(-1)^n Ln(n)/n Converge

Jasonbradley

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Jun 16, 2024

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Does the Series (-1)^n ln(n)/n Converge?

This article explores the convergence of the series ∑ (-1)^n ln(n)/n. We will determine if this series converges, and if so, we will analyze its convergence behavior.

The Alternating Series Test

The series in question is an alternating series because the terms alternate in sign due to the presence of (-1)^n. To investigate convergence, we apply the Alternating Series Test:

1. Non-increasing terms: We need to show that the absolute value of the terms, |ln(n)/n|, is decreasing for sufficiently large n.
   • Since ln(x) grows slower than x, as n increases, the ratio ln(n)/n decreases.
2. Limit approaches zero: We must ensure that the limit as n approaches infinity of |ln(n)/n| is equal to zero.
   • Using L'Hopital's Rule, we find that lim (n→∞) ln(n)/n = lim (n→∞) 1/n = 0.

Since both conditions of the Alternating Series Test are satisfied, we conclude that the series ∑ (-1)^n ln(n)/n converges.

Conditional Convergence

The series converges due to the alternating nature of its terms. However, if we consider the series of absolute values, ∑ |ln(n)/n|, we find that it diverges. This can be shown using the Integral Test:

1. Positive and decreasing terms: We know that |ln(n)/n| is positive and decreasing.
2. Integral comparison: We compare the series to the integral of |ln(x)/x| from 1 to infinity. This integral diverges, indicating that the series of absolute values also diverges.

Therefore, the series ∑ (-1)^n ln(n)/n converges conditionally. This means that the series converges, but only because of the alternating signs. If the signs were all positive or all negative, the series would diverge.

Conclusion

The series ∑ (-1)^n ln(n)/n converges conditionally. It satisfies the conditions of the Alternating Series Test, confirming its convergence. However, the series of absolute values diverges, indicating that the convergence is solely due to the alternating nature of the series.