5.3 The Integral Test and Estimates of Sums
5.3
Brian E. Veitch
The Integral Test and Estimates of Sums
The next few sections we learn techniques that help determine if a series converges. In the
last section we were able to find the sum of the series. It’s difficult to find the sum of a
series. We’ll spend most of our time now just determining if the series converges.
Consider the following series
∞
X
1
1
1
1
1
=
+
+
+
+ ...
2
1
2
2
2
n
1
2
3
4
n=1
We have no simple formula for the sequence of partial sums Sn . We’re going to compare
1
the terms of a sequence to the graph of y = 2 .
x
So the area of each of these rectangles is a term in the series. If we were to add up the area
of all these rectangles, we would have the sum for the series. Since we’re only concerned now
with knowing if it converges to a finite number, we’re only going to focus on its convergence.
Notice that the area of rectangles is less than the area
241
5.3 The Integral Test and Estimates of Sums
1
+
12
Z
Brian E. Veitch
∞
1
1
dx
x2
therefore, we have the following relation
Z ∞
∞
X
1
1
1
< 2+
dx
2
n
1
x2
1
n=1
Z
∞
The natural question at this point is, does
1
1
dx converge? It’s an improper integral
x2
and we know how to evaluate those.
Z
1
∞
Z
t
1
dx
2
1 x
t
1 = lim − t→∞
x
1
dx = lim
t→∞
x2
1
1 1
= lim − +
t→∞
t 1
= 0+1
= 1
So
Z ∞
∞
X
1
1
1
1
< 2+
dx < 2 + 1 = 2
2
2
n
1
x
1
1
n=1
Since we know our series is made up of all positive terms and we just showed the sum
can’t be more than 2, we conclude the series converges.
∞
X
1
√ converges.
Example 5.31. Let’s try to use this idea to determine if
n
n=1
1
Let’s compare it to the following graph of y = √ .
x
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5.3 The Integral Test and Estimates of Sums
Brian E. Veitch
1
Notice how the area of the each rectangle is larger than the area under y = √ . This
x
means
Z ∞
∞
X
1
1
√ >
√ dx
n
x
1
n=1
and
Z
1
∞
√
√
1
√ dx = lim 2 t − 2 1 = ∞
t→∞
x
So our series must add to more than what the integral sums to and we just showed that
it diverges to ∞. Therefore, our series
∞
X
1
√ diverges to ∞
x
n=1
1
Why did we choose rectangles above the graph of y = √ . Suppose we choose to make
x
1
our rectangles below y = √ , as shown below.
x
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5.3 The Integral Test and Estimates of Sums
This shows that the series
Brian E. Veitch
∞
X
1
√
n
n=1
sums to a number less than
Z
1
Z
∞
1
√ dx
x
∞
1
√ dx = ∞. So we really can’t say anything about the series. Sumx
1
ming to a number less than ∞ doesn’t mean anything.
The problem is
The method we used is called the Integral Test. It involves relating the terms of a
series to an improper integral.
5.3.1
The Integral Test
Suppose f is continuous, positive, decreasing function on [1, ∞) and let an = f (n). Then
the series
X
Z
converges if and only if
an
∞
f (x) dx converges. This means
1
Z
1. If
∞
f (x) dx converges, then
P
an converges.
1
244
5.3 The Integral Test and Estimates of Sums
Z
∞
f (x) dx diverges, then
2. If
Brian E. Veitch
P
an diverges.
1
Example 5.32. Does the following series converge?
∞
X
n=1
(n2
n
1
2
3
4
= 2 + 2 + 2 + 2 + ...
2
+ 1)
2
5
10
17
1. Let’s first determine if the sequence an =
lim
n→∞ (n2
(n2
n
converges to 0.
+ 1)2
n
n
≈ lim 4 = 0
2
n→∞ n
+ 1)
Since the sequence an → 0, it’s possible the series converges.
2. Check to make sure it’s decreasing. In order to use the integral test, the sequence
must always be decreasing. We figure this out by looking at the derivative.
Let f (x) =
(x2
x
. Usin the quotient rule to find f 0 (x), we get
+ 1)2
f 0 (x) =
1 − 3x2
(x2 + 1)2
For sufficiently large x values, f 0 (x) < 0, meaning the original function f (x) is always
decreasing.
3. The conditions to use the integral test have been met. Let’s use the integral test now.
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5.3 The Integral Test and Estimates of Sums
Z
1
∞
Brian E. Veitch
Z t
x
x
dx = lim
dx
2
2
2
2
t→∞
(x + 1)
1 (x + 1)
Let u = x2 + 1, and du = 2x dx
Z t −2
u
= lim
dx
t→∞ 2
2
t
1 = lim − t→∞
2u
2
1
1
= lim − +
t→∞
2t 4
1
= 0+
4
1
=
4
Z
Since
1
∞
(x2
x
dx converges, the integral test concludes the series
+ 1)2
∞
X
1
n
converges
(n2 + 1)2
.
5.3.2
Convergence of p-series
This will be an extremely useful series. In later sections, we use this series quite a bit.
Question: What values of p is
X 1
convergent?
np
It’s a pretty simple check. We use the integral test to determine the convergence. First,
we need to make sure the conditions have been satisfied to use the integral test.
1
> 0 for n ≥ 1? Yes, an > 0 for all values of p as long as n ≥ 1.
np
1
2. Is an decreasing? Let an = f (n) = p .
n
−p
Since f 0 (n) = p+1 < 0 for n ≥ 1, we conclude an is always decreasing.
n
1. Is an =
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5.3 The Integral Test and Estimates of Sums
Brian E. Veitch
3. Z
The conditions are met, so let’s use the integral test. For what vaules of p does
∞
1
dx converge?
xp
1
We showed back in the Improper Integral section that
Z
1
∞

