Convergence Tests(for Series with Non-negative Terms)

Math 55 - Elementary Analysis III

Institute of MathematicsUniversity of the Philippines

Diliman

Math 55 Convergence Tests I 1/ 21

Recall

R The seriesn=1

an converges if its sequence of partial sums

{Sn} is convergent. Otherwise, the series diverges.

R Divergence Test

If limn an 6= 0, then the series

n=1

an is divergent.

Math 55 Convergence Tests I 2/ 21

Recall

R The seriesn=1

an converges if its sequence of partial sums

{Sn} is convergent. Otherwise, the series diverges.

R Divergence Test

If limn an 6= 0, then the series

n=1

an is divergent.

Math 55 Convergence Tests I 2/ 21

Test for Convergence/Divergence

Consider a seriesn=1

an and a function f , continuous and

positive on [1,) such that an = f(n) for all n N.

. . .

n1f(x) dx > Sn a1

If limn

n1f(x) dx = L, then

n=1

an = limnSn < a1 + limn

n1f(x) dx = a1 + L

and thus the series converges.

Math 55 Convergence Tests I 3/ 21

Test for Convergence/Divergence

Consider a seriesn=1

an and a function f , continuous and

positive on [1,) such that an = f(n) for all n N.

. . .

n1f(x) dx > Sn a1

If limn

n1f(x) dx = L, then

n=1

an = limnSn < a1 + limn

n1f(x) dx = a1 + L

and thus the series converges.

Math 55 Convergence Tests I 3/ 21

Test for Convergence/Divergence

Consider a seriesn=1

an and a function f , continuous and

positive on [1,) such that an = f(n) for all n N.

. . .

n1f(x) dx > Sn a1

If limn

n1f(x) dx = L, then

n=1

an = limnSn < a1 + limn

n1f(x) dx = a1 + L

and thus the series converges.

Math 55 Convergence Tests I 3/ 21

Test for Convergence/Divergence

Consider a seriesn=1

an and a function f , continuous and

positive on [1,) such that an = f(n) for all n N.

. . .

n1f(x) dx > Sn a1

If limn

n1f(x) dx = L, then

n=1

an = limnSn < a1 + limn

n1f(x) dx = a1 + L

and thus the series converges.

Math 55 Convergence Tests I 3/ 21

Test for Convergence/Divergence

Consider a seriesn=1

an and a function f , continuous and

positive on [1,) such that an = f(n) for all n N.

. . .

n1f(x) dx > Sn a1

If limn

n1f(x) dx = L, then

n=1

an = limnSn

< a1 + limn

n1f(x) dx = a1 + L

and thus the series converges.

Math 55 Convergence Tests I 3/ 21

Test for Convergence/Divergence

Consider a seriesn=1

an and a function f , continuous and

positive on [1,) such that an = f(n) for all n N.

. . .

n1f(x) dx > Sn a1

If limn

n1f(x) dx = L, then

n=1

an = limnSn < a1 + limn

n1f(x) dx

= a1 + L

and thus the series converges.

Math 55 Convergence Tests I 3/ 21

Test for Convergence/Divergence

Consider a seriesn=1

an and a function f , continuous and

positive on [1,) such that an = f(n) for all n N.

. . .

n1f(x) dx > Sn a1

If limn

n1f(x) dx = L, then

n=1

an = limnSn < a1 + limn

n1f(x) dx = a1 + L

and thus the series converges.

Math 55 Convergence Tests I 3/ 21

Test for Convergence/Divergence

Consider a seriesn=1

an and a function f , continuous and

positive on [1,) such that an = f(n) for all n N.

. . .

n1f(x) dx < Sn an

Now, suppose limn

n1f(x) dx =. Then

n=1

an = limnSn > limn

n1f(x) dx + an =

and thus the series diverges.

Math 55 Convergence Tests I 4/ 21

Test for Convergence/Divergence

Consider a seriesn=1

an and a function f , continuous and

positive on [1,) such that an = f(n) for all n N.

. . .

n1f(x) dx < Sn an

Now, suppose limn

n1f(x) dx =. Then

n=1

an = limnSn > limn

n1f(x) dx + an =

and thus the series diverges.

