Convergence Tests, Absolute vs Conditional Convergence & Series Sums

A p-series has the form Σ(n=1 to ∞) 1/n^p where p > 0 is a constant. Option A is a p-series with p=1 (the harmonic series). Option C is a p-series with p=2. Option B is an alternating series, not a p-series, because it has the (-1)ⁿ term. While the absolute value of the terms in option B forms a p-series, the series itself is not a p-series due to the alternating signs.

A geometric series has a constant ratio between successive terms. The general form is Σ∞n=0 arⁿ, where a is the first term and r is the common ratio. For the series Σ∞n=0 3(1/4)ⁿ, when n=0: 3(1/4)⁰ = 3(1) = 3. When n=1: 3(1/4)¹ = 3/4. When n=2: 3(1/4)² = 3/16. When n=3: 3(1/4)³ = 3/64. The ratio between consecutive terms is consistently 1/4, confirming this is a geometric series with first term 3 and common ratio 1/4. Option C correctly shows the first four terms of this series.

For a geometric series Σ∞n=0 arⁿ with |r| < 1, the sum is given by a/(1-r). Here, a = 5 and r = 2/3. Since |2/3| = 2/3 < 1, the series converges. Using the formula, the sum = 5/(1 - 2/3) = 5/(⅓) = 5 × 3 = 15. We can verify this by expanding the series: 5(⅔)⁰ + 5(⅔)¹ + 5(⅔)² + 5(⅔)³ + ... = 5(1) + 5(⅔) + 5(4/9) + 5(8/27) + ... = 5 + 10/3 + 20/9 + 40/27 + ... ≈ 5 + 3.333 + 2.222 + 1.481 + ... which approaches 15.

The nth term test (also called the divergence test) states that if lim(n→∞) aₙ ≠ 0, then the series Σaₙ diverges. For the series Σ∞n=1 n/(n+1), we examine lim(n→∞) n/(n+1). We can simplify this by dividing numerator and denominator by n: limn→∞ (n/n)/((n+1)/n) = limn→∞ 1/(1 + 1/n) = 1/(1 + 0) = 1. Since the limit equals 1, which is not equal to 0, by the nth term test, the series diverges.

For the integral test, we consider the function f(x) = 1/(x² + 1), which is positive, continuous, and decreasing for x ≥ 1. We evaluate the improper integral ∫(1 to ∞) 1/(x² + 1) dx. The antiderivative of 1/(x² + 1) is arctan(x). So, ∫∞1 1/(x² + 1) dx = limb→∞ [arctan(x)](1 to b) = limb→∞ [arctan(b) - arctan(1)]. Since arctan(b) approaches π/2 as b→∞ and arctan(1) = π/4, we get π/2 - π/4 = π/4. This improper integral converges to π/4, a finite value. Therefore, by the integral test, the series Σ∞n=1 1/(n² + 1) also converges.

A p-series has the form Σ∞n=1 1/n^p where p is a positive constant. The p-series test states that a p-series converges if and only if p is greater than 1, and diverges if p is less than or equal to 1. In this problem, we can identify that the series Σ∞n=1 1/n³ is a p-series with p = 3. Since 3 is greater than 1, we can conclude by the p-series test that this series converges. The intuition behind this test is that when p is large enough (specifically, when p is greater than 1), the terms 1/np decrease rapidly enough that their sum approaches a finite limit. Option B incorrectly states that the series diverges. Options C and D incorrectly identify either the value of p or the conclusion of the test. This question tests the fundamental application of the p-series test, which is one of the most important convergence tests for series with positive terms.

For an alternating series Σ(-1)ⁿ⁺¹bₙ where bₙ > 0, the alternating series test requires that: (1) bₙ is decreasing, and (2) lim(n→∞) bₙ = 0. Here, bₙ = 1/(n + 2). First, bₙ is decreasing because as n increases, the denominator increases, making the fraction smaller. Second, lim(n→∞) 1/(n + 2) = 0. Both conditions are satisfied, so by the alternating series test, the series converges.

An infinite series of numbers converges to a real number S (or has sum S) if and only if the limit of its sequence of partial sums exists and equals S. This is the fundamental definition of convergence for infinite series. Here, we're given that lim(n→∞) sₙ = S, where sₙ is the nth partial sum sₙ = a₁ + a₂ + ... + aₙ. Since this limit exists and equals S, by definition, the infinite series Σ(n=1 to ∞) aₙ converges and has sum equal to S. This is a biconditional statement: convergence of the series is equivalent to the existence of the limit of partial sums, and when the limit exists, it equals the sum of the series.

