The problem presented is the infinite series:

$\sum_{n=1}^\infty \frac{\arctan n}{n}.$

This is a series involving the arctangent function and division by $n$. Let us analyze whether the series converges or diverges.

Step 1: Check Convergence

To determine convergence, we can use the Comparison Test, the Ratio Test, or related methods. First, observe the growth of the terms $\frac{\arctan n}{n}$ as $n \to \infty$.

• The arctangent function, $\arctan n$, grows logarithmically for large $n$, i.e., much slower than $n$. However, dividing it by $n$ might still lead to a divergence.

Asymptotic Behavior of $\frac{\arctan n}{n}$

For large $n$, the dominant term of $\arctan n$ approaches $\frac{\pi}{2}$. Thus, for large $n$, $\frac{\arctan n}{n} \sim \frac{\frac{\pi}{2}}{n}.$ This resembles the harmonic series scaled by a constant, $\frac{\pi}{2n}$, which diverges because $\sum \frac{1}{n}$ is a well-known divergent series.

Step 2: Formal Test

We apply the Limit Comparison Test with $a_n = \frac{\arctan n}{n}$ and $b_n = \frac{1}{n}$.

$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{\arctan n}{n}}{\frac{1}{n}} = \lim_{n \to \infty} \arctan n = \frac{\pi}{2}.$

Since the limit is a positive finite constant ($\frac{\pi}{2}$), and $\sum \frac{1}{n}$ diverges, the series $\sum \frac{\arctan n}{n}$ also diverges.

Final Answer:

The infinite series: $\sum_{n=1}^\infty \frac{\arctan n}{n}$ diverges.

Would you like to go deeper into why this series diverges, or do you have additional questions?

Related Questions:

1. What is the formal definition of the Limit Comparison Test?
2. Why does $\sum \frac{1}{n}$ (harmonic series) diverge?
3. How does the growth of $\arctan n$ compare to logarithmic growth for large $n$?
4. Can the Ratio Test be used here to prove divergence?
5. Are there any conditions where a series involving $\arctan n$ would converge?

Tip:

When analyzing infinite series, comparing the asymptotic growth of the terms with a known series (e.g., geometric or harmonic series) is a quick way to determine convergence.