Math 55 Convergence Tests I 4/ 21

Test for Convergence/Divergence

Consider a seriesn=1

an and a function f , continuous and

positive on [1,) such that an = f(n) for all n N.

. . .

n1f(x) dx < Sn an

Now, suppose limn

n1f(x) dx =. Then

n=1

an = limnSn > limn

n1f(x) dx + an =

and thus the series diverges.

Math 55 Convergence Tests I 4/ 21

Test for Convergence/Divergence

Consider a seriesn=1

an and a function f , continuous and

positive on [1,) such that an = f(n) for all n N.

. . .

n1f(x) dx < Sn an

Now, suppose limn

n1f(x) dx =.

Then

n=1

an = limnSn > limn

n1f(x) dx + an =

and thus the series diverges.

Math 55 Convergence Tests I 4/ 21

Test for Convergence/Divergence

Consider a seriesn=1

an and a function f , continuous and

positive on [1,) such that an = f(n) for all n N.

. . .

n1f(x) dx < Sn an

Now, suppose limn

n1f(x) dx =. Then

n=1

an = limnSn

> limn

n1f(x) dx + an =

and thus the series diverges.

Math 55 Convergence Tests I 4/ 21

Test for Convergence/Divergence

Consider a seriesn=1

an and a function f , continuous and

positive on [1,) such that an = f(n) for all n N.

. . .

n1f(x) dx < Sn an

Now, suppose limn

n1f(x) dx =. Then

n=1

an = limnSn > limn

n1f(x) dx + an

=

and thus the series diverges.

Math 55 Convergence Tests I 4/ 21

Test for Convergence/Divergence

Consider a seriesn=1

an and a function f , continuous and

positive on [1,) such that an = f(n) for all n N.

. . .

n1f(x) dx < Sn an

Now, suppose limn

n1f(x) dx =. Then

n=1

an = limnSn > limn

n1f(x) dx + an =

and thus the series diverges.

Math 55 Convergence Tests I 4/ 21

Integral Test

Theorem

Letn=1

an be a series with nonnegative terms. Suppose f is a

continuous, positive, and decreasing function in [1,) andf(n) = an for all n N. Then the series(i) converges if the improper integral

1

f(x) dx converges.

(ii) diverges if the improper integral

1

f(x) dx diverges.

Math 55 Convergence Tests I 5/ 21

Examples

Example

Show by the Integral Test that the harmonic series

n=1

1

nis

divergent.

Solution. Let f(x) =1

x. On [1,), f is continuous, positive

and decreasing. Hence, 1

1

xdx = lim

t

t1

1

xdx = lim

t ln |x|t1

= limt ln t =

So the harmonic series diverges by the Integral Test.

Math 55 Convergence Tests I 6/ 21

Examples

Example

Show by the Integral Test that the harmonic series

n=1

1

nis

divergent.

Solution. Let f(x) =1

x. On [1,), f is continuous, positive

and decreasing.

Hence, 1

1

xdx = lim

t

t1

1

xdx = lim

t ln |x|t1

= limt ln t =

So the harmonic series diverges by the Integral Test.

Math 55 Convergence Tests I 6/ 21

Examples

Example

Show by the Integral Test that the harmonic series

n=1

1

nis

divergent.

Solution. Let f(x) =1

x. On [1,), f is continuous, positive

and decreasing. Hence, 1

1

xdx = lim

t

t1

1

xdx

= limt ln |x|

t1

= limt ln t =

So the harmonic series diverges by the Integral Test.

Math 55 Convergence Tests I 6/ 21

Examples

Example

Show by the Integral Test that the harmonic series

n=1

1

nis

divergent.

Solution. Let f(x) =1

x. On [1,), f is continuous, positive

and decreasing. Hence, 1

1

xdx = lim

t

t1

1

xdx = lim

t ln |x|t1

= limt ln t =

So the harmonic series diverges by the Integral Test.

Math 55 Convergence Tests I 6/ 21

Examples

Example

Show by the Integral Test that the harmonic series

n=1

1

nis

divergent.

