Solved problems on integral test and harmonic series

Solved problems on integral test and harmonic

series

Mika Seppälä: Solved Problems on Integral Test and Harmonic Series

integral testLet f be a positive decreasing function, and

let ak = f(k).

I f the improper integral f x( )dx converges, 1

∞

∫then the series a

kk=1

∞∑ converges.

I f the improper integral f x( )dx diverges, 1

∞

∫then the series a

kk=1

∞∑ diverges.


Harmonic seriesThe series is called

the Harmonic Series.

1

kk=1

∞∑ =1 +12+13+L

Using the Integral Test for the function

we prove that the Harmonic Series

diverges. f (x) =

1x


error estimate by the integral test

Let f be a positive decreasing function, and

let ak = f(k).

If the series converges by the integral

test, then a

kk=1

∞∑

ak

k=1

M

∑ − ak

k=1

∞

∑1 244 34 4

≤ f (x) dxM

∞

∫

Error of the approximation by the partial sum. Mth


1

k ln k( )k=2

∞

∑

k

ekk=1

∞

∑

n2−n2

n=1

∞

∑

Determine whether the following series converge or diverge.

OVERVIEW OF PROBLEMS

arctan(n)

n2 +1n=1

∞

∑

1 2

3 4

5


OVERVIEW OF PROBLEMS

Show that 1

k2k=1

∞∑ converges and

estimate the sum with error < 0 .001 .

Show that 1

nln2 n( )n=2

∞∑ converges and

estimate the sum with error < 0 .5 .

For which values of the parameter p the

series 1

k pk=1

∞∑ converges?

6

7

8


Problem 1

INTEGRAL TEST

1

k ln k( )k=2

∞

∑

Solution

The integral dx

x ln x( )2

∞

∫ diverges since

dx

x ln x( )2

∞

∫ = limM→ ∞

dx

x ln x( )2

M

∫ =limx→ ∞

ln ln x( )( )⎤⎦2

M

= limM→ ∞

ln ln M( )( ) −ln ln 2( )( )( ) =∞


INTEGRAL TEST

Hence, by the Integral Test the series

1

k ln k( )k=2

∞∑ also diverges.

Solution(cont’d)


Problem 2

INTEGRAL TEST

Solution

To use the Integral Test, we have to compute

dx

x ln x( )ln ln x( )( )= lim

M→ ∞

dx

k ln k( )ln ln k( )( )4

M

∫4

∞

∫ .


INTEGRAL TEST

To compute dx

x ln x( )ln ln x( )( )4

M

∫ substitute

t =ln ln x( )( ). Then dt =dx

x ln x( ) and we get

dx

x ln x( )ln ln x( )( )4

M

∫ =dtt

ln ln 4( )( )

ln ln M( )( )

∫ =ln t( )⎤⎦ln ln 4( )( )

ln ln M( )( ).

Solution(cont’d)


INTEGRAL TEST

dx

k ln x( )ln ln x( )( )4

M

∫ =ln ln ln M( )( )( ) −ln ln ln 4( )( )( )

Since limM→ ∞

ln ln ln M( )( )( ) −ln ln ln 4( )( )( )( ) =∞ the

improper integral dx

x ln x( )ln ln x( )( )4

∞

∫ diverges and

by the Integral Test so does the series.

Solution(cont’d)


Problem 3

INTEGRAL TEST

k

ekk=1

∞

∑

Solution

Compute the improper integral xe−x dx1

∞

∫ by

integration by parts with u =x and dv =e−x

xe−x dx1

∞

∫ = limM→ ∞

xe−x dx1

M

∫ = limM→ ∞

−xe−x

1

M+ e−x dx

1

M

∫⎛

⎝⎜

⎞

⎠⎟


INTEGRAL TEST

xe−x dx1

∞

∫ = limM→ ∞

−xe−x

1

M+ −e−x( )

1

M⎛⎝⎜

⎞⎠⎟

= limM→ ∞

e−1 −Me−M + e−1 −e−M( ) =

2e.

Since the integral xe−x dx1

∞

∫ converges, by the

Integral Test so does the series.

