Solved problems on comparison theorem for series.

Solved problems on comparison theorem for

series

Mika Seppälä: Solved Problems on Comparison Test

comparison test

Let 0 ≤ ak ≤ bk for all k.

b

kk=1

∞

∑ converges ⇒ ak

k=1

∞

∑ converges.

a

kk=1

∞

∑ diverges ⇒ bk

k=1

∞

∑ diverges.

Mika Seppälä: Solved Problems on Comparison Test

1

OVERVIEW OF PROBLEMS

Let 0 ≤ ak ≤ bk for all k. Assume that

the series and both converges.

Show that the series converges.

a

kk=1

∞

∑

a

kb

kk=1

∞

∑

bk

k=1

∞

∑

Mika Seppälä: Solved Problems on Comparison Test

2

OVERVIEW OF PROBLEMS

Let ak and bk positive for all k. Assume

that the series converges and that

Show that the series converges.

a

kk=1

∞

∑

b

kk=1

∞

∑ limk→ ∞

bk

ak

=L < ∞.

Mika Seppälä: Solved Problems on Comparison Test

4 5

6 7

OVERVIEW OF PROBLEMSUse Comparison Test to determine whether the series converge or diverge.

9n

1 +10 nn=1

∞

∑

4 + 2 n

3 nn=1

∞

∑

2 + cos(n)3 n

n=1

∞

∑

ln 1 +

12 n

⎛

⎝⎜⎞

⎠⎟n=1

∞

∑3

Mika Seppälä: Solved Problems on Comparison Test

Let 0 ≤ ak ≤ bk for all k. Assume that

the series and both converges.

Show that the series converges.

a

kk=1

∞

∑

a

kb

kk=1

∞

∑

bk

k=1

∞

∑Problem 1

COMPARISON TEST

Mika Seppälä: Solved Problems on Comparison Test

COMPARISON TESTSolution

∀ε > 0 ∃m

1∈• such that k > m

1⇒ b

k≤ε

Since converges, . By definition

of limit this means,

b

kk=1

∞

∑ limk→ ∞

bk=0

Assume . Since is positive for all k, we have

bk ε < 1

Mika Seppälä: Solved Problems on Comparison Test

COMPARISON TESTSolution(cont’d)

there is a number m1 such that k > m

1⇒ b

k<1 .

Recall also that by assumptions . Then the Comparison Theorem implies that

akb

k> 0

a

kb

kk=m1

∞∑ converges.

Mika Seppälä: Solved Problems on Comparison Test

COMPARISON TESTSolution(cont’d)

Remark that is suffices to show that

akb

kk=m1

∞∑ converges because

akb

kk=1

∞

∑ = akb

kk=1

m1

∑ + akb

kk=m1

∞

∑

and the sum akb

kk=1

m1∑ is finite.

Mika Seppälä: Solved Problems on Comparison Test

Let ak and bk positive for all k. Assume that

the series converges and that

Show that the series converges.

a

kk=1

∞

∑

b

kk=1

∞

∑ limk→ ∞

bk

ak

=L < ∞.Problem 2

COMPARISON TEST

Mika Seppälä: Solved Problems on Comparison Test

COMPARISON TESTSolution

Since , there is a number such

that limk→ ∞

bk

ak

=L m1

k > m

1⇒

bk

ak

≤L +1 .

Therefore

k > m

1⇒ b

k≤ L +1( )ak

.

Since , so does

akk=1

∞∑ L +1( )akk=1

∞∑ .

Mika Seppälä: Solved Problems on Comparison Test

COMPARISON TESTSolution(cont’d)

I t suffices to show that bkk=m1

∞∑ converges

because bk= b

kk=1

m1∑ +k=1

∞∑ bkk=m1

∞∑ and the

sum bkk=1

m1∑ is finite.

This implies by the comparison theorem that

b

kk=m1

∞∑ converges.

Mika Seppälä: Solved Problems on Comparison Test

9n

1 +10 nn=1

∞

∑

COMPARISON TESTProblem 3

Solution

For all n ≥1, 9 n

1 +10 n<

9 n

10 n=

910

⎛

⎝⎜⎞

⎠⎟

n

. We know

that 910

⎛

⎝⎜⎞

⎠⎟

n

n=1

∞∑ is a convergent geometric

series.

Mika Seppälä: Solved Problems on Comparison Test

COMPARISON TESTSolution(cont’d)

By Comparison Theorem, we conclude that

the series 9n

1 +10 nn=1

∞∑ converges.

Mika Seppälä: Solved Problems on Comparison Test

COMPARISON TESTProblem 4

Solution

4 + 2 n

3 nn=1

∞

∑

By rewriting,

4 + 2 n

3 n=

43 n

+2 n

3 n=4.

13

⎛

⎝⎜⎞

⎠⎟

n

+23

⎛

⎝⎜⎞

⎠⎟

n

We know that the geometric series

1

3

⎛

⎝⎜⎞

⎠⎟

n

n=1

∞∑ and 23

⎛

⎝⎜⎞

⎠⎟

n

n=1

∞∑ converge.

Mika Seppälä: Solved Problems on Comparison Test

COMPARISON TESTSolution(cont’d)

Therefore we can write

4 + 2 n

3 nn=1

∞

∑ =4.13

⎛

⎝⎜⎞

⎠⎟

n

n=1

∞

∑ +23

⎛

⎝⎜⎞

⎠⎟

n

n=1

∞

∑

Hence converges.

4 + 2 n

3 nn=1

∞∑

Mika Seppälä: Solved Problems on Comparison Test

COMPARISON TESTProblem 5

Solution

ln 1 +

12 n

⎛

⎝⎜⎞

⎠⎟n=1

∞

∑

Let an=

12 n

and bn=ln 1 +

12 n

⎛

⎝⎜⎞

⎠⎟. We know that

the geometric series 12 nn=1

∞∑ converges.

Mika Seppälä: Solved Problems on Comparison Test

COMPARISON TESTSolution(cont’d)

Also limn→ ∞

bn

an

=1. Then from Problem 2 , it follows

that the series ln 1 +12 n

⎛

⎝⎜⎞

⎠⎟n=1

∞∑ converges.

Mika Seppälä: Solved Problems on Comparison Test

COMPARISON TESTProblem 6

Solution

Let an=

13 n

and bn=

3n2 + 5n

3 n n2 +1( ). We know that

the geometric sequence 13 nn=1

∞∑ converges.

Mika Seppälä: Solved Problems on Comparison Test

COMPARISON TESTSolution(cont’d)

Also limn→ ∞

bn

an

=limn→ ∞

3n2 + 5nn2 +1

=3.

Then from Problem 2 , it follows that the series

3n2 + 5n

2 n n2 +1( )n=1

∞∑ converges.

Mika Seppälä: Solved Problems on Comparison Test

COMPARISON TESTProblem 7

Solution

2 + cos(n)3 n

n=1

∞

∑

Since for all n , we obtain −1 ≤cos(n) ≤1

1

3n≤2 + cos(n)

3 n≤

33 n

=1

3 n−1.

Mika Seppälä: Solved Problems on Comparison Test

COMPARISON TESTSolution(cont’d)

We know that the geometric series

1

3n−1n=1

∞∑

converges.

By Comparison Theorem, we conclude

that the series 2 + cos(n)

3 nn=1

∞∑ converges.