Assessing convergence types—absolute, conditional, or divergent—is key for physics models using series. Distinguish by checking if ∑|a_n| converges (absolute) or diverges while the alternating converges (conditional). Apply AST: (ln n)/n decreases to zero for n≥3, confirming convergence. The absolute series ∑ (ln n)/n diverges by Integral Test, as ∫ (ln x)/x dx = (ln $x)^2$/2 → ∞. Claiming absolute convergence by comparison to ∑1/n² fails because (ln n)/n grows slower than 1/n but still diverges. A transferable approach: test absolute with integral or comparison, falling back to AST for conditional if needed.

This problem tests your ability to classify series convergence as absolute, conditional, or divergent. The series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$ is the alternating harmonic series, which converges by the alternating series test since $\frac{1}{n}$ decreases to 0. To check absolute convergence, we examine $\sum_{n=1}^{\infty} \left|\frac{(-1)^{n-1}}{n}\right| = \sum_{n=1}^{\infty} \frac{1}{n}$, which is the harmonic series and diverges. Since the original series converges but the absolute value series diverges, we have conditional convergence. Choice E incorrectly claims the alternating series test proves absolute convergence, but this test only establishes convergence of the alternating series itself. Remember: a series converges conditionally when it converges but its absolute value series diverges.

This problem examines $\sum_{n=1}^{\infty} (-1)^n\frac{1}{n^3+n}$. The alternating series test confirms convergence since $\frac{1}{n^3+n}$ decreases to 0. For absolute convergence, we analyze $\sum_{n=1}^{\infty} \frac{1}{n^3+n} = \sum_{n=1}^{\infty} \frac{1}{n(n^2+1)}$. For large $n$, this behaves like $\frac{1}{n^3}$, and by limit comparison with the convergent p-series $\sum \frac{1}{n^3}$, our series converges. Since the absolute value series converges, the original series converges absolutely. Choice A incorrectly suggests only conditional convergence without properly checking the absolute value series. The key: when terms behave like $\frac{1}{n^p}$ with $p>1$ for large $n$, expect absolute convergence.

This problem asks about the convergence of $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^2}$. The alternating series test confirms convergence since $\frac{1}{n^2}$ decreases to 0. For absolute convergence, we check $\sum_{n=1}^{\infty} \frac{1}{n^2}$, which is a p-series with $p=2>1$, so it converges. Since the absolute value series converges, the original series converges absolutely. Choice D incorrectly claims divergence for a p-series with $p=2$, but p-series converge when $p>1$. The fundamental principle: a series converges absolutely when its absolute value series converges, making the alternating nature irrelevant to the convergence classification.

This question requires determining the convergence type of $\sum_{n=2}^{\infty} \frac{(-1)^n}{n\ln n}$. For the alternating series test, we need $\frac{1}{n\ln n}$ to be decreasing and approach 0, which it does since both $n$ and $\ln n$ increase. To check absolute convergence, we examine $\sum_{n=2}^{\infty} \frac{1}{n\ln n}$, which diverges by the integral test since $\int_2^{\infty} \frac{1}{x\ln x}dx = \ln(\ln x)|_2^{\infty} = \infty$. Therefore, the series converges conditionally. Choice B incorrectly suggests that slow growth of $\ln n$ causes divergence, when actually the alternating nature ensures convergence. The key insight: when an alternating series converges but its absolute value series diverges, you have conditional convergence.

Classifying alternating perturbations like ∑ $(-1)^n$ / (√n +1) involves convergence skills. Absolute: ∑ 1/(√n +1) ≈ ∑ 1/√n diverges (p=1/2<1). Conditional: AST as 1/(√n +1) decreases to 0. Converges conditionally. The distractor of absolute by comparison to ∑ $1/n^{3/2}$ fails, as it diverges like 1/√n, not converges. Strategy: approximate for large n; if absolute like divergent p-series, test AST for conditional.

In Fourier-type sums, classifying convergence like for ∑ $(-1)^n$ (ln n)/n is vital. Absolute: ∑ (ln n)/n diverges by integral test, as ∫ (ln x)/x dx = (1/2)(ln $x)^2$ → ∞. Conditional: AST applies because (ln n)/n decreases to 0 for large n. Thus, conditional. The distractor of divergence because (ln n)/n does not decrease fails, as it does decrease for n > e. Tip: assess absolute with integrals; if divergent, verify decreasing and limit 0 for conditional via AST.

Determining absolute or conditional convergence is essential for analyzing infinite series in calculus. Absolute convergence requires the series of absolute terms to converge, while conditional means only the original converges. Tests involve checking ∑ |a_n| with comparison or integral, and if divergent, applying the alternating series test to the original. Here, the absolute series ∑ ln(n+1)/√n diverges by comparison to ∑ 1/√n (p=1/2<1), but the original converges by AST since ln(n+1)/√n decreases to 0. A common distractor is thinking it converges absolutely by ratio test, but the ratio limit is 1, inconclusive. Always verify absolute convergence first; if it fails, assess conditional via AST for alternating series with decreasing terms to 0.

Determining whether a series converges absolutely, conditionally, or diverges is a key skill in series analysis. Absolute convergence occurs when the series of absolute values converges, while conditional convergence means the original series converges but the absolute series diverges. To test, first check if the absolute series ∑ 1/(√n + √(n+1)) converges, which it does not as it behaves like ∑ 1/(2√n) (p=1/2<1). Since it alternates with terms 1/(√n + √(n+1)) decreasing to 0, it converges by AST, hence conditionally. A tempting distractor claims divergence because terms do not go to 0, but they do approach 0. Remember, always check absolute convergence first, and if it fails, then check for conditional convergence using AST or other methods.

Determining whether a series converges absolutely, conditionally, or diverges is a key skill in series analysis. Absolute convergence occurs when the series of absolute values converges, while conditional convergence means the original series converges but the absolute series diverges. To test, first check if the absolute series ∑ 1/(n $(ln(n+1))^2$) converges, which it does by integral test as ∫ dx/(x (ln $x)^2$) converges. Thus, the original series converges absolutely. A tempting distractor suggests convergence by comparison to ∑ 1/(n ln n) so diverges, but the extra (ln $n)^2$ in denominator makes it converge. Remember, always check absolute convergence first, and if it fails, then check for conditional convergence using AST or other methods.