This question tests the ratio test on an alternating series with polynomial and exponential terms. For $\sum_{n=1}^{\infty} \frac{(-1)^n n^2}{6^n}$, we apply the ratio test to the absolute values: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)^2/6^{n+1}}{n^2/6^n} = \frac{(n+1)^2}{6n^2} = \frac{1}{6} \cdot \left(\frac{n+1}{n}\right)^2$. Taking the limit: $\lim_{n \to \infty} \frac{1}{6} \cdot 1^2 = \frac{1}{6} < 1$, so the series converges absolutely. Choice B incorrectly states the limit is 6, inverting the result. The alternating sign doesn't affect the ratio test calculation when we use absolute values.

The ratio test is a key method for determining the convergence of infinite series, particularly those involving factorials and exponential-like terms. To apply it here, compute the limit of the absolute value of the ratio of consecutive terms: for a_n = n! / $n^n$, the ratio $|a_{n+1}$/a_n| = $(n/(n+1))^n$. As n approaches infinity, this limit is 1/e, which is less than 1. Therefore, the ratio test indicates that the series converges. A tempting distractor might be choice A, which claims divergence with limit 1, but this overlooks the precise limit involving e. Always remember that in the ratio test, recognizing limits like 1/e can confirm convergence for series balancing factorials and powers.

The ratio test is a key method for determining the convergence of infinite series, particularly those involving factorials or exponentials, even with alternating signs since we use absolute values. To apply it here, compute the limit of the absolute value of the ratio of consecutive terms: for a_n = $(-2)^n$ / n!, the ratio $|a_{n+1}$/a_n| = 2/(n+1). As n approaches infinity, this limit is 0, which is less than 1. Therefore, the ratio test indicates that the series converges absolutely. A tempting distractor might be choice A, which claims divergence with limit 2 > 1, but this neglects the factorial's growth. Always remember that in the ratio test, a limit of 0 strongly suggests convergence, especially for exponential series with factorial denominators.

The ratio test is a key method for determining the convergence of infinite series, particularly those involving linear terms over exponentials. To apply it here, compute the limit of the absolute value of the ratio of consecutive terms: for a_n = n / $2^n$, the ratio $|a_{n+1}$/a_n| = ((n+1)/n) / 2. As n approaches infinity, this limit is 1/2, which is less than 1. Therefore, the ratio test indicates that the series converges. A tempting distractor might be choice A, which claims divergence with limit 2, but this misapplies the ratio by inverting it. Always remember that in the ratio test, for series like geometric with polynomial adjustments, a limit less than 1 ensures convergence.

The ratio test is a key method for determining the convergence of infinite series, particularly those involving factorials and polynomials. To apply it here, compute the limit of the absolute value of the ratio of consecutive terms: for a_n = (n+2) $2^n$ / n!, the ratio $|a_{n+1}$/a_n| = 2 (n+3) / ((n+1)(n+2)). As n approaches infinity, this limit is 0, which is less than 1. Therefore, the ratio test indicates that the series converges. A tempting distractor might be choice B, which claims divergence with limit 2 > 1, but this ignores the factorial's overpowering effect. Always remember that in the ratio test, when polynomials are multiplied by exponentials over factorials, the limit often approaches 0, confirming convergence.

The skill here is using the ratio test to determine the convergence of infinite series. To apply the ratio test, compute the limit L = $lim_{n→∞}$ $|a_{n+1}$/a_n| where a_n = $4^n$ / (2n)!, so $|a_{n+1}$/a_n| = 4 / ((2n+1)(2n+2)). The limit is 0 as n approaches infinity. Since L = 0 < 1, the series converges absolutely. A tempting distractor is choice E, which claims convergence with limit 1/4, but this fails because the double factorial growth in the denominator produces 0, not 1/4. A transferable ratio-test strategy is to simplify the expression for $|a_{n+1}$/a_n| carefully, evaluate the limit, and conclude absolute convergence if less than 1, divergence if greater than 1, or inconclusive if equal to 1.

The skill here is using the ratio test to determine the convergence of infinite series. To apply the ratio test, compute the limit L = $lim_{n→∞}$ $|a_{n+1}$/a_n| where a_n = $3^n$ / n!, so $|a_{n+1}$/a_n| = 3/(n+1). The limit is 0 as n approaches infinity. Since L = 0 < 1, the series converges absolutely. A tempting distractor is choice E, which suggests convergence with limit 1/3, but this fails because it incorrectly places the 3 in the denominator instead of the numerator in the ratio. A transferable ratio-test strategy is to simplify the expression for $|a_{n+1}$/a_n| carefully, evaluate the limit, and conclude absolute convergence if less than 1, divergence if greater than 1, or inconclusive if equal to 1.

This problem requires applying the ratio test to determine series convergence. For $\sum_{n=1}^{\infty} \frac{n!}{5^n}$, we calculate $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!/5^{n+1}}{n!/5^n} = \frac{(n+1)! cdot 5^n}{n! cdot 5^{n+1}} = \frac{n+1}{5}$. As $n \to \infty$, this ratio approaches $\infty$, which is greater than 1, so the series diverges. Choice A incorrectly suggests the limit is $\frac{1}{5}$, likely by mishandling the factorial terms. The key insight is that factorial growth in the numerator eventually overwhelms any exponential base in the denominator.

This problem requires applying the ratio test to a series with linear and exponential terms. For $\sum_{n=1}^{\infty} \frac{n}{2^n}$, we compute $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)/2^{n+1}}{n/2^n} = \frac{n+1}{2n} = \frac{1}{2} \cdot \frac{n+1}{n}$. As $n \to \infty$, this approaches $\frac{1}{2} \cdot 1 = \frac{1}{2} < 1$, so the series converges. Choice A incorrectly claims the limit is 2, likely by mishandling the exponential terms. The exponential denominator with base 2 grows fast enough to overcome linear growth in the numerator.

This problem requires applying the ratio test to a series with polynomial numerator and factorial denominator. For $\sum_{n=1}^{\infty} \frac{n^4}{n!}$, we calculate $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)^4/(n+1)!}{n^4/n!} = \frac{(n+1)^4 \cdot n!}{n^4 \cdot(n+1)!} = \frac{(n+1)^4}{n^4(n+1)} = \frac{(n+1)^3}{n^4} = \frac{1}{n} \cdot \left(\frac{n+1}{n}\right)^3$. As $n \to \infty$, this approaches $0 < 1$, so the series converges. Choice B incorrectly suggests the limit is infinity and divergence, missing that factorial growth dominates polynomial growth. Factorials in denominators almost always lead to convergence.