1


, p>1
1
p−1
dx
=

xp
0,
p≤1
∞
X
1
converges when p > 1. Like I said...simple.
Therfore, the p-series,
np
1
Example 5.33. Consider the following p series. Which ones converge?
∞
X
1
1.
. This series converges since p = 16 > 1.
n16
1
2.
∞
X
n−1/4 . First, rewrite it as a p-series.
1
∞
X
1
1/4
n
1
which diverges because p =
1
< 1.
4
∞
X
1
3.
. This series diverges because p = −3 < 0.
−3
n
n=1
4.
∞
X
n−1.001 + n−3/2 . First, rewrite it as a p-series.
n=1
∞
X
1
1
n1.001
+
1
n3/2
Since the first p-series has p = 1.001 > 1 and the second series has p = 3/2, they both
converge.
247
5.3 The Integral Test and Estimates of Sums
Brian E. Veitch
Example 5.34. Determine if the following series converges
∞
X
ne−n
n=1
1. First, let’s determine if ne−n is decreasing and converges to 0.
f 0 (n) = e−n − ne−n = e−n (1 − n) < 0 for all n > 2
So the sequence is decreasing. Now we check to see if it converges to 0.
LH 1
n
= lim n = 0
n
n→∞ e
n=1 e
lim ne−n = lim
n→∞
2. The sequence an = ne−n satisfies the requirements to use the integral test. Let’s use
it now.
∞
Z
xe−x dx
1
We use Integration by Parts
Let u = x, dv = e−x dx
du = dx, v = −e−x
We’ll drop the bounds for now
Z
xe
−x
dx = −xe
−x
= −xe
−x
Z
−e−x dx
Z
e−x dx
−
+
= −xe−x − e−x
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5.3 The Integral Test and Estimates of Sums
Brian E. Veitch
Therefore,
Z
∞
xe
−x
xe−x dx
dx = lim
t→∞
1
t
Z
1
t
= lim −xe−x − e−x 1
t→∞
= lim −te−t − e−t − −1e−1 − e−1
t→∞
= 0 − 0 − −2e−1
= 2e−1
You can verify
lim −te−t = 0
t→∞
by using L’Hospitals Rule.
3. Now that we showed the integral converges, the integral test concludes the series
∞
X
ne−n converges
n=1
Example 5.35. Determine if the following series converges
∞
X
2
1
n ln n
1
decreases and converges to 0. At this point, something like
n ln n
this should be easy to identify.
1. The sequence an =
2. Since the sequence satisfies the requirements for the integral test, let’s use it now.
∞
Z
2
Let u = ln x, du =
1
dx
x ln x
1
dx. Using this substitution we have
x
Z
∞
ln 2
1
du = ln u|∞
ln 2 = ∞ − ln(ln 2) = ∞
u
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5.3 The Integral Test and Estimates of Sums
Brian E. Veitch
3. Since the integral diverges, the Integral Test concludes the following series
∞
X
n=2
1
diverges
n ln n
The last part of this section is to discuss a way to approximate a series. Suppose you
find the n-th partial sum Sn . Since this isn’t the true sum, there is a remainder amount Rn
such that
S = a1 + a2 + a3 + a4 + . . . + an + an+1 + an+2 + an+3 + . . .
|
{z
} |
{z
}
Sn
Rn
Sn is probably a decent approximation, but sometimes we want a bit of accuracy. If the
sequence an satisfies the integral test, then
Z
Rn = an+1 + an+2 + an+3 + . . . ≤
∞
f (x) dx
n
Example 5.36. Find the 5-th partial sum of
∞
X
n2
n=1
en
We add up the first 5 terms of the sequence an =
S5 =
n2
.
en
1
4
9
16
25
+ 2 + 3 + 4 + 25 = 1.818803087
1
e
e
e
e
e
Suppose I want a partial sum that’s accurate up to 0.00001. This means
Rn ≤ 0.00001
∞
Z
Z
f (x) =
n
Z
We need to find
n
∞
x2
=
ex
Z
∞
x2 e−x x ≤ 0.00001
n
∞
x2 e−x dx. Using integration by parts, you can verify that
n
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5.3 The Integral Test and Estimates of Sums
Brian E. Veitch
∞
Z
x2 e−x dx = e−n (n2 + 2n + 2)
n
Now we just have to solve for n
en (n2 + 2n + 2) < 0.00001
There really isn’t an easy way to solve this. I recommend just try plugging in some values
for n.
1. n = 5
e5 (52 + 2(5) + 2) = 0.249304 > 0.00001
2. n = 10
e10 (102 + 2(10) + 2) = 0.005539 > 0.00001
3. n = 15
e15 (152 + 2(15) + 2) = 0.000079 > 0.00001
4. n = 16
e16 (162 + 2(15) + 2) = 0.000033 > 0.00001
5. n = 17
e17 (172 + 2(17) + 2) = 0.000013 > 0.00001
6. n = 18
e18 (182 + 2(18) + 2) = 0.000006 < 0.00001
Therefore, we need to add up the first 18 terms to be accurate to within 0.00001.
18
X
n2
n=1
en
=
1
4
9
18
+ 2 + 3 + ... + 18 = 1.99229
1
e
e
e
e
.
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