To analyze the nature of the series

$\sum_{n \geq 1} \arctan\left(\frac{1}{n}\right),$

we determine whether it converges or diverges. Here's how:

1. Expression Behavior Analysis:

The terms of the series are given by: $a_n = \arctan\left(\frac{1}{n}\right).$

• For large $n$, $\frac{1}{n} \to 0$, and using the approximation $\arctan(x) \approx x$ for small $x$, we have: $a_n \sim \frac{1}{n}.$

2. Comparison with a Known Series:

The series $\sum \frac{1}{n}$ is the harmonic series, which is divergent. Since $a_n \sim \frac{1}{n}$, this suggests the given series diverges.

3. Formal Test Using Limit Comparison Test:

We compare $a_n = \arctan\left(\frac{1}{n}\right)$ with $b_n = \frac{1}{n}$.

The limit comparison test evaluates: $\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\arctan\left(\frac{1}{n}\right)}{\frac{1}{n}}.$

Using the approximation $\arctan(x) \approx x$ for small $x$: $\lim_{n \to \infty} \frac{\arctan\left(\frac{1}{n}\right)}{\frac{1}{n}} = \lim_{n \to \infty} 1 = 1.$

Since the limit is finite and nonzero, the series $\sum_{n \geq 1} a_n$ has the same nature as $\sum_{n \geq 1} \frac{1}{n}$, which diverges.

Conclusion:

$\sum_{n \geq 1} \arctan\left(\frac{1}{n}\right) \quad \text{is divergent.}$

Would you like me to explain any part in more detail or clarify the convergence tests?

Related Questions:

1. What is the general form of the limit comparison test, and when is it useful?
2. How does the approximation $\arctan(x) \approx x$ influence the analysis?
3. Can you apply the integral test to this series to verify divergence?
4. How does this series compare to $\sum_{n=1}^\infty \ln(1+\frac{1}{n})$?
5. What happens if the terms involve $\arctan(1/n^2)$ instead?

Tip:

For series involving $\arctan(x)$, always consider its asymptotic behavior for large $x$ or small $x$ to simplify calculations.