Solution(cont’d)


Problem 4

Solution

INTEGRAL TEST

n2−n2

n=1

∞

∑

Compute the improper integral x2−x2

dx 1

∞

∫ by

substituting u =2 −x2

, du =−2 ln 2( ) x2 −x2

dx,

x2 −x2

dx1

∞

∫ = limM→ ∞

x2 −x2

dx1

M

∫ = limM→ ∞

du

−2 ln 2( )2−1

2−M

∫


INTEGRAL TESTSolution(cont’d)

x2−x2

dx1

∞

∫ =−limM→ ∞

u2−1

2−M

2 ln 2( )=−

limM→ ∞

2 −M −2 −1( )

2 ln 2( )=

1

4 ln 2( ).

Hence, the integral x2 −x2

dx1

∞

∫ converges and by

the Integral Test so does the series n2 −n2

n=1

∞∑ .


arctan(n)

n2 +1n=1

∞

∑Problem 5

Solution

INTEGRAL TEST

Compute the improper integral arctan(x)

x2 +1dx

1

∞

∫

by substituting u =arctan(x), du =dx

x2 +1.


INTEGRAL TEST

arctan(x)

x2 +1dx

1

∞

∫ = limM→ ∞

arctan(x)x2 +1

dx1

M

∫ = limM→ ∞

uduπ 2

arctan(M)

∫

= limM→ ∞

u2

2π 2

arctan(M)

= limM→ ∞

12

arctan2 (M) −12

π2

⎛

⎝⎜⎞

⎠⎟

2⎛

⎝⎜⎜

⎞

⎠⎟⎟=π 2

4−π 2

4=0 .

Since arctan(x)

x2 +1dx

1

∞

∫ converges, by the Integral

Test so does the series.

Solution(cont’d)


Problem 6

INTEGRAL TEST

For which values of the parameter p the series

1

k pk=1

∞∑ converges?


INTEGRAL TESTSolution

The improper integral dx

xp1

∞

∫ converges if and

only if p >1 .

Hence, by the Integral Test the series 1k pk=1

∞∑converges if and only if p >1 .


Problem 7

INTEGRAL TEST

Show that 1

k2k=1

∞∑ converges and estimate the

sum with error < 0 .001 .




x21

∞

∫ converges since

dx

x21

∞

∫ = limM→ ∞

dx

x21

M

∫ = limM→ ∞

−1x

⎛

⎝⎜⎞

⎠⎟1

M

= limM→ ∞

−1M

− −1( )⎛

⎝⎜⎞

⎠⎟=1.

Therefore by the Integral Test so does the series.


dx

x2M

∞

∫ < 0 .001

INTEGRAL TEST

In order to estimate the sum with error <0.001, we have to find out how many terms we need to take in our approximation. In other words, we need to find out M so that

Solution(cont’d)


Computing as before, we find dx

x2M

∞

∫ =1M

< 0 .001 .

Hence M >1000 . This means 1000 th partial sum estimates the sum with error < 0 .001 .



Problem 8

INTEGRAL TEST

Show that 1

nln2 n( )n=2

∞∑ converges and estimate

the sum with error < 0 .5 .




x ln2 x( )2

∞

∫ converges since

by substituting u =ln x( ) , du =1 x we find that

dx

x ln2 x( )2

∞

∫ = limM→ ∞

dx

x ln2 x( )2

M

∫ = limM→ ∞

du

u2ln 2( )

ln M( )

∫ = limM→ ∞

−1u

ln 2( )

ln M( )⎛

⎝⎜⎜

⎞

⎠⎟⎟

= limM→ ∞

1 ln 2( ) −1 ln M( )( ) =1 ln 2( ).


Therefore by the Integral Test the series

1

nln2 n( )n=2

∞∑ converges. To estimate the sum

with error < 0 .5 , we need to find M such that

dx

x ln2 x( )M

∞

∫ < 0 .5 .




Computing as before, dx

x ln2 x( )M

∞

∫ =1

ln M( )< 0 .5 .

Hence, ln M( ) > 1 0 .5 =2 . That is M > e2 =7.4 .

This means 8 th partial sum estimates the sum with error < 0 .5 .

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Solved problems on integral test and harmonic series