Question
The series
s
=
∑
n
=
1
∞
1
(
2
n
+
1
)
(
3
n
+
1
)
s=
n=1
∑
∞
(2n+1)(3n+1)
1
converges.
We use
s
7
=
∑
n
=
1
7
1
(
2
n
+
1
)
(
3
n
+
1
)
≈
0.147498...
s
7
=
n=1
∑
7
(2n+1)(3n+1)
1
≈0.147498... to approximate
s
s.
Use the integral test to give a lower bound and an upper bound for the error
s
−
s
7
s−s
7
.
(You may use the fact that
∫
1
(
2
x
+
1
)
(
3
x
+
1
)
d
x
=
ln
(
6
x
+
2
6
x
+
3
)
+
C
∫
(2x+1)(3x+1)
1
dx=ln(
6x+3
6x+2
)+C.)
Give exact values and separate your answers by a comma.
Formatting example:
π
,
17
−
π
π,17−π
L
o
w
e
r
b
o
u
n
d
,
U
p
p
e
r
b
o
u
n
d
=
Lower bound, Upper bound=
s
=
∑
n
=
1
∞
1
(
2
n
+
1
)
(
3
n
+
1
)
s=
n=1
∑
∞
(2n+1)(3n+1)
1
converges.
We use
s
7
=
∑
n
=
1
7
1
(
2
n
+
1
)
(
3
n
+
1
)
≈
0.147498...
s
7
=
n=1
∑
7
(2n+1)(3n+1)
1
≈0.147498... to approximate
s
s.
Use the integral test to give a lower bound and an upper bound for the error
s
−
s
7
s−s
7
.
(You may use the fact that
∫
1
(
2
x
+
1
)
(
3
x
+
1
)
d
x
=
ln
(
6
x
+
2
6
x
+
3
)
+
C
∫
(2x+1)(3x+1)
1
dx=ln(
6x+3
6x+2
)+C.)
Give exact values and separate your answers by a comma.
Formatting example:
π
,
17
−
π
π,17−π
L
o
w
e
r
b
o
u
n
d
,
U
p
p
e
r
b
o
u
n
d
=
Lower bound, Upper bound=
Answers
GPT-4o mini
To use the integral test to estimate the error \( s - s_7 \), we need to evaluate the integral of the function \( f(x) = \frac{1}{(2x+1)(3x+1)} \) from \( n = 7 \) to \( \infty \).
### Step 1: Set up the bounds for the error
The integral test states that if \( f(x) \) is decreasing, then:
\[
s - s_7 \approx \int_7^\infty f(x) \, dx
\]
The error \( s - s_7 \) can be bounded as follows:
\[
f(8) < s - s_7 < \int_7^\infty f(x) \, dx
\]
### Step 2: Calculate the upper bound using the integral
We evaluate the integral:
\[
\int_7^\infty \frac{1}{(2x+1)(3x+1)} \, dx
\]
Using the provided antiderivative:
\[
\int \frac{1}{(2x+1)(3x+1)} \, dx = \ln\left(\frac{6x+3}{6x+2}\right) + C
\]
we can compute the definite integral as follows:
\[
\int_7^\infty \frac{1}{(2x+1)(3x+1)} \, dx = \lim_{b \to \infty} \left[ \ln\left(\frac{6x+3}{6x+2}\right) \right]_{7}^{b}
\]
### Step 3: Evaluate at the bounds
1. As \( x \to \infty \):
\[
\ln\left(\frac{6b + 3}{6b + 2}\right) \to \ln(1) = 0
\]
2. At \( x = 7 \):
\[
\ln\left(\frac{6(7) + 3}{6(7) + 2}\right) = \ln\left(\frac{42 + 3}{42 + 2}\right) = \ln\left(\frac{45}{44}\right)
\]
Thus, the upper bound for the error is:
\[
\int_7^\infty \frac{1}{(2x+1)(3x+1)} \, dx = 0 - \ln\left(\frac{45}{44}\right) = -\ln\left(\frac{45}{44}\right)
\]
### Step 4: Compute the lower bound using \( f(8) \)
Next, we compute \( f(8) \):
\[
s - s_7 > f(8) = \frac{1}{(2(8)+1)(3(8)+1)} = \frac{1}{(16+1)(24+1)} = \frac{1}{17 \times 25} = \frac{1}{425}
\]
### Final Bounds
Thus, we have:
\[
\frac{1}{425} < s - s_7 < -\ln\left(\frac{45}{44}\right)
\]
### Final Values
The exact bounds for the error are:
\[
\text{Lower bound} = \frac{1}{425}, \quad \text{Upper bound} = -\ln\left(\frac{45}{44}\right)
\]
Expressing this in the required format:
\[
\frac{1}{425}, -\ln\left(\frac{45}{44}\right)
\]
Thus the final answer is:
\[
\frac{1}{425}, -\ln\left(\frac{45}{44}\right)
\]
### Step 1: Set up the bounds for the error
The integral test states that if \( f(x) \) is decreasing, then:
\[
s - s_7 \approx \int_7^\infty f(x) \, dx
\]
The error \( s - s_7 \) can be bounded as follows:
\[
f(8) < s - s_7 < \int_7^\infty f(x) \, dx
\]
### Step 2: Calculate the upper bound using the integral
We evaluate the integral:
\[
\int_7^\infty \frac{1}{(2x+1)(3x+1)} \, dx
\]
Using the provided antiderivative:
\[
\int \frac{1}{(2x+1)(3x+1)} \, dx = \ln\left(\frac{6x+3}{6x+2}\right) + C
\]
we can compute the definite integral as follows:
\[
\int_7^\infty \frac{1}{(2x+1)(3x+1)} \, dx = \lim_{b \to \infty} \left[ \ln\left(\frac{6x+3}{6x+2}\right) \right]_{7}^{b}
\]
### Step 3: Evaluate at the bounds
1. As \( x \to \infty \):
\[
\ln\left(\frac{6b + 3}{6b + 2}\right) \to \ln(1) = 0
\]
2. At \( x = 7 \):
\[
\ln\left(\frac{6(7) + 3}{6(7) + 2}\right) = \ln\left(\frac{42 + 3}{42 + 2}\right) = \ln\left(\frac{45}{44}\right)
\]
Thus, the upper bound for the error is:
\[
\int_7^\infty \frac{1}{(2x+1)(3x+1)} \, dx = 0 - \ln\left(\frac{45}{44}\right) = -\ln\left(\frac{45}{44}\right)
\]
### Step 4: Compute the lower bound using \( f(8) \)
Next, we compute \( f(8) \):
\[
s - s_7 > f(8) = \frac{1}{(2(8)+1)(3(8)+1)} = \frac{1}{(16+1)(24+1)} = \frac{1}{17 \times 25} = \frac{1}{425}
\]
### Final Bounds
Thus, we have:
\[
\frac{1}{425} < s - s_7 < -\ln\left(\frac{45}{44}\right)
\]
### Final Values
The exact bounds for the error are:
\[
\text{Lower bound} = \frac{1}{425}, \quad \text{Upper bound} = -\ln\left(\frac{45}{44}\right)
\]
Expressing this in the required format:
\[
\frac{1}{425}, -\ln\left(\frac{45}{44}\right)
\]
Thus the final answer is:
\[
\frac{1}{425}, -\ln\left(\frac{45}{44}\right)
\]