Convergence & Divergence of Infinite Series

We need to examine what happens to the terms as n gets very large. Each term is (n²+1)/n, which simplifies to n + 1/n. As n approaches infinity, this expression grows without bound because the n term dominates. The nth term test states that if the limit of the terms does not equal zero, the series must diverge. Since our terms approach infinity, not zero, the series fails the nth term test and therefore diverges. This is the most direct way to see the divergence without needing more complex tests.

A geometric series has the form a + ar + ar² + ar³ + ... where a is the first term and r is the common ratio. Here, the first term a is 4. To find the ratio r, we divide any term by the previous term: 2/4 = 1/2, and 1/2 = 1/2, confirming r = 1/2. Since |r| < 1, the series converges and its sum is given by the formula a/(1-r). Substituting our values: sum = 4/(1 - 1/2) = 4/(½) = 8. Therefore, the series converges to 8. Option A is incorrect because it uses the wrong formula. Option C is a plausible miscalculation. Option D is incorrect because geometric series with |r| < 1 always converges.

A p-series has the form sum of 1/np. The p-series test tells us that if p > 1, the series converges, and if p ≤ 1, the series diverges. In this case, p = 3, which is greater than 1. Therefore, by the p-series test, this series converges. The reasoning behind this test involves comparing the series to an integral, but for our purposes, we can simply apply the rule: p = 3 > 1, so convergence is guaranteed. Options B and D incorrectly state the conclusion, while option C incorrectly identifies the condition.

The alternating series test requires three conditions: the terms must alternate in sign, the absolute value of terms must be decreasing, and the limit of the terms must be zero. This series has terms that alternate in sign due to (-1)ⁿ. The absolute values are 1/n, which form a decreasing sequence. The limit of 1/n as n approaches infinity is zero. All three conditions are satisfied, so the alternating series test confirms convergence. Option B is incorrect because the nth term test for divergence doesn't apply when terms do approach zero. Option C is incorrect because this is not a geometric series. Option D is incorrect because the terms clearly do alternate.

The ratio test examines the limit of the absolute value of aₙ₊₁/aₙ. For this series, aₙ = 2ⁿ/n! and aₙ₊₁ = 2ⁿ⁺¹/(n+1)!. The ratio is [2ⁿ⁺¹/(n+1)!] / [2ⁿ/n!] = [2ⁿ⁺¹/2ⁿ] * [n!/(n+1)!] = 2 * [1/(n+1)]. As n approaches infinity, this ratio approaches 0. Since the limit is less than 1, the ratio test guarantees absolute convergence. This makes sense because factorial growth dominates exponential growth. Option A is incorrect because factorials grow faster than exponentials. Option C misapplies the ratio test conclusion. Option D is incorrect because comparing to 2ⁿ alone would suggest divergence, but the factorial in the denominator changes everything.

The integral test applies when we have a positive, decreasing function. Consider f(x) = 1/(x²+1), which is positive and decreasing for x >= 1. The integral from 1 to infinity of 1/(x²+1) dx can be evaluated using the arctangent function. This integral equals lim(b->infinity) [arctan(b) - arctan(1)] = π/2 - π/4 = π/4. Since this improper integral converges to a finite value, the series also converges by the integral test. Option A incorrectly states the conclusion. Option B is wrong because comparing to the divergent harmonic series 1/n doesn't help prove convergence. Option D is incorrect because the terms do approach zero.

For large n, the term 1/(n²+5) behaves similarly to 1/n² because the +5 becomes negligible. We can use the direct comparison test: since n²+5 > n², we have 1/(n²+5) < 1/n² for all n. The series sum of 1/n² is a convergent p-series (p = 2 > 1). Since our series has terms that are smaller than a convergent series term-by-term, our series must also converge. The comparison test provides a clean, direct proof that avoids more complex calculations. Option B has the wrong conclusion. Options C and D use an inappropriate comparison series.

For large n, the term (2n+1)/(n²) behaves like 2n/n² = 2/n. We apply the limit comparison test with the harmonic series 1/n. Compute the lim_{n 🠒∞}  [(2n+1)/(n²)] / [1/n] = limit of [(2n+1)/(n²)] * [n/1] = limit of (2n+1)/n = limit of 2 + 1/n = 2. Since this limit is a positive finite number, both series have the same behavior. The harmonic series diverges, so our series also diverges. This test is more precise than direct comparison when terms are asymptotically similar. Option A has the wrong conclusion. Options C and D use an inappropriate comparison series.

