Math Problem Statement

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.

Solution

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

n=1an=n=19n7+7tan1(n)3n82\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 n=1bn\sum_{n=1}^{\infty} b_n of the form

n=1bn=n=1c1np,\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} c \cdot \frac{1}{n^p},

and identify constants cc and pp. Let’s work through it step-by-step.


Step 1: Simplify the Given Series as nn \to \infty

We are given:

an=9n7+7tan1(n)3n82.a_n = \frac{9n^7 + 7 \tan^{-1}(n)}{3n^8 - 2}.

For large nn, the terms 9n79n^7 and 3n83n^8 dominate in the numerator and denominator respectively. So, the behavior of the sequence ana_n can be approximated by:

an9n73n8=931n=31n.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:

bn=31np.b_n = 3 \cdot \frac{1}{n^p}.


Step 2: Identify the Comparison Series

Thus, the comparison series has:

bn=31n1=3n.b_n = 3 \cdot \frac{1}{n^1} = \frac{3}{n}.

This is a p-series with p=1p = 1.


Step 3: Apply the Direct Comparison Test

  • p-series test: The series n=11n\sum_{n=1}^{\infty} \frac{1}{n} diverges because p=1p = 1.

  • Now, we check if anbna_n \leq b_n for all n1n \geq 1. For large nn, we already saw that:

    an3n.a_n \approx \frac{3}{n}.

    So, for sufficiently large nn, we have anbna_n \leq b_n.


Step 4: Conclusion

Since n=1bn=n=13n\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} \frac{3}{n} diverges, and anbna_n \leq b_n for all n1n \geq 1, the given series n=1an\sum_{n=1}^{\infty} a_n also diverges by the Direct Comparison Test.


Final Answer

  • c=3c = 3
  • p=1p = 1
  • n=1bn\sum_{n=1}^{\infty} b_n diverges.
  • Therefore, n=1an\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>1p > 1 for p-series?
  5. How does tan1(n)\tan^{-1}(n) behave asymptotically for large nn?

Tip: When comparing series, always focus on the dominant terms as nn \to \infty.

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Math Problem Analysis

Mathematical Concepts

Series
Direct Comparison Test
Asymptotic Analysis
p-series

Formulas

a_n = (9n^7 + 7tan^−1(n)) / (3n^8 − 2)
b_n = c/n^p
p-series formula: ∑n=1∞ 1/n^p

Theorems

Direct Comparison Test
p-series Test

Suitable Grade Level

Undergraduate Calculus