Solution. Let f(x) =1

x. On [1,), f is continuous, positive

and decreasing. Hence, 1

1

xdx = lim

t

t1

1

xdx = lim

t ln |x|t1

= limt ln t

=

So the harmonic series diverges by the Integral Test.

Math 55 Convergence Tests I 6/ 21

Examples

Example

Show by the Integral Test that the harmonic series

n=1

1

nis

divergent.

Solution. Let f(x) =1

x. On [1,), f is continuous, positive

and decreasing. Hence, 1

1

xdx = lim

t

t1

1

xdx = lim

t ln |x|t1

= limt ln t =

So the harmonic series diverges by the Integral Test.

Math 55 Convergence Tests I 6/ 21

Examples

Example

Determine if the series

n=2

1

n(lnn)2converges or diverges.

Solution. Let f(x) =1

x(lnx)2. Clearly, f is continuous,

positive and decreasing on [2,). Then 2

1

x(lnx)2dx = lim

t

t2

1

x(lnx)2dx

= limt

ln tln 2

du

u2(letting u = lnx)

= limt

1

u

ln tln 2

= limt

( 1

ln t+

1

ln 2

)=

1

ln 2.

Hence, the series converges by the Integral Test.

Math 55 Convergence Tests I 7/ 21

Examples

Example

Determine if the series

n=2

1

n(lnn)2converges or diverges.

Solution. Let f(x) =1

x(lnx)2. Clearly, f is continuous,

positive and decreasing on [2,).

Then 2

1

x(lnx)2dx = lim

t

t2

1

x(lnx)2dx

= limt

ln tln 2

du

u2(letting u = lnx)

= limt

1

u

ln tln 2

= limt

( 1

ln t+

1

ln 2

)=

1

ln 2.

Hence, the series converges by the Integral Test.

Math 55 Convergence Tests I 7/ 21

Examples

Example

Determine if the series

n=2

1

n(lnn)2converges or diverges.

Solution. Let f(x) =1

x(lnx)2. Clearly, f is continuous,

positive and decreasing on [2,). Then 2

1

x(lnx)2dx = lim

t

t2

1

x(lnx)2dx

= limt

ln tln 2

du

u2(letting u = lnx)

= limt

1

u

ln tln 2

= limt

( 1

ln t+

1

ln 2

)=

1

ln 2.

Hence, the series converges by the Integral Test.

Math 55 Convergence Tests I 7/ 21

Examples

Example

Determine if the series

n=2

1

n(lnn)2converges or diverges.

Solution. Let f(x) =1

x(lnx)2. Clearly, f is continuous,

positive and decreasing on [2,). Then 2

1

x(lnx)2dx = lim

t

t2

1

x(lnx)2dx

= limt

ln tln 2

du

u2(letting u = lnx)

= limt

1

u

ln tln 2

= limt

( 1

ln t+

1

ln 2

)=

1

ln 2.

Hence, the series converges by the Integral Test.

Math 55 Convergence Tests I 7/ 21

Examples

Example

Determine if the series

n=2

1

n(lnn)2converges or diverges.

Solution. Let f(x) =1

x(lnx)2. Clearly, f is continuous,

positive and decreasing on [2,). Then 2

1

x(lnx)2dx = lim

t

t2

1

x(lnx)2dx

= limt

ln tln 2

du

u2(letting u = lnx)

= limt

1

u

ln tln 2

= limt

( 1

ln t+

1

ln 2

)=

1

ln 2.

Hence, the series converges by the Integral Test.

Math 55 Convergence Tests I 7/ 21

Examples

Example

Determine if the series

n=2

1

n(lnn)2converges or diverges.

Solution. Let f(x) =1

x(lnx)2. Clearly, f is continuous,

positive and decreasing on [2,). Then 2

1

x(lnx)2dx = lim

t

t2

1

x(lnx)2dx

= limt

ln tln 2

du

u2(letting u = lnx)

= limt

1

u

ln tln 2

= limt

( 1

ln t+

1

ln 2

)

=1

ln 2.

Hence, the series converges by the Integral Test.