This is an alternating series where the terms are (-1)ⁿ⁺¹/(2n-1), which gives the sequence 1/1, -1/3, 1/5, -1/7, 1/9, and so on. To apply the alternating series test, we need to verify three conditions. First, the terms must alternate in sign, which they clearly do due to the factor (-1)ⁿ⁺¹. Second, the absolute values of the terms must form a decreasing sequence. The absolute values are 1/(2n-1), which gives 1, 1/3, 1/5, 1/7, and these values decrease since the denominator increases with n. Third, the limit of the absolute values must equal zero. We have the limit as n approaches infinity of 1/(2n-1) equals zero, since the denominator grows without bound. All three conditions are satisfied, so by the alternating series test, the series converges. Option B is incorrect because the nth term test for divergence only applies when terms do not approach zero, but here they do approach zero. Option C is incorrect because the ratio test would be inconclusive for this series. Option D is false because the terms do approach zero.

The harmonic series is the classic example of a series whose terms approach zero but still diverges. While 1/n approaches zero as n increases, the sum grows without bound, just very slowly. We can prove this by grouping terms: 1 + 1/2 + (1/3+1/4) + (1/5+1/6+1/7+1/8) + ... where each group sums to more than 1/2. Since we have infinitely many such groups, the total sum exceeds any finite bound. This is a special case of the p-series with p = 1, and the p-series test tells us it diverges when p ≤ 1. The series grows slowly but definitely diverges to positive infinity.

To determine absolute versus conditional convergence, we first test the series of absolute values: sum of 1/n. This is the harmonic series, which diverges. Therefore, the original series does not converge absolutely. However, the original series with alternating signs does converge by the alternating series test (terms decrease in absolute value and approach zero). When a series converges but does not converge absolutely, we say it converges conditionally. This is a classic example of conditional convergence. Option A is incorrect because absolute convergence fails. Options C and D are incorrect because the alternating series does converge.

Absolute convergence is a very strong property. When a series converges absolutely, the sum is independent of the order in which you add the terms. This means you can rearrange, regroup, or reorder the terms in any way you like, and the series will still converge to the same sum. This is not true for conditionally convergent series, where rearrangements can change the sum or even cause divergence. The reason is that with absolute convergence, the positive and negative terms each form convergent series separately, so the order doesn't matter. This property is sometimes called the Riemann rearrangement theorem for absolutely convergent series.

A series is absolutely convergent if Σ|aₙ| converges, and it's conditionally convergent if Σaₙ converges but Σ|aₙ| diverges. First, we examine the original series Σ∞n=1 (-1)ⁿ⁺¹/n. This is an alternating series with bₙ = 1/n, which is decreasing and has limit 0. By the alternating series test, this series converges. Now we examine the absolute series Σ|aₙ| = Σ|(-1)ⁿ⁺¹/n| = Σ1/n. The series Σ1/n is the harmonic series, which is a p-series with p = 1, and p-series converge only when p > 1. Since p = 1, the harmonic series diverges. Therefore, Σ|aₙ| diverges while Σaₙ converges, which means the series is conditionally convergent.

By definition, an infinite series converges to a number S if and only if the sequence of its partial sums converges to S. The partial sums are the sums of the first n terms: Sₙ = a₁ + a_2 + ... + aₙ. If Sₙ approaches 5 as n gets larger and larger, this means adding more terms to the sum gets us arbitrarily close to 5. Therefore, the infinite series sum from n=1 to infinity of aₙ converges to exactly 5. This is the fundamental definition of series convergence. Option A is incorrect because approaching a finite limit means convergence. Option B is incorrect because the limit is 5, not 0. Option D is incorrect because 5 is a finite number, not infinity.

The harmonic series is a classic divergent series. We can visualize this by comparing the sum to the integral of 1/x from 1 to infinity, which results in ln(x). Since the natural logarithm function ln(x) goes to infinity as x goes to infinity, the partial sums of the harmonic series also go to infinity. However, the growth is very slow; for example, it takes millions of terms for the sum to reach moderate values like 20. Therefore, Sₙ grows without bound.