Use the Direct Comparison Test to determine whether the series converges or diverges.

∑
n
=
1
∞
a
n
=
∑
n
=
1
∞
9
n
7
+
7
tan
−
1
⁡
(
n
)
3
n
8
−
2

The comparison series is

∑
n
=
1
∞
b
n
=
∑
n
=
1
∞
c
(
1
n
p
)
where
c
=
and
p
=
, which means
a
n

b
n
for all
n
≥
1
.

∑
n
=
1
∞
b
n
is a , therefore

∑
n
=
1
∞
a
n
by the Direct Comparison Test.

We are asked to use the **Direct Comparison Test** to determine whether the given series 

\[
\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{9n^7 + 7 \tan^{-1}(n)}{3n^8 - 2}
\]

**converges** or **diverges**. To use the **Direct Comparison Test**, we need to find an appropriate comparison series \( \sum_{n=1}^{\infty} b_n \) of the form

\[
\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} c \cdot \frac{1}{n^p},
\]

and identify constants \( c \) and \( p \). Let’s work through it step-by-step.

---

### Step 1: Simplify the Given Series as \( n \to \infty \)

We are given:

\[
a_n = \frac{9n^7 + 7 \tan^{-1}(n)}{3n^8 - 2}.
\]

For large \( n \), the terms \( 9n^7 \) and \( 3n^8 \) dominate in the numerator and denominator respectively. So, the behavior of the sequence \( a_n \) can be approximated by:

\[
a_n \approx \frac{9n^7}{3n^8} = \frac{9}{3} \cdot \frac{1}{n} = 3 \cdot \frac{1}{n}.
\]

This suggests that our comparison series might look like:

\[
b_n = 3 \cdot \frac{1}{n^p}.
\]

---

### Step 2: Identify the Comparison Series

Thus, the comparison series has:

\[
b_n = 3 \cdot \frac{1}{n^1} = \frac{3}{n}.
\]

This is a **p-series** with \( p = 1 \). 

---

### Step 3: Apply the Direct Comparison Test

- **p-series test**: The series \( \sum_{n=1}^{\infty} \frac{1}{n} \) **diverges** because \( p = 1 \).  

- Now, we check if \( a_n \leq b_n \) for all \( n \geq 1 \). For large \( n \), we already saw that:

  \[
  a_n \approx \frac{3}{n}.
  \]

  So, for sufficiently large \( n \), we have \( a_n \leq b_n \).  

---

### Step 4: Conclusion

Since \( \sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} \frac{3}{n} \) **diverges**, and \( a_n \leq b_n \) for all \( n \geq 1 \), the given series \( \sum_{n=1}^{\infty} a_n \) also **diverges** by the Direct Comparison Test.

---

### Final Answer

- \( c = 3 \)
- \( p = 1 \)
- \( \sum_{n=1}^{\infty} b_n \) diverges.
- Therefore, \( \sum_{n=1}^{\infty} a_n \) also diverges by the **Direct Comparison Test**.

---

Do you want further details or have any questions?  

Here are five related questions you may find useful:  
1. How does the p-series test determine convergence?  
2. What are the conditions for using the Direct Comparison Test?  
3. Can the Limit Comparison Test be applied here instead?  
4. What happens when \( p > 1 \) for p-series?  
5. How does \( \tan^{-1}(n) \) behave asymptotically for large \( n \)?  

**Tip:** When comparing series, always focus on the dominant terms as \( n \to \infty \).

Use the Direct Comparison Test to determine whether the series converges or diverges. ∑n=1∞a_n=∑n=1∞9n^7+7tan^−1(n)/3n^8−2. The comparison series is ∑n=1∞b_n=∑n=1∞c(1/n^p) where c= and p=, which means a_n/b_n for all n≥1. ∑n=1∞b_n is a , therefore ∑n=1∞a_n by the Direct Comparison Test.

This example applies the Direct Comparison Test to determine whether the series ∑n=1∞a_n=∑n=1∞(9n^7 + 7tan^−1(n)) / (3n^8 − 2) converges or diverges. It includes detailed steps and comparisons to a p-series ∑n=1∞b_n = c(1/n^p), with constants c = 3 and p = 1, showing how the series diverges.

Math Problem Statement

∑ n

1 ∞ a n

∑ n

∑ n

1 ∞ b n

∑ n

1 ∞ c ( 1 n p ) where c

and p

∑ n

∑ n

Solution

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