This question tests your ability to classify infinite series as convergent or divergent. The series $\sum_{n=1}^{\infty} 3\left(\frac{2}{3}\right)^n$ is a geometric series with first term $a = 3 \cdot \frac{2}{3} = 2$ and common ratio $r = \frac{2}{3}$. Since $|r| = \frac{2}{3} < 1$, the geometric series converges to $\frac{a}{1-r} = \frac{2}{1-\frac{2}{3}} = 6$. Choice E incorrectly claims $|r| > 1$, which would be true if the ratio were $\frac{3}{2}$ instead of $\frac{2}{3}$. Remember: geometric series $\sum ar^n$ converge when $|r| < 1$ and diverge when $|r| \geq 1$.

This question tests your ability to classify infinite series as convergent or divergent. For the series $\sum_{n=1}^{\infty} \frac{n}{n+1}$, we first check if the terms approach zero: $\lim_{n\to\infty} \frac{n}{n+1} = \lim_{n\to\infty} \frac{1}{1+\frac{1}{n}} = 1 \neq 0$. By the divergence test (nth term test), if $\lim_{n\to\infty} a_n \neq 0$, then $\sum a_n$ must diverge. The series diverges because we're essentially adding terms that approach 1, giving an infinite sum. Choice D incorrectly suggests that having $\lim a_n = 1$ would make the series converge. Remember: if terms don't approach zero, the series must diverge—this is the first test to try.

This question tests your ability to classify infinite series as convergent or divergent. The series $\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ can be analyzed using the integral test since $f(x) = \frac{1}{x\ln x}$ is positive, continuous, and decreasing for $x \geq 2$. The integral $\int_2^{\infty} \frac{1}{x\ln x}dx$ can be evaluated by substitution: let $u = \ln x$, then $du = \frac{1}{x}dx$, giving $\int \frac{1}{u}du = \ln|u| = \ln(\ln x)$. Since $\lim_{x\to\infty} \ln(\ln x) = \infty$, the integral diverges, so the series diverges. Choice E incorrectly assumes that terms approaching zero guarantees convergence. Remember: the integral test is powerful for series involving logarithms—if $\int f(x)dx$ diverges, then $\sum f(n)$ diverges.

This problem requires analyzing the series $\sum_{n=1}^{\infty} \frac{n}{n^2+4}$. For large $n$, the terms behave like $\frac{n}{n^2} = \frac{1}{n}$, suggesting comparison with the harmonic series. We can verify that $\lim_{n\to\infty} \frac{n/(n^2+4)}{1/n} = \lim_{n\to\infty} \frac{n^2}{n^2+4} = 1$. Since this limit is positive and finite, and the harmonic series $\sum \frac{1}{n}$ diverges, our series also diverges by the limit comparison test. The comparison with a p-series (option A) would be incorrect because the series behaves like $p=1$, not $p>1$. When the highest powers of $n$ in numerator and denominator differ by exactly 1, the series typically behaves like the harmonic series.

This question tests the skill of classifying infinite series as convergent or divergent. The series ∑ $(-1)^n$ / n² has absolute series ∑ 1/n², which converges as a p-series with p=2>1. Since the absolute series converges, the original converges absolutely by the absolute convergence theorem. This implies conditional convergence as well, but absolute is stronger. A tempting distractor is choice D, which claims conditional convergence by the alternating series test, but this fails because absolute convergence is proven, making it absolutely convergent. A transferable strategy for classifying series is to check absolute convergence first for alternating series using tests like comparison or p-series, as it provides more information than conditional alone.

This question tests the skill of classifying infinite series as convergent or divergent. The series ∑ $(-1)^{n+1}$/n is the alternating harmonic series, which converges by the alternating series test since the absolute terms 1/n decrease monotonically to 0. However, the absolute series ∑ 1/n is the divergent harmonic series, indicating conditional convergence. This distinction requires checking both conditional and absolute convergence for alternating series. A tempting distractor is choice A, which claims divergence because it is a harmonic series, but this fails as it ignores the alternating signs that enable conditional convergence. A transferable strategy for classifying series is to first check if terms go to 0, then apply tests like alternating series for signs or comparison for positives, and verify absolute convergence if needed.

This problem involves determining convergence of an alternating series. The series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ is the alternating harmonic series, where terms alternate in sign and have magnitude $1/n$. To apply the alternating series test, we check: (1) terms decrease in magnitude: $\frac{1}{n+1} < \frac{1}{n}$, and (2) $\lim_{n \to \infty} \frac{1}{n} = 0$. Both conditions are satisfied, so the series converges by the alternating series test. Choice A incorrectly claims divergence—the alternating series test never proves divergence, only convergence. For alternating series $\sum(-1)^n a_n$, check if $a_n$ decreases to 0 for convergence.

This problem involves the series $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$, which can be analyzed using partial fractions. We decompose $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$, creating a telescoping series. The partial sums are $S_N = \sum_{n=1}^{N} \left(\frac{1}{n} - \frac{1}{n+1}\right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + ... + \left(\frac{1}{N} - \frac{1}{N+1}\right) = 1 - \frac{1}{N+1}$. As $N \to \infty$, we get $\lim_{N\to\infty} S_N = 1$, so the series converges to 1. The harmonic series comparison (option A) is incorrect because this series converges while the harmonic series diverges. When you see products of consecutive integers in the denominator, always check for telescoping by partial fractions.

This question tests convergence of alternating series. The series $\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$ alternates in sign with terms of magnitude $\frac{1}{\sqrt{n}}$. For the alternating series test: (1) terms decrease: $\frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}}$, and (2) $\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0$. Both conditions hold, so the series converges by the alternating series test. Note that the series of absolute values $\sum \frac{1}{\sqrt{n}}$ diverges (p-series with p = 1/2), making this conditionally convergent. For alternating series, always check the two conditions of the alternating series test before considering other methods.

This problem requires classifying a series combining polynomial and exponential terms. The series $\sum_{n=1}^{\infty}\frac{n^2}{2^n}$ is ideal for the Ratio Test due to the exponential denominator. Applying the Ratio Test: $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\frac{(n+1)^2/2^{n+1}}{n^2/2^n} = \lim_{n\to\infty}\frac{(n+1)^2}{2n^2} = \frac{1}{2}\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^2 = \frac{1}{2} < 1$. Since the limit is less than 1, the series converges by the Ratio Test. Choice C suggesting the Root Test could also work, but the Ratio Test is more straightforward here; choice E incorrectly focuses on the numerator growing while ignoring that $2^n$ grows much faster. When exponential terms appear in denominators, they typically dominate polynomial numerators, leading to convergence via the Ratio Test.