Determine whether the series converges or diverges: $\displaystyle\sum_{n=1}^\infty \frac{\arctan(8n)}{n^{1.8}}$

Neetesh Kumar | December 25, 2024

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This is the solution to Math 1c
Assignment: 11.4 Question Number 17
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Step-by-step solution:

We aim to determine whether the given series converges or diverges.

Step 1: Analyze the general term:

The general term of the series is:

$a_n = \frac{\arctan(8n)}{n^{1.8}}$

Behavior of $\arctan(8n)$

For large $n$, $\arctan(8n)$ approaches $\frac{\pi}{2}$ because the arctangent function has a horizontal asymptote at $\frac{\pi}{2}$. Therefore, for large $n$:

$\arctan(8n) \sim \frac{\pi}{2}$

Substituting this approximation into $a_n$:

$a_n \sim \frac{\frac{\pi}{2}}{n^{1.8}} = \frac{C}{n^{1.8}}, \quad \text{where } C = \frac{\pi}{2}.$

Step 2: Compare with a $p$-series:

The term $\frac{C}{n^{1.8}}$ is a constant multiple of the $p$-series $\frac{1}{n^{1.8}}$, where $p = 1.8$. For a $p$-series $\displaystyle\sum \frac{1}{n^p}$, convergence occurs if $p > 1$. Since $p = 1.8 > 1$, the $p$-series $\displaystyle\sum \frac{1}{n^{1.8}}$ converges.

Step 3: Use the Limit Comparison Test:

To rigorously confirm convergence, we apply the limit comparison test with the convergent $p$-series $\displaystyle\sum \frac{1}{n^{1.8}}$:

Let $b_n = \frac{1}{n^{1.8}}$. Compute:

$\displaystyle\lim_{n \to \infty} \frac{a_n}{b_n} = \displaystyle\lim_{n \to \infty} \frac{\frac{\arctan(8n)}{n^{1.8}}}{\frac{1}{n^{1.8}}}$

Simplify:

$\displaystyle\lim_{n \to \infty} \arctan(8n)$

As $n \to \infty$, $\arctan(8n) \to \frac{\pi}{2}$, so:

$\displaystyle\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{\pi}{2}.$

Since the limit is a finite positive constant, and $\displaystyle\sum \frac{1}{n^{1.8}}$ converges, the given series also converges.

Conclusion: The series converges.

Final Answer:

$\boxed{\text{converges}}$

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