Math 55 Convergence Tests I 7/ 21

Examples

Example

Determine if the series

n=2

1

n(lnn)2converges or diverges.

Solution. Let f(x) =1

x(lnx)2. Clearly, f is continuous,

positive and decreasing on [2,). Then 2

1

x(lnx)2dx = lim

t

t2

1

x(lnx)2dx

= limt

ln tln 2

du

u2(letting u = lnx)

= limt

1

u

ln tln 2

= limt

( 1

ln t+

1

ln 2

)=

1

ln 2.

Hence, the series converges by the Integral Test.

Math 55 Convergence Tests I 7/ 21

p-Series

Theorem

The series

n=1

1

np, called a p-series for a positive constant p, is

convergent if p > 1 and divergent if p 1.

Proof. If p = 1, the series becomes the harmonic seriesn=1

1

n,

which is divergent. If p 6= 1, f(x) = 1xp

is positive, decreasing

and continuous on [1,). Therefore, 1

1

xpdx = lim

t

t1

xp dx = limt

xp+1

p + 1t1

= limt

(tp+1

p + 1 1

p + 1)

=

, 0 < p < 11p 1 , p > 1

and the result follows, by the Integral Test.

Math 55 Convergence Tests I 8/ 21

p-Series

Theorem

The series

n=1

1

np, called a p-series for a positive constant p, is

convergent if p > 1 and divergent if p 1.

Proof. If p = 1, the series becomes the harmonic seriesn=1

1

n,

which is divergent.

If p 6= 1, f(x) = 1xp

is positive, decreasing

and continuous on [1,). Therefore, 1

1

xpdx = lim

t

t1

xp dx = limt

xp+1

p + 1t1

= limt

(tp+1

p + 1 1

p + 1)

=

, 0 < p < 11p 1 , p > 1

and the result follows, by the Integral Test.

Math 55 Convergence Tests I 8/ 21

p-Series

Theorem

The series

n=1

1

np, called a p-series for a positive constant p, is

convergent if p > 1 and divergent if p 1.

Proof. If p = 1, the series becomes the harmonic seriesn=1

1

n,

which is divergent. If p 6= 1, f(x) = 1xp

is positive, decreasing

and continuous on [1,).

Therefore, 1

1

xpdx = lim

t

t1

xp dx = limt

xp+1

p + 1t1

= limt

(tp+1

p + 1 1

p + 1)

=

, 0 < p < 11p 1 , p > 1

and the result follows, by the Integral Test.

Math 55 Convergence Tests I 8/ 21

p-Series

Theorem

The series

n=1

1

np, called a p-series for a positive constant p, is

convergent if p > 1 and divergent if p 1.

Proof. If p = 1, the series becomes the harmonic seriesn=1

1

n,

which is divergent. If p 6= 1, f(x) = 1xp

is positive, decreasing

and continuous on [1,). Therefore, 1

1

xpdx = lim

t

t1

xp dx

= limt

xp+1

p + 1t1

= limt

(tp+1

p + 1 1

p + 1)

=

, 0 < p < 11p 1 , p > 1

and the result follows, by the Integral Test.

Math 55 Convergence Tests I 8/ 21

p-Series

Theorem

The series

n=1

1

np, called a p-series for a positive constant p, is

convergent if p > 1 and divergent if p 1.

Proof. If p = 1, the series becomes the harmonic seriesn=1

1

n,

which is divergent. If p 6= 1, f(x) = 1xp

is positive, decreasing

and continuous on [1,). Therefore, 1

1

xpdx = lim

t

t1

xp dx = limt

xp+1

p + 1t1

= limt

(tp+1

p + 1 1

p + 1)

=

, 0 < p < 11p 1 , p > 1

and the result follows, by the Integral Test.

Math 55 Convergence Tests I 8/ 21

p-Series

Theorem

The series

n=1

1

np, called a p-series for a positive constant p, is

convergent if p > 1 and divergent if p 1.

