Math 152 - Lecture Notes # 7 -
David Maslanka
The Integral Test and Remainder Theorem
Suppose that we have been able to use the Integral Test to verify that the series
∞
S
=
∑ f( n )
converges.
n=1
∞
Now let
Rn
=
⎞ ⎛ n
⎞
⎛
⎜
f( k ) ⎟⎟⎟ − ⎜⎜⎜
f( k ) ⎟⎟⎟ .
⎜⎜
⎠ ⎝k= 1
⎠
⎝k = 1
∑
∑
Then Rn is the error in approximating the infinite sum, S, with the nth partial sum.
Since the function f is nonincreasing on [ 1 , ∞ ) it follows that
∞
Rn
⌠
⎮ f( x ) d x .
⌡
= f(n+1) + f(n+2) + ... <
n
Similarly, we have that
Rn = f ( n + 1 ) + f ( n + 2 ) + . . . >
⌠
⎮
⌡
∞
n+1
So we have verified the following theorem:
f ( x ) dx .
Remainder Estimate for the Integral Test
∞
If
∑ f( n )
converges by the Integral Test and Rn = S - sn , then
n=1
⌠
⎮
⌡
(1)
∞
∞
f ( x ) dx
<
⌠
⎮ f( x ) d x .
⌡
Rn <
n+1
n
Moreover, if we add sn to each side of the inequality ( 1 ) we find that
(2)
⌠
⎮
⌡
sn +
∞
∞
f( x ) d x
S
<
<
⌠
⎮ f ( x ) dx .
⌡
sn +
n+1
n
_______________________________________________________________________________________________
Example
∞
Obtain the best estimate for S =
∑n
1
n=1
using the tenth partial sum, s10 , of this series and the Remainder Theorem.
2
Solution:
Since f ( x ) =
∞
1
is positive, continuous and decreasing on [ 1 , +∞ ) and
x2
⌠
⎮ f( x ) dx = 1,
⌡
1
∞
∑n
then it follows that
1
n=1
converges to a real number, S , by the Integral Test.
2
Now by the Remainder Theorem:
∞
∞
⌠
= ⎮ f( x ) d x
⌡
11
11
1
<
R10
1
⌠
⎮ f( x ) d x =
⌡
10
10
<
and upon adding s10 to each side of this inequality we find that
∞
1
11
where
s10
=
1
1
2
+ s10
∑n
<
n=1
1
+
2
2
+
1
3
2
+
1
2
1
<
. . .
10
+
1
∑n
n=1
Thus,
=
102
∞
Given only this information, our best guess for S =
+ s10
1
2
[
,
1
+ s10
11
S A pprox
SApprox
= s10 +
1.54977 rounded to six significant figures.
is the average of its upper and lower bounds.
]
1
+ s10
10
1 ⎡ 1
1 ⎤
21
⎥⎥ = s10 +
⎢⎢
+
= 1.64522 .
2 ⎣ 10 11 ⎦
220
The maximum error in this estimate is given by:
⎛ ∞ 1 ⎞
⎟−S
⎜
⎜⎜
Approx
2⎟
⎟
n
⎝n = 1 ⎠
∑
<
1 ⎡ 1
1 ⎤
1
⎥⎥ =
⎢⎢
−
.
2 ⎣ 10 11 ⎦
220