 2 + 3 + 4 + 5 +:::+ 98 + 99 + 100:

Instead we utilizesigma notationand write the above as

S=100X

n=1n:

The variablenis referred to as theindexof the summation. The numbers 1 and 100 are called the bounds.

The notation tells us to evaluate the expression inside the sum at all whole number values of the indexn

inclusively between the lower and upper numbers. More generally for example we have 1

Math 123 - Shields Innite Series Week 6

1 X n=72an=a7+a8+a9+a10+a11+a12: Notice that the indexing is somewhat arbitrary and can be manipulated. For example both expressions below dene the same sum. 3 X n=012 n= 1 +12 +14 +18 5 X n=212 n2= 1 +12 +14 +18

How do we bridge the gap to the innite? Thinking about how we handled innitely nested roots is produc-

tive. An innite sum is naturally realized as a sequence of nite sums which continually grow in length.

DenitionLetfangn=ibe a sequence. We dene thekthpartial sumof the sequence as S k=i+k1X n=ia n=ai+ai+1+ai+2+:::+ai+k2+ai+k2

ExampleThe 4thpartial sum of the sequencef1n

gn=1is S 4=4X n=11n = 1 +12 +13 +14

Note:Thekthpartial sum will containkterms.

DenitionLetfangn=ibe a sequence. By the innite sum1X n=ia nwe mean the sequencen S ko k=1where S

krepresents thekthpartial sum. If the sequence of partial sums converges, that is if limk!1Skexists and is

equal to sum nite valueSthen we say that the sumconvergestoS. Otherwise we say that itdiverges.

ExampleDoes the series1X

n=1nconverge?

Examine the partial sums.

S 1= 1 S

2= 1 + 2 = 3

S

3= 1 + 2 + 3 = 6

S

4= 1 + 2 + 3 + 4 = 10

S n=n(n+ 1)2

We have that limn!1Sn=1so the series diverges.

2

Math 123 - Shields Innite Series Week 6

ExampleDoes the seriesX

n=11n

1n+ 1converge?

Again, look at the sequence of partial sums.

S

1= 112

=12 S

2= 112

+12 13 =23 S

3= 112

+12 13 +13 14 =34 S n= 11n+ 1

We have that lim

n!1Sn= 1. Therefore the series converges to 1.

A large part of the remainder of the course will be devoted to developing and applying tests to determine if

series converge or diverge.

6.2 Divergence Test

Our rst test gives us a condition which guarantees if a given series will diverge. It does not allows us to

ever conclude that a series converges however.

Divergence TestThe the series1X

n=ia nconverges then limn!1an= 0. In particular, if limn!1an6= 0 then the series diverges.

ProofFor convenience assume that the series index begins at 1 and the series converges to the valueS. In

that case S n=a1+a2+:::+ +an1+an S n1=a1+a2+:::+ +an1+an andan=SnSn1. lim n!1an= limn!1SnSn1=SS= 0ExampleDoes the seriesX n=0nn+ 1converge? lim n!1an= limn!1nn+ 1= 16= 0 so the series diverges. 3

Math 123 - Shields Innite Series Week 6

6.3 The Harmonic Series Diverges

As an important example of what the divergence test does not imply we have the harmonic series.

DenitionTheharmonic seriesis the series

1 X n=11n = 1 +12 +13 +14

TheoremThe harmonic series diverges.

ProofLook at the partial sums given by the powers of 2. S 1= 1 S

2=S1+12

S

4= 1 +12

+13 +14 >1 +12 +14 +14 = 1 +12 + 214 = 1 + 212 S

8=S3+15

+16 +17 +18 > S3+ 418 >1 + 312

Continuing this style of estimation we have that

S

2n>1 +n12

This shows that the sequenceSnis not bounded above. SinceSn=Sn1+1n , we can see that the sequence is also increasing. Therefore, it must diverge.Note:limn!11n = 0 and yet the harmonic series diverges. The divergent series test only gives us a condition for when a series will diverge. It cannot tell us that a series converges. 4

Math 123 - Shields Innite Series Week 6

6.4 Geometric Series

DenitionAgeometric seriesis any series which can be written in the form 1 X i=0ar n

Usually we do not care about the value at which our index begins. However, for geometric series we will

assume that the series begin at 0. Geometric series are notably because, unlike most series we will encounter,

they allow us to precisely determine the value of the sum.

Geometric Series TheoremThe geometric series1X

n=0ar nconverges ifjrj<1 and otherwise diverges. The value of the convergent sum is a1r. Proof S n=a+ar+ar2+:::+arn rS n=ar+ar2+:::+arn+arn+1 =)SnrSn=aarn+1=)Sn(1r) =a(1rn+1) =)Sn=a1r(1rn+1)

Now look at the limit, lim

n!1Sn. The only term dependent onnis thern. This converges if and only if

1< r1 as we saw in the sequences section. However, the partial sum is not valid whenr= 1. When

r= 1, the divergence test shows us the series diverges. Therefore the series converges exactly whenjrj<1.

With that assumption, taking the limit we have that

S= limn!1Sn=a1r(10) =a1rExamplesDetermine if the following sums converge or diverge. If they converge, then nd the value.

(i) 1X i=0 12 n

This is geometric witha= 1 andr=12

.jrj<1 so the sum converges to1112 = 2. (ii) 1X i=135 n

This is geometric witha= 3 andr=15

.jrj<1 so the series will converge. However, we cannot use the formula to determine its value. The formula assumes that the sum is indexed beginning at 0. We must rst reindex. 1 X i=135 n=1X i=115 35
n1=15 1 X i=035 n=15 3115
=34 (iii) 1X i=0 nen

This can be rewritten as

1X i=0 e n. Therefore the series is geometric witha= 1 andr=e . Here jrj>1 so the series diverges. 5

Math 123 - Shields Innite Series Week 6

(iv)

Mak esense of 0 :999.

We can view this repeating decimal as the sequence 1X i=1910 n. After reindexing it can be written 1 X i=0110 910
n. This is geometric witha=910 andr=110 . Therefore it converges toquotesdbs_dbs13.pdfusesText_19

[PDF] convergence et divergence optique

[PDF] convergence et divergence définition

[PDF] convergence et divergence suite

[PDF] suite convergente définition

[PDF] dialogue entre un vendeur et un client en anglais

[PDF] guide de conversation espagnol pdf

[PDF] la conversation amoureuse pdf

[PDF] dialogue d amour entre deux amoureux

[PDF] dialogue tragique entre deux amoureux

[PDF] dialogue d'amour triste

[PDF] dialogue d'amour entre une fille et un garcon

[PDF] poème conversation jean tardieu

[PDF] exercice appel téléphonique

[PDF] simulation accueil téléphonique

[PDF] accueil téléphonique exercices