Proof. If p = 1, the series becomes the harmonic seriesn=1

1

n,

which is divergent. If p 6= 1, f(x) = 1xp

is positive, decreasing

and continuous on [1,). Therefore, 1

1

xpdx = lim

t

t1

xp dx = limt

xp+1

p + 1t1

= limt

(tp+1

p + 1 1

p + 1)

=

, 0 < p < 11p 1 , p > 1

and the result follows, by the Integral Test.

Math 55 Convergence Tests I 8/ 21

p-Series

Theorem

The series

n=1

1

np, called a p-series for a positive constant p, is

convergent if p > 1 and divergent if p 1.

Proof. If p = 1, the series becomes the harmonic seriesn=1

1

n,

which is divergent. If p 6= 1, f(x) = 1xp

is positive, decreasing

and continuous on [1,). Therefore, 1

1

xpdx = lim

t

t1

xp dx = limt

xp+1

p + 1t1

= limt

(tp+1

p + 1 1

p + 1)

=

, 0 < p < 11p 1 , p > 1

and the result follows, by the Integral Test.Math 55 Convergence Tests I 8/ 21

Examples of p-Series

Example

1 The series

n=1

1

n2is a p-series with p = 2 > 1 and hence,

convergent.

2 The seriesn=1

1n

is a p-series with p = 12 < 1 and hence,

divergent.

Math 55 Convergence Tests I 9/ 21

Examples of p-Series

Example

1 The series

n=1

1

n2is a p-series with p = 2 > 1 and hence,

convergent.

2 The seriesn=1

1n

is a p-series with p = 12 < 1 and hence,

divergent.

Math 55 Convergence Tests I 9/ 21

Comparing Series

We know that

n=1

1

n2is convergent. How about

n=1

1

n2 + 1?

Notice that for any n N,1

n2 + 1 1 and solnn

n>

1

nfor n 3.

Since the harmonic series

n=1

1

ndiverges,

the seriesn=1

lnn

n

diverges by Comparison Test.

Math 55 Convergence Tests I 14/ 21

Examples

Example

Determine whether the series

n=1

lnn

nconverges or diverges.

Solution. Note that for n 3, lnn > 1 and solnn

n>

1

nfor n 3.

Since the harmonic series

n=1

1

ndiverges, the series

n=1

lnn

n

diverges by Comparison Test.

Math 55 Convergence Tests I 14/ 21

Comparing Series

In some cases a useless comparison arises. For example,

consider

n=2

1

n2 1.

Notice that for n 2,

1

n2 0 such that m < c < M .

For sufficientlylarge n,

m < anbn < M

mbn < an < Mbn

If Mn=1

bn converges, so doesn=1

an by Comparison Test.

Similarly, if mn=1

bn diverges, so doesn=1

an.

Math 55 Convergence Tests I 16/ 21

Limit Comparison Test

Theorem

Supposen=1

an andn=1

bn are series with positive terms. If

limn

anbn

= c for some constant c > 0, then both series converge

or both diverge.

Proof. Let m,M > 0 such that m < c < M . For sufficientlylarge n,

m < anbn < M

mbn < an < Mbn

If Mn=1

bn converges, so doesn=1

an by Comparison Test.

Similarly, if mn=1

bn diverges, so doesn=1

an.

Math 55 Convergence Tests I 16/ 21

Limit Comparison Test

Theorem

Supposen=1

an andn=1

bn are series with positive terms. If

limn

anbn

= c for some constant c > 0, then both series converge

or both diverge.

Proof. Let m,M > 0 such that m < c < M . For sufficientlylarge n,

m < anbn < M

mbn < an < Mbn

If Mn=1

bn converges, so doesn=1

an by Comparison Test.

Similarly, if mn=1

bn diverges, so doesn=1

an.

Math 55 Convergence Tests I 16/ 21

Limit Comparison Test

Theorem

Supposen=1

an andn=1

bn are series with positive terms. If

limn

anbn

= c for some constant c > 0, then both series converge

or both diverge.

Proof. Let m,M > 0 such that m < c < M . For sufficientlylarge n,

m < anbn < M

mbn < an < Mbn

If Mn=1

bn converges, so doesn=1

an by Comparison Test.

Similarly, if mn=1

bn diverges, so doesn=1

an.