For comparison test, we need to find a series with known convergence behavior that our series can be compared to. Notice that 1/(n(n+1)) = 1/n - 1/(n+1), which is a telescoping series. However, for comparison, observe that for n ≥ 1, n(n+1) > n², so 1/(n(n+1)) < 1/n². The series Σ∞n=1 1/n² is a convergent p-series with p = 2 > 1. Since 0 < 1/(n(n+1)) < 1/n² for all n ≥ 1, and Σ1/n² converges, by the direct comparison test, Σ1/(n(n+1)) also converges. We can also verify this by noting that the partial sums approach 1, since it's a telescoping series: sₙ = (1 - 1/2) + (1/2 -⅓) + ... + (1/n - 1/(n+1)) = 1 - 1/(n+1), which approaches 1 as n→∞.

For limit comparison test, we compare our series to a series with known behavior. Let aₙ = (2n+1)/(n²+n) and bₙ = 1/n. We compute lim(n→∞) aₙ/bₙ = lim(n→∞) [(2n+1)/(n²+n)] / (1/n) = lim(n→∞) [(2n+1)/(n²+n)] × [n/1] = lim(n→∞) (2n+1)n/(n²+n) = lim(n→∞) (2n²+n)/(n²+n) = lim(n→∞) (2 + 1/n)/(1 + 1/n) = 2/1 = 2. Since the limit equals 2, which is a positive finite number, the limit comparison test tells us that Σaₙ and Σbₙ have the same convergence behavior. The series Σ∞n=1 1/n is the harmonic series, which diverges. Therefore, Σ∞n=1 (2n+1)/(n²+n) also diverges.

For the ratio test, we compute lim(n→∞) |aₙ₊₁/aₙ|. Let aₙ = n²/2ⁿ. Then aₙ₊₁ = (n+1)²/2ⁿ⁺¹. We compute limn→∞ |aₙ₊₁/aₙ| = limn→∞ [(n+1)²/2ⁿ⁺¹] / [n²/2ⁿ] = limn→∞ [(n+1)²/2ⁿ⁺¹] × [2ⁿ/n²] = limn→∞ (n+1)²/(2 · n²) = limn→∞ (n² + 2n + 1)/(2n²) = limn→∞ (1 + 2/n + 1/n²)/2 = 1/2. Since the limit equals 1/2, which is less than 1, by the ratio test, the series converges absolutely. This is a geometric-like series where the terms decrease exponentially, making it converge even though the numerator grows polynomially.

By the wording of the question and the choices, this has to do with conditional convergence of the given series.  The series Σ(-1)ⁿ⁺¹/n converges by the alternating series test because the terms 1/n are decreasing in absolute value and approach zero. However, when we test for absolute convergence by examining Σ|(-1)ⁿ⁺¹/n| = Σ(1/n), we get the harmonic series, which diverges. Since the series converges but does not converge absolutely, it is conditionally convergent. For conditionally convergent series, the Riemann rearrangement theorem states that by rearranging terms, you can make the series converge to any real number you want, or even make it diverge. This is because the positive terms (1 + 1/3 + 1/5 + ...) diverge to infinity, and the negative terms (-1/2 - 1/4 - 1/6 - ...) diverge to negative infinity. By choosing the order, you can control the partial sums. This shows why absolute convergence is an important concept. Option A is true only for absolutely convergent series. Option C is incorrect because rearranging terms does not change the fundamental form of the terms themselves. Option D is incorrect because we have already established that the alternating harmonic series does not converge absolutely.

For absolutely convergent series, the rearrangement theorem states that if a series converges absolutely, then any series obtained from it by regrouping or rearranging the terms has the same value. This property distinguishes absolutely convergent series from conditionally convergent series. If a series is absolutely convergent with sum S, then no matter how we rearrange the terms, the new series will also converge and will converge to the same sum S. This is because the absolute convergence ensures that the series behaves like a finite series in terms of term order - the sum depends only on which terms are included, not on the order in which they are added.

The sum of the series is the limit of Sₙ as n→∞. limn→∞ arctan(n) = π/2. So the series converges to π/2.

This is a fundamental theorem regarding partial sums (Monotonic Sequence Theorem). Since all terms aₙ are positive, the sequence of partial sums Sₙ is strictly increasing. We are also given that Sₙ is bounded above (by 100). A sequence that is both increasing and bounded must converge to a limit. Therefore, the infinite series converges.