1. Given the series \[\sum_{n=1}^{\infty} \frac{3n^2 - 2}{\sqrt{n^8 + n^3 + n + 4}}:\](a) Let \( a_n = \frac{3n^2 - 2}{\sqrt{n^8 + n^3 + n + 4}} \). Define a series \( b_n \) with which to compare it.(b) Plot the first 50 terms of \( a_n \) and \( b_n \) on the same graph to determine which is larger. If the graph is not clear, use the logical test \( a_n < b_n \) to test the logical value comparing each term.(c) State whether \( \sum_{n=1}^{\infty} b_n \) converges or not (you should be able to tell by looking), and state whether any conclusion can be made about the convergence of \( \sum_{n=1}^{\infty} a_n \) as a result.(d) If (c) is inconclusive, determine whether \( \frac{a_n}{b_n} \) converges or not, and state your conclusion about the convergence of \( \sum_{n=3}^{\infty} a_n \) (NOTE: If you still cannot conclude anything, start over with a different \( b_n! \)).### Explanation of the Task- **Part (a):** You need to define a comparison series \( b_n \) for the given series \( a_n \).- **Part (b):** Create a plot with the first 50 terms of \( a_n \) and \( b_n \) to visually compare their sizes.- **Part (c):** Determine if the series \( \sum b_n \) converges, and then decide if this helps in concluding the behavior of the series \( \sum a_n \).- **Part (d):** Use the ratio \( \frac{a_n}{b_n} \) if part (c) is not conclusive, to further explore convergence. If this step also fails, begin with a new series \( b_n \).

Given: Here the given series is ∑n=1∞3n2-2n8+n3+n+4.

3n2 – 2 - 1. Given the series /n® + n³ + n + 4 n=1 3n2 – 2 (a) Let an Define a series b, with which to compare it. Vn8 + n³ + n +4 (b) Plot the first 50 terms of an and b, on the same graph to determine which is larger. If the graph is not clear, use the logical test a, < b, to test the logical value comparing each term. (c) State whether b, converges or not (you should be able to tell by looking), and n=1 state whether any conclusion can be made about the convergence of an as a n=1

3n2 – 2 - 1. Given the series /n® + n³ + n + 4 n=1 3n2 – 2 (a) Let an Define a series b, with which to compare it. Vn8 + n³ + n +4 (b) Plot the first 50 terms of an and b, on the same graph to determine which is larger. If the graph is not clear, use the logical test a, < b, to test the logical value comparing each term. (c) State whether b, converges or not (you should be able to tell by looking), and n=1 state whether any conclusion can be made about the convergence of an as a n=1

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

1. Given the series

\[
\sum_{n=1}^{\infty} \frac{3n^2 - 2}{\sqrt{n^8 + n^3 + n + 4}}:
\]

(a) Let \( a_n = \frac{3n^2 - 2}{\sqrt{n^8 + n^3 + n + 4}} \). Define a series \( b_n \) with which to compare it.

(b) Plot the first 50 terms of \( a_n \) and \( b_n \) on the same graph to determine which is larger. If the graph is not clear, use the logical test \( a_n < b_n \) to test the logical value comparing each term.

(c) State whether \( \sum_{n=1}^{\infty} b_n \) converges or not (you should be able to tell by looking), and state whether any conclusion can be made about the convergence of \( \sum_{n=1}^{\infty} a_n \) as a result.

(d) If (c) is inconclusive, determine whether \( \frac{a_n}{b_n} \) converges or not, and state your conclusion about the convergence of \( \sum_{n=3}^{\infty} a_n \) (NOTE: If you still cannot conclude anything, start over with a different \( b_n! \)).

### Explanation of the Task
- **Part (a):** You need to define a comparison series \( b_n \) for the given series \( a_n \).
- **Part (b):** Create a plot with the first 50 terms of \( a_n \) and \( b_n \) to visually compare their sizes.
- **Part (c):** Determine if the series \( \sum b_n \) converges, and then decide if this helps in concluding the behavior of the series \( \sum a_n \).
- **Part (d):** Use the ratio \( \frac{a_n}{b_n} \) if part (c) is not conclusive, to further explore convergence. If this step also fails, begin with a new series \( b_n \).

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