Math 55 Convergence Tests I 16/ 21

Limit Comparison Test

Theorem

Supposen=1

an andn=1

bn are series with positive terms. If

limn

anbn

= c for some constant c > 0, then both series converge

or both diverge.

Proof. Let m,M > 0 such that m < c < M . For sufficientlylarge n,

m < anbn < M

mbn < an < Mbn

If Mn=1

bn converges, so doesn=1

an by Comparison Test.

Similarly, if mn=1

bn diverges, so doesn=1

an.

Math 55 Convergence Tests I 16/ 21

Limit Comparison Test

The following corollary takes care of the case when limn

anbn

= 0

or .

Corollary

Suppose thatn=1

an andn=1

bn are series with positive terms.

i If limn

anbn

= 0 andn=1

bn converges, thenn=1

an also

converges.

ii If limn

anbn

= andn=1

bn diverges, thenn=1

an also

diverges.

Math 55 Convergence Tests I 17/ 21

Examples

Example

Show that the series

n=2

1

n2 1 is convergent.

Solution. Let an =1

n2 1 and bn =1

n2. Then

limn

anbn

= limn

n2

n2 1 = 1 > 0.

Sincen=2

1

n2converges, the series

n=2

1

n2 1 also converges, bythe Limit Comparison Test.

Math 55 Convergence Tests I 18/ 21

Examples

Example

Show that the series

n=2

1

n2 1 is convergent.

Solution. Let an =1

n2 1 and bn =1

n2.

Then

limn

anbn

= limn

n2

n2 1 = 1 > 0.

Sincen=2

1

n2converges, the series

n=2

1

n2 1 also converges, bythe Limit Comparison Test.

Math 55 Convergence Tests I 18/ 21

Examples

Example

Show that the series

n=2

1

n2 1 is convergent.

Solution. Let an =1

n2 1 and bn =1

n2. Then

limn

anbn

= limn

n2

n2 1

= 1 > 0.

Sincen=2

1

n2converges, the series

n=2

1

n2 1 also converges, bythe Limit Comparison Test.

Math 55 Convergence Tests I 18/ 21

Examples

Example

Show that the series

n=2

1

n2 1 is convergent.

Solution. Let an =1

n2 1 and bn =1

n2. Then

limn

anbn

= limn

n2

n2 1 = 1

> 0.

Sincen=2

1

n2converges, the series

n=2

1

n2 1 also converges, bythe Limit Comparison Test.

Math 55 Convergence Tests I 18/ 21

Examples

Example

Show that the series

n=2

1

n2 1 is convergent.

Solution. Let an =1

n2 1 and bn =1

n2. Then

limn

anbn

= limn

n2

n2 1 = 1 > 0.

Since

n=2

1

n2converges, the series

n=2

1

n2 1 also converges, bythe Limit Comparison Test.

Math 55 Convergence Tests I 18/ 21

Examples

Example

Determine whether the series

n=2

3n3 + 4nn7 + 4

converges or

diverges.

Solution. Let an =3n3 + 4nn7 + 4

and bn =3n3n7

=3n

. Then

limn

anbn

= limn

n(3n3 + 4n)

3n7 + 4

= limn

3n72 + 4n

32

3(n7 + 4)12

= 1 > 0.

Sincen=2

3n

diverges, the seriesn=2

3n4 + 4nn6 + 4

also diverges, by

the Limit Comparison Test.

Math 55 Convergence Tests I 19/ 21

Examples

Example

Determine whether the series

n=2

3n3 + 4nn7 + 4

converges or

diverges.

Solution. Let an =3n3 + 4nn7 + 4

and bn =3n3n7

=3n

.

Then

limn

anbn

= limn

n(3n3 + 4n)

3n7 + 4

= limn

3n72 + 4n

32

3(n7 + 4)12

= 1 > 0.

Sincen=2

3n

diverges, the seriesn=2

3n4 + 4nn6 + 4

also diverges, by

the Limit Comparison Test.

Math 55 Convergence Tests I 19/ 21

Examples

Example

Determine whether the series

n=2

3n3 + 4nn7 + 4

converges or

diverges.

