To determine the convergence of the series

$\sum_{n=1}^{\infty} \frac{\cos(n\pi)}{n},$

we proceed as follows:

Step 1: Simplify $\cos(n\pi)$

The term $\cos(n\pi)$ alternates between $-1$ and $1$ depending on whether $n$ is odd or even: $\cos(n\pi) = (-1)^n.$

Thus, the series becomes $\sum_{n=1}^\infty \frac{\cos(n\pi)}{n} = \sum_{n=1}^\infty \frac{(-1)^n}{n}.$

Step 2: Recognize the Alternating Series

The series is now of the form $\sum_{n=1}^\infty \frac{(-1)^n}{n},$ which is the alternating harmonic series.

Step 3: Apply the Alternating Series Test

The alternating series test (Leibniz criterion) states that an alternating series converges if:

1. The terms decrease in absolute value: $\left|\frac{1}{n+1}\right| < \left|\frac{1}{n}\right|$.
2. The terms approach zero: $\lim_{n \to \infty} \frac{1}{n} = 0$.

Both conditions are satisfied here, so the series converges.

Step 4: Determine the Nature of Convergence

The alternating harmonic series is known to converge conditionally because the harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges. The alternating factor $(-1)^n$ makes the series conditionally convergent.

Conclusion:

The series $\sum_{n=1}^\infty \frac{\cos(n\pi)}{n}$ converges conditionally.

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Would you like me to expand this explanation or compute the sum explicitly?

Related Questions:

1. What is the difference between conditional and absolute convergence?
2. How does the Alternating Series Test work for other series?
3. Can you compute the exact value of the alternating harmonic series?
4. What happens if we consider $\sum_{n=1}^\infty \frac{1}{n}$ instead of the alternating series?
5. How does the behavior of $\cos(n\pi)$ affect the series' convergence?

Tip:

Always check whether a series converges absolutely or conditionally by considering both the absolute value and any alternating terms.