math 1xx3 tutorial 5 - mcmaster university...2018/02/01 · the alternating series test the...

Math 1XX3 Tutorial 5

TA: Lee van Brussel

McMaster University

TA: Lee van Brussel Math 1XX3 Tutorial 5

Info

All slides will be posted (so no need to worry about copyingdown everything).

PLEASE ask questions if you do not understand something!

Today we will cover: The Alternating Series Test, Absolute &Conditional convergence, the Ratio Test, the Root Test andstrategies for testing series.


The Alternating Series Test


Let∑∞

n=1 an be an alternating series, i.e., a series such thatan = (−1)nbn or an = (−1)n−1bn with bn = |an| > 0 for every n.If the alternating series satisfies

bn+1 ≤ bn for all n (1)

limn→∞

bn = 0 (2)

then the series is convergent.



Example

Does the series

∞∑n=1

(−1)n+1 2

n3 + 5

converge or diverge?

Here, bn = 2n3+5

. We want to check first that limn→∞ bn = 0(since its the easiest to do):

limn→∞

bn = limn→∞

2

n3 + 5= lim

n→∞

2n3

1 + 5n3

=0

1 + 0= 0.



Next, we check that bn+1 ≤ bn for all n:

bn+1 =2

(n + 1)3 + 5≤ 2

n3 + 5= bn for all n.

Conditions (1) and (2) are satisfied. By the Alternating SeriesTest,

∑∞n=1(−1)n+1 2

n3+5converges.


Absolute Convergence

Absolute Convergence

A series∑

an is called absolutely convergent if∑|an| is

convergent.

Important Remark: Every absolutely convergent series isconvergent!

In other words, if∑|an| converges, then so does

∑an.

There are serveral ways to check if a series is absolutely convergent:

(1) The Ratio Test,

(2) The Root Test.


The Ratio Test

The Ratio Test

(i) If limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = L < 1, then the series∑∞

n=1 an is absolutely

convergent (and thus, convergent).

(ii) If limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = L > 1, then the series∑∞

n=1 an is divergent.

(iii) If limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = 1, then the Ratio Test is inconclusive (can’t

use it!)


The Ratio Test

Example

Does the series

∞∑n=1

n45n−1

(−7)n


We will apply the Ratio Test with an =n45n−1

(−7)n.


The Ratio Test

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

∣∣∣∣(n + 1)45n

(−7)n+1· (−7)n

n45n−1

∣∣∣∣= lim

n→∞

5(n + 1)4

7n4

=5

7limn→∞

(n + 1)4

n4

=5

7· 1

< 1.

We conclude by the Ratio Test that the series∑∞

n=1n45n−1

(−7)n isabsolutely convergent, and therefore convergent.


The Root Test

The Root Test

(i) If limn→∞

n√|an| = L < 1, then the series

∑∞n=1 an is absolutely

convergent (and thus convergent).

(ii) If limn→∞

n√|an| = L > 1 or lim

n→∞n√|an| =∞, then the series∑∞

n=1 an is divergent.

(iii) If limn→∞

n√|an| = 1, the Root Test is inconclusive.


The Root Test

Example

Does the series

∞∑n=1

(−2n

n + 1

)5n


We will apply the Root Test with an =

(−2n

n + 1

)5n

.


The Root Test

limn→∞

n√|an| = lim

n→∞

∣∣∣∣∣(−2n

n + 1

)5n∣∣∣∣∣1/n

= limn→∞

(2n

n + 1

)5n/n

= 25 limn→∞

(n

n + 1

)5

= 25 · 1

> 1.

We conclude by the Root Test that∑∞

n=1

(−2nn+1

)5ndiverges.


Conditional Convergence


A series∑

an is called conditionally convergent if it isconvergent but not absolutely convergent.

Example

Show that the series

∞∑n=2

(−1)n

ln n

is conditionally convergent.

We want to show that∑∞

n=2(−1)nln n converges but the series∑∞

n=2

∣∣∣ (−1)nln n

∣∣∣ =∑∞

n=21

ln n diverges.



To show that the series converges, we can use the AlternatingSeries Test. First, we have

limn→∞

bn = limn→∞

1

ln n= 0.

Also, since the logarithm is an increasing function, we immediatelyhave ln n ≤ ln(n + 1) for every n. Therefore

bn+1 =1

ln(n + 1)≤ 1

ln n= bn for every n.

By the Alternating Series Test,∑∞

n=2(−1)nln n converges.

Now we will show∑∞

n=21

ln n diverges.



Notice that for every n, we have ln n ≤ n. Therefore 1n ≤

1ln n for

every n which gives

∞∑n=2

1

n≤∞∑n=2

1

ln n.

But the series on the left is the Harmonic Series with the first termmissing.

Therefore, the series on the left diverges and so by the comparisontest,

∑∞n=2

1ln n diverges.

This shows that the series∑∞

n=2(−1)nln n is conditionally convergent.


Strategy for Testing Series

Say you’re given some series∑

an and asked whether it convergesor diverges.

Without being told what test to use, how do we know where tostart?

Unfortunately there are no “rules” to tell us. We have to be clever!

However, there are some useful suggestions (please see section11.7 in the textbook).

Let’s try figuring out some problems without being told what touse.


Problem #1

Determine whether the series

∞∑n=1

1π2 + arctan(πn)

converges or diverges.