Solution. Let an =3n3 + 4nn7 + 4

and bn =3n3n7

=3n

. Then

limn

anbn

= limn

n(3n3 + 4n)

3n7 + 4

= limn

3n72 + 4n

32

3(n7 + 4)12

= 1 > 0.

Sincen=2

3n

diverges, the seriesn=2

3n4 + 4nn6 + 4

also diverges, by

the Limit Comparison Test.

Math 55 Convergence Tests I 19/ 21

Examples

Example

Determine whether the series

n=2

3n3 + 4nn7 + 4

converges or

diverges.

Solution. Let an =3n3 + 4nn7 + 4

and bn =3n3n7

=3n

. Then

limn

anbn

= limn

n(3n3 + 4n)

3n7 + 4

= limn

3n72 + 4n

32

3(n7 + 4)12

= 1 > 0.

Sincen=2

3n

diverges, the seriesn=2

3n4 + 4nn6 + 4

also diverges, by

the Limit Comparison Test.

Math 55 Convergence Tests I 19/ 21

Examples

Example

Determine whether the series

n=2

3n3 + 4nn7 + 4

converges or

diverges.

Solution. Let an =3n3 + 4nn7 + 4

and bn =3n3n7

=3n

. Then

limn

anbn

= limn

n(3n3 + 4n)

3n7 + 4

= limn

3n72 + 4n

32

3(n7 + 4)12

= 1

> 0.

Sincen=2

3n

diverges, the seriesn=2

3n4 + 4nn6 + 4

also diverges, by

the Limit Comparison Test.

Math 55 Convergence Tests I 19/ 21

Examples

Example

Determine whether the series

n=2

3n3 + 4nn7 + 4

converges or

diverges.

Solution. Let an =3n3 + 4nn7 + 4

and bn =3n3n7

=3n

. Then

limn

anbn

= limn

n(3n3 + 4n)

3n7 + 4

= limn

3n72 + 4n

32

3(n7 + 4)12

= 1 > 0.

Since

n=2

3n

diverges, the series

n=2

3n4 + 4nn6 + 4

also diverges, by

the Limit Comparison Test.

Math 55 Convergence Tests I 19/ 21

Exercises

1 Determine whether each of the following series converges ordiverges.

a.

n=1

lnn

n3

b.

n=1

e1n

n2

c.

n=1

n2 13n4 + 1

d.

n=1

n + 2

(n + 1)3

e.

n=1

4 + 3n

2n

f.

n=1

n!

nn

g.

n=1

sin2 n

n3

h.

n=1

sin

(1

n

)

2 For what values of p is the seriesn=1

n(1 + n2)p convergent?

3 Prove the corollary to the Limit Comparison Test.

Math 55 Convergence Tests I 20/ 21

Exercises

1 Determine whether each of the following series converges ordiverges.

a.

n=1

lnn

n3

b.

n=1

e1n

n2

c.

n=1

n2 13n4 + 1

d.

n=1

n + 2

(n + 1)3

e.

n=1

4 + 3n

2n

f.

n=1

n!

nn

g.

n=1

sin2 n

n3

h.

n=1

sin

(1

n

)

2 For what values of p is the seriesn=1

n(1 + n2)p convergent?

3 Prove the corollary to the Limit Comparison Test.

Math 55 Convergence Tests I 20/ 21

Exercises

1 Determine whether each of the following series converges ordiverges.

a.

n=1

lnn

n3

b.

n=1

e1n

n2

c.

n=1

n2 13n4 + 1

d.

n=1

n + 2

(n + 1)3

e.

n=1

4 + 3n

2n

f.

n=1

n!

nn

g.

n=1

sin2 n

n3

h.

n=1

sin

(1

n

)

2 For what values of p is the seriesn=1

n(1 + n2)p convergent?

3 Prove the corollary to the Limit Comparison Test.

Math 55 Convergence Tests I 20/ 21

Exercises

1 Determine whether each of the following series converges ordiverges.

a.

n=1

lnn

n3

b.

n=1

e1n

n2

c.

n=1

n2 13n4 + 1

d.

n=1

n + 2

(n + 1)3

e.

n=1

4 + 3n

2n

f.

n=1

n!

nn

g.

n=1

sin2 n

n3

h.

n=1

sin

(1

n

)

2 For what values of p is the seriesn=1

n(1 + n2)p convergent?

