This question tests your ability to classify infinite series as convergent or divergent. For the series $\sum_{n=1}^{\infty}\frac{3n}{n^2+1}$, we analyze the behavior for large $n$: $\frac{3n}{n^2+1} \sim \frac{3n}{n^2} = \frac{3}{n}$. By limit comparison with $\sum\frac{1}{n}$ (harmonic series), we compute $\lim_{n\to\infty} \frac{\frac{3n}{n^2+1}}{\frac{1}{n}} = \lim_{n\to\infty} \frac{3n^2}{n^2+1} = 3$. Since this limit is positive and finite, and $\sum\frac{1}{n}$ diverges, our series also diverges. Choice C incorrectly assumes that terms approaching zero guarantees convergence. Remember: use limit comparison when terms behave like a known series for large $n$.

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 $\sum_{n=1}^{\infty}\frac{1}{n}$ is the harmonic series, which is a p-series with $p = 1$. For p-series $\sum\frac{1}{n^p}$, the series converges when $p > 1$ and diverges when $p \leq 1$. Since $p = 1$, the harmonic series diverges. Choice B incorrectly assumes that because the terms $\frac{1}{n} \to 0$, the series must converge, but this is only a necessary condition, not sufficient. Remember: the harmonic series is the classic example of a divergent series whose terms approach zero.

This question tests your ability to classify infinite series as convergent or divergent. For the series $\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^n$, we need to find $\lim_{n\to\infty} \left(\frac{n}{n+1}\right)^n$. Rewriting as $\left(\frac{1}{1+\frac{1}{n}}\right)^n = \frac{1}{\left(1+\frac{1}{n}\right)^n}$, and since $\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e$, we get $\lim_{n\to\infty} \left(\frac{n}{n+1}\right)^n = \frac{1}{e} \approx 0.368 \neq 0$. By the divergence test, since terms don't approach zero, the series diverges. Choice E incorrectly interprets the ratio test—having a limit less than 1 would indicate convergence, not divergence. Remember: always check if $\lim a_n = 0$ first; if not, the series must diverge.

This question tests your ability to classify infinite series as convergent or divergent. The series $\sum_{n=1}^{\infty}\frac{5}{2^n} = 5\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^n$ is a geometric series with first term $a = \frac{5}{2}$ and common ratio $r = \frac{1}{2}$. Since $|r| = \frac{1}{2} < 1$, the geometric series converges to $5 \cdot \frac{\frac{1}{2}}{1-\frac{1}{2}} = 5$. Choice A incorrectly claims $|r| > 1$, perhaps confusing the base 2 in the denominator with the ratio. Remember: rewrite series in the form $\sum ar^n$ to identify geometric series and check if $|r| < 1$.

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 determining convergence of the series $\sum_{n=2}^{\infty} \frac{1}{n\ln n}$. To analyze this series, we can apply the integral test by evaluating $\int_2^{\infty} \frac{1}{x\ln x} dx$. Using the substitution $u = \ln x$, we get $du = \frac{1}{x}dx$, transforming the integral to $\int_{\ln 2}^{\infty} \frac{1}{u} du = \ln|u|\big|_{\ln 2}^{\infty}$. This integral diverges to infinity, so by the integral test, the series diverges. The ratio test (option A) would be inconclusive here as the limit equals 1. Remember that $\frac{1}{n\ln n}$ is a classic example of a divergent series that decreases more slowly than the harmonic series.

This problem involves classifying the geometric series $\sum_{n=1}^{\infty} \left(\frac{5}{6}\right)^n$. A geometric series $\sum ar^n$ converges when $|r| < 1$ and diverges when $|r| \geq 1$. Here, the common ratio is $r = \frac{5}{6}$, and since $|\frac{5}{6}| = \frac{5}{6} < 1$, the series converges. The sum of this convergent geometric series is $\frac{a}{1-r} = \frac{5/6}{1-5/6} = \frac{5/6}{1/6} = 5$. The nth-term test (option A) only tells us when a series must diverge, not when it converges. For any series with constant ratio between consecutive terms, immediately recognize it as geometric and check if $|r| < 1$.

This problem involves the series $\sum_{n=1}^{\infty} \frac{n!}{n^n}$, which requires the ratio test. Computing $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{(n+1)!/(n+1)^{n+1}}{n!/n^n} = \lim_{n\to\infty} \frac{(n+1) \cdot n^n}{(n+1)^{n+1}} = \lim_{n\to\infty} \frac{n^n}{(n+1)^n}$. This simplifies to $\lim_{n\to\infty} \left(\frac{n}{n+1}\right)^n = \lim_{n\to\infty} \left(1 - \frac{1}{n+1}\right)^n = \frac{1}{e} < 1$. Since the ratio test limit is less than 1, the series converges. The nth-term test (option A) would only check if terms go to zero, not prove convergence. For series involving factorials and exponentials, the ratio test is typically the most effective approach.