To determine the convergence or divergence of the series

$\sum_{n=2}^{\infty} \frac{n}{\ln(n)},$

we can apply the Comparison Test or the Integral Test, as these are useful for series involving terms with logarithmic functions.

Reasoning for Test Selection:

1. Comparison Test: We can compare $\frac{n}{\ln(n)}$ with a known divergent series, as it resembles a harmonic-like series (which tends to diverge). We know that $\sum \frac{1}{n}$ diverges, and $\frac{n}{\ln(n)}$ grows even faster than $\frac{1}{n}$, suggesting divergence.

2. Integral Test: By converting the sum into an integral, $\int \frac{x}{\ln(x)} dx$, and examining whether this integral converges or diverges, we can determine the behavior of the series.

For this specific case, the Integral Test would be quite suitable because we can analyze the integral:

$\int_{2}^{\infty} \frac{x}{\ln(x)} \, dx.$

This integral diverges, implying that the series $\sum_{n=2}^{\infty} \frac{n}{\ln(n)}$ also diverges.

Would you like a detailed explanation of the calculations, or do you have further questions?

Related Questions

1. How does the Comparison Test work for determining convergence or divergence?
2. Can the Ratio Test be applied to this series, and if so, what does it show?
3. Why does the series $\sum \frac{1}{n}$ diverge?
4. What other types of series typically use the Integral Test?
5. How does logarithmic growth affect the convergence of series?

Tip

When dealing with logarithmic terms in series, consider using the Integral Test or Comparison Test with a similar simpler series for quicker insights into convergence behavior.