3 Prove the corollary to the Limit Comparison Test.

Math 55 Convergence Tests I 20/ 21

Exercises

1 Determine whether each of the following series converges ordiverges.

a.

n=1

lnn

n3

b.

n=1

e1n

n2

c.

n=1

n2 13n4 + 1

d.

n=1

n + 2

(n + 1)3

e.

n=1

4 + 3n

2n

f.

n=1

n!

nn

g.

n=1

sin2 n

n3

h.

n=1

sin

(1

n

)

2 For what values of p is the seriesn=1

n(1 + n2)p convergent?

3 Prove the corollary to the Limit Comparison Test.

Math 55 Convergence Tests I 20/ 21

Exercises

1 Determine whether each of the following series converges ordiverges.

a.

n=1

lnn

n3

b.

n=1

e1n

n2

c.

n=1

n2 13n4 + 1

d.

n=1

n + 2

(n + 1)3

e.

n=1

4 + 3n

2n

f.

n=1

n!

nn

g.

n=1

sin2 n

n3

h.

n=1

sin

(1

n

)

2 For what values of p is the seriesn=1

n(1 + n2)p convergent?

3 Prove the corollary to the Limit Comparison Test.

Math 55 Convergence Tests I 20/ 21

Exercises

1 Determine whether each of the following series converges ordiverges.

a.

n=1

lnn

n3

b.

n=1

e1n

n2

c.

n=1

n2 13n4 + 1

d.

n=1

n + 2

(n + 1)3

e.

n=1

4 + 3n

2n

f.

n=1

n!

nn

g.

n=1

sin2 n

n3

h.

n=1

sin

(1

n

)

2 For what values of p is the seriesn=1

n(1 + n2)p convergent?

3 Prove the corollary to the Limit Comparison Test.

Math 55 Convergence Tests I 20/ 21

Exercises

1 Determine whether each of the following series converges ordiverges.

a.

n=1

lnn

n3

b.

n=1

e1n

n2

c.

n=1

n2 13n4 + 1

d.

n=1

n + 2

(n + 1)3

e.

n=1

4 + 3n

2n

f.

n=1

n!

nn

g.

n=1

sin2 n

n3

h.

n=1

sin

(1

n

)

2 For what values of p is the seriesn=1

n(1 + n2)p convergent?

3 Prove the corollary to the Limit Comparison Test.

Math 55 Convergence Tests I 20/ 21

Exercises

1 Determine whether each of the following series converges ordiverges.

a.

n=1

lnn

n3

b.

n=1

e1n

n2

c.

n=1

n2 13n4 + 1

d.

n=1

n + 2

(n + 1)3

e.

n=1

4 + 3n

2n

f.

n=1

n!

nn

g.

n=1

sin2 n

n3

h.

n=1

sin

(1

n

)

2 For what values of p is the series

n=1

n(1 + n2)p convergent?

3 Prove the corollary to the Limit Comparison Test.

Math 55 Convergence Tests I 20/ 21

Exercises

1 Determine whether each of the following series converges ordiverges.

a.

n=1

lnn

n3

b.

n=1

e1n

n2

c.

n=1

n2 13n4 + 1

d.

n=1

n + 2

(n + 1)3

e.

n=1

4 + 3n

2n

f.

n=1

n!

nn

g.

n=1

sin2 n

n3

h.

n=1

sin

(1

n

)

2 For what values of p is the series

n=1

n(1 + n2)p convergent?

3 Prove the corollary to the Limit Comparison Test.

Math 55 Convergence Tests I 20/ 21

References

1 Stewart, J., Calculus, Early Transcendentals, 6 ed., ThomsonBrooks/Cole, 2008

2 Dawkins, P., Calculus 3, online notes available athttp://tutorial.math.lamar.edu/

Math 55 Convergence Tests I 21/ 21