solved problems on integral test and harmonic series
TRANSCRIPT
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.
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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( )( )( ) =∞
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
INTEGRAL TEST
Hence, by the Integral Test the series
1
k ln k( )k=2
∞∑ also diverges.
Solution(cont’d)
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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
∞
∫ .
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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)
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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)
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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
∫⎛
⎝⎜
⎞
⎠⎟
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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)
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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
∫
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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
∞∑ .
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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.
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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)
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
Problem 6
INTEGRAL TEST
For which values of the parameter p the series
1
k pk=1
∞∑ converges?
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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 .
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
Problem 7
INTEGRAL TEST
Show that 1
k2k=1
∞∑ converges and estimate the
sum with error < 0 .001 .
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
INTEGRAL TESTSolution
The improper integral dx
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.
Mika Seppälä: Solved Problems on Integral Test and Harmonic 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)
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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 .
INTEGRAL TESTSolution(cont’d)
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
Problem 8
INTEGRAL TEST
Show that 1
nln2 n( )n=2
∞∑ converges and estimate
the sum with error < 0 .5 .
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
INTEGRAL TESTSolution
The improper integral dx
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( ).
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
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 .
INTEGRAL TESTSolution(cont’d)
Mika Seppälä: Solved Problems on Integral Test and Harmonic Series
INTEGRAL TESTSolution(cont’d)
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 .