This problem requires using the comparison test to determine convergence of the series. For large n, the dominant terms are 3n in the numerator and n³ in the denominator, so the series behaves like 3/n². Since n³ + 2 > n³, we have (3n+1)/(n³+2) < (3n+1)/n³ < 4n/n³ = 4/n² for n ≥ 1. The series ∑4/n² = 4∑1/n² converges (p-series with p = 2 > 1), so by the comparison test, our series converges. Choice A incorrectly reverses the inequality direction, which would require the divergence of the comparison series to conclude anything. When using the comparison test, always verify your inequality direction matches your conclusion: if your series is less than a convergent series, it converges.

This problem requires comparing an exponential series for convergence. Since 3ⁿ + n > 3ⁿ, we have 2ⁿ/(3ⁿ+n) < 2ⁿ/3ⁿ = (2/3)ⁿ. The geometric series ∑(2/3)ⁿ converges because |2/3| < 1. By the comparison test, since our series terms are smaller than those of a convergent series, our series converges. Choice A incorrectly states the inequality as ≥ instead of ≤, which would not allow us to conclude convergence. For exponential series, identify the dominant exponential term and compare with the corresponding geometric series, remembering that ∑rⁿ converges if and only if |r| < 1.

This problem requires careful analysis of the dominant terms. For large n, the numerator behaves like n and the denominator like n², giving behavior similar to 1/n. More precisely, (n+2)/(n²+√n) > n/(n²+n²) = n/(2n²) = 1/(2n) for large n. Since ∑1/n diverges (harmonic series), and our series terms are eventually larger than a constant multiple of 1/n, our series diverges by comparison. Choice A incorrectly concludes convergence from comparison with a divergent series. When the dominant behavior suggests divergence, look for a divergent series that is smaller than yours to confirm divergence.

This problem tests applying the comparison test to a series involving logarithms. Since ln n grows slower than any positive power of n, we need to find an appropriate bound for ln n/n^(3/2). For large n, ln n < n^(1/4) (this can be verified by noting that ln n/n^(1/4) → 0 as n → ∞). Therefore, ln n/n^(3/2) < n^(1/4)/n^(3/2) = 1/n^(5/4) for sufficiently large n. Since ∑1/n^(5/4) converges (p-series with p = 5/4 > 1), our series converges by the comparison test. Choice B incorrectly uses a divergent comparison series with the wrong inequality direction. Remember that logarithmic growth is slower than any positive power growth, making comparisons with n^α for small positive α useful.

This problem tests comparison for a rational function series. The dominant behavior is $n/n^3 = 1/n^2$ for large n. Since $n^3 + 7n + 1 > n^3$, we have $n/(n^3+7n+1) < n/n^3 = 1/n^2$. The series $\sum 1/n^2$ converges (p-series with p = 2 > 1), so our series converges by comparison. Choice A incorrectly states the inequality in the wrong direction, and even if it were correct, we couldn't conclude convergence from our series being larger than a convergent series. For rational functions, identify the highest degree terms in numerator and denominator to determine the comparison series, then verify the inequality direction.

This problem tests comparison with a logarithmic perturbation. Since $\ln n > 0$ for $n \geq 2$, we have $n^2 + \ln n > n^2$, which means $\frac{1}{n^2 + \ln n} < \frac{1}{n^2}$. The series $\sum \frac{1}{n^2}$ converges (p-series with $p = 2 > 1$), so by comparison, our series converges. Choice A incorrectly reverses the inequality—adding positive terms to the denominator decreases the fraction value. When logarithmic terms appear with polynomial terms, the polynomial behavior dominates for large n, so you can often ignore the logarithmic term when choosing your comparison series.

The skill here is using comparison tests to determine the convergence of infinite series. To analyze the series $\sum \frac{\ln(n+1)}{n^2}$, note that $\ln(n+1)$ grows slower than any positive power of $n$, such as $n^{1/2}$. For large $n$, $\ln(n+1) < n^{1/2}$, so $\frac{\ln(n+1)}{n^2} < \frac{n^{1/2}}{n^2} = \frac{1}{n^{3/2}}$. Apply the direct comparison test with $b_n = \frac{1}{n^{3/2}}$, which converges ($p=\frac{3}{2}>1$), so the series converges. A tempting distractor is claiming divergence by limit comparison with $\sum \frac{1}{n^2}$ where the limit is $\infty$ (actually $\lim \frac{\ln(n+1)}{n^2} \cdot n^2 = \infty$, but with convergent reference, it does not imply divergence). A transferable comparison strategy is to bound logarithmic terms with polynomial terms to enable direct comparison with a known p-series.

The skill here is using comparison tests to determine the convergence of infinite series. To analyze the series ∑ $√n/(n^2$ + 1), observe that for large n, it behaves like $√n/n^2$ = $1/n^{3/2}$. Apply the limit comparison test with b_n = $1/n^{3/2}$, computing lim $[√n/(n^2$ + 1)] / $(1/n^{3/2}$) = lim $n^2$$/(n^2$ + 1) = 1, a positive finite value. Since ∑ $1/n^{3/2}$ converges (p = 3/2 > 1), the given series converges. A tempting distractor is claiming divergence by limit comparison with ∑ 1/n where the limit is 0, but a limit of 0 with a divergent series does not prove divergence. A transferable comparison strategy is to match the degree of the polynomial terms in numerator and denominator to identify the equivalent p-series for limit comparison.

The skill being tested is the comparison test for determining series convergence. To apply it to sum √n $/(n^2$ +1), observe that the term behaves like $1/n^{3/2}$ for large n. Choose to compare with sum $1/n^{3/2}$, verifying that √n $/(n^2$ +1) ≤ √n $/n^2$ = $1/n^{3/2}$ for n ≥ 1 since the denominator is larger. Since sum $1/n^{3/2}$ converges as a p-series with p=3/2>1, the original series converges by comparison. A tempting distractor is choice C, which claims a_n ≥ $1/n^{3/2}$ and concludes divergence, but the inequality is actually the opposite, making it invalid. A transferable comparison strategy is to bound the term by ignoring lower-order terms in the denominator to establish an upper bound with a known convergent series.

The skill being tested is the comparison test for determining series convergence. To apply it to sum $n^2$ $/(n^4$ +5n), note the term behaves like $1/n^2$ for large n. Choose to compare with sum $1/n^2$, verifying that $n^2$ $/(n^4$ +5n) ≤ $n^2$ $/n^4$ = $1/n^2$ for n ≥ 1 since the denominator is larger. Since sum $1/n^2$ converges (p=2>1), the original series converges by comparison. A tempting distractor is choice E, which uses a_n ≤ 1/n with the divergent harmonic series to conclude convergence, but this is inconclusive for convergence. A transferable comparison strategy is to find an upper bound by minimizing the denominator with dominant terms for convergence conclusions.