Whenever you see a nasty looking series like this, check to see iflimn→∞ an 6= 0 first. In our case,

limn→∞

an = limn→∞

1π2 + arctan(πn)

=1

π2 + π

2

=1

π6= 0.

By the Test for Divergence, we get that the series diverges.


Problem #2

Does the series

∞∑n=1

(−1)nπ2n

(2n)!


The above is an alternating series, so we could try the AlternatingSeries Test.

However, showing that bn+1 ≤ bn for every n could get a littlemessy and take some time.

But since the terms of the series involve powers of n and factorials,this is our clue to use the Ratio Test.


Problem #2

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

∣∣∣∣∣(−1)n+1 π2(n+1)

(2(n + 1))!· (−1)n

(2n)!

π2n

∣∣∣∣∣= lim

n→∞

π2n+2

(2n + 2)!· (2n)!

π2n

= limn→∞

π2

(2n + 2)(2n + 1)

= 0

< 1.

By the Ratio Test, the series∑∞

n=1(−1)n π2n

(2n)! converges absolutelyand thus, converges.


Problem #3

Determine whether the series

∞∑n=1

√n2 + 1

n3

converges or diverges.

It is easy to see that

limn→∞

√n2 + 1

n3= 0.

So the Test for Divergence tells us nothing.


Problem #3

By pulling out the largest factor of n from the top and bottom, wehave:

√n2 + 1

n3=

n

n3

√1 +

1

n2=

1

n2

√1 +

1

n2.

This is our clue to use the Limit Comparison Test with the

convergent p-series∑∞

n=1 1/n2 since limn→∞

√1 + 1

n2= 1.

Taking an =√n2+1n3

and bn = 1n2

,

limn→∞

anbn

= limn→∞

√n2+1n3

1n2

= limn→∞

√n2 + 1

n3· n2 = lim

n→∞

√n2 + 1

n= 1 > 0.

By the Limit Comparison Test,∑∞

n=1

√n2+1n3

converges.


Problem #4

Does the series

∞∑n=1

ne−n2


Our clue for this one comes from noticing that the function

f (x) = xe−x2

is very easy to integrate.

If we can show that f is continuous, positive and decreasing on theinterval [1,∞) we may use the Integral Test.


Problem #4

Since x and e−x2

are continuous and positive on the interval[1,∞) their product f (x) = xe−x

2is continuous and positive on

the interval [1,∞).

Decreasing Check:

f ′(x) = e−x2(1− 2x2) < 0 for all x ≥ 1.

So we may use the Integral test.


Problem #4

∫ ∞1

xe−x2dx = lim

t→∞

∫ t

1xe−x

2dx

= limt→∞

1

2

(1

e− 1

et2

)=

1

2e

where we used u-substitution with u = −x2.

Since the improper integral converges, the series∑∞

n=1 ne−n2

converges by the Integral Test.


Problem #4

How large must N be so that

RN =∞∑n=1

ne−n2 −

N∑n=1

ne−n2< 10−5?

Recall the useful estimation∫ ∞N+1

xe−x2dx ≤ RN ≤

∫ ∞N

xe−x2dx .

We will set∫∞N xe−x

2dx < 10−5 and solve for N.


Problem #4

∫ ∞N

xe−x2dx =

1

2e−N

2< 10−5 =⇒ N >

√ln(50000) ≈ 3.3

so we need N ≥ 4.

Using N = 4, show that 0.4 ≤∑∞

n=1 ne−n2 .

Here, we use the other side of the inequality,∫ ∞N+1

xe−x2dx ≤ RN =

∞∑n=1

ne−n2 −

N∑n=1

ne−n2.

Plugging in N = 4 and isolating for∑∞

n=1 ne−n2 we get:


Problem #4

∞∑n=1

ne−n2 ≥

∫ ∞4+1

xe−x2

+4∑

n=1

ne−n2

=1

2e25+ e−1

2+ 2e−2

2+ 3e−3

2+ 4e−4

2

≈ 0.40488

≥ 0.4


Problem #5

Show that the series

∞∑n=1

((−1)n

nn+

n − 1

n3 + 1

)converges.

Let us first consider the series

∞∑n=1

(−1)n

nnand

∞∑n=1

n − 1

n3 + 1

separately.

The first series has a simple form with powers of n. This is a goodhint to use the Root Test.


Problem #5

limn→∞

n√|an| = lim

n→∞

∣∣∣∣(−1)n

nn

∣∣∣∣1/n

= limn→∞

(1

nn

)1/n

= limn→∞

1

n

= 0

< 1.

By the Root Test,∑∞

n=1(−1)nnn is absolutely convergent, and

therefore convergent.


Problem #5

For the second series, notice that for all n:

n − 1

n3 + 1≤ n

n3=

1

n2.

Therefore,

∞∑n=1

n − 1

n3 + 1≤∞∑n=1

1

n2.

Since the series∑∞

n=11n2

is a p-series with p = 2 > 1, it converges.

By the Comparison Test, the series∑∞

n=1n−1n3+1

converges.

So both∑∞

n=1(−1)nnn and

∑∞n=1

n−1n3+1

are convergent. By the sumproperty of convergent series:

∞∑n=1

(−1)n

nn+∞∑n=1

n − 1

n3 + 1=∞∑n=1

((−1)n

nn+

n − 1

n3 + 1

)converges.


math 1xx3 tutorial 5 - mcmaster university...2018/02/01 · the alternating series test the...

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