Alternating Series & Error Bounds Through Partial Sums

To find the third partial sum S_3, we need to calculate the sum of the first three terms of the sequence. First, we determine the first term a₁ by plugging n=1 into the formula 2n - 1, which gives us 2(1) - 1 = 1. Next, we find the second term a_2 by plugging in n=2, which yields 2(2) - 1 = 3. Then, we find the third term a_3 by plugging in n=3, resulting in 2(3) - 1 = 5. Finally, we add these three terms together to get the partial sum: 1 + 3 + 5 equals 9.

The partial sum Sₙ of a geometric series is the sum of the first n terms: a + ar + ar² + ... + arⁿ⁻¹. To derive the closed form, we multiply Sₙ by r to get rSₙ = ar + ar² + ... + arⁿ. We then subtract the equation for rSₙ from the equation for Sₙ. The intermediate terms cancel out, leaving us with Sₙ - rSₙ = a - arⁿ. Factoring out Sₙ on the left and a on the right gives Sₙ(1 - r) = a(1 - rⁿ). Dividing both sides by (1 - r) yields the correct formula Sₙ = a(1 - rⁿ) / (1 - r).

We need to find the sum of the first two terms corresponding to n=0 and n=1. First, we calculate the term for n=0, which is 3(½)⁰= 3(1) = 3. Next, we calculate the term for n=1, which is 3(½)^1 = 3(0.5) = 1.5. Finally, we add these two terms together to get the partial sum S_2. So, 3 + 1.5 equals 4.5.

According to the Alternating Series Estimation Theorem, the absolute error |S - Sₙ| is less than the magnitude of the next term, aₙ₊₁. We want the error to be less than 0.0001, so we set the inequality 1 / (n+1)⁴ < 0.0001. Taking the reciprocal of both sides reverses the inequality, giving (n+1)⁴ > 10,000. Taking the fourth root of both sides, we get n+1 > 10. Subtracting 1 from both sides, we find n > 9. The smallest integer n that satisfies this inequality is n = 10. When n = 10, the error is guaranteed to be less than |a_11| = 1/11^4 = 1/14641 ≈ 0.0000683, which is indeed less than 0.0001. If we used n = 9, the error would be less than |a_10| = 1/10^4 = 0.0001, but not guaranteed to be strictly less than 0.0001. Therefore, n = 10 is the smallest value that guarantees the error is less than 0.0001.

We know that the n-th term of a series can be found by taking the difference between the n-th partial sum and the (n-1)-th partial sum: aₙ = Sₙ - S_(n-1). Substituting the given formula, we get aₙ = log₁0(n + 1) - log₁0((n-1) + 1) = log₁0(n + 1) - log₁0(n). Using the quotient property of logarithms, log(a) - log(b) = log(a/b), we can combine these into a single logarithm: log₁0((n + 1) / n). Simplifying the fraction inside gives log₁0(1 + 1/n).

For a convergent alternating series with decreasing terms, the error when using the nth partial sum is at most the absolute value of the next term. Here, S₄ uses the first four terms (n=1 to 4). The next term is the fifth term: a₅ = (-1)^(4) / 5² = 1/25. So the maximum error is 1/25.

We need 1/(2N²) < 0.001. Multiply both sides: 1/(2N²) < 0.001 => 1/N² < 0.002 => N² > 1/0.002 = 500 => N > √500 ≈ 22.36. Therefore, the smallest integer N that satisfies the condition is 23. However, among the given choices, 10 and 22 are too small, while 100 and 500 are larger. To ensure the error bound is less than 0.001, we must choose N at least 23, so from the given options, 100 is the smallest that works.

The interval [0,2] divided into 4 equal subintervals gives Δx = (2-0)/4 = 0.5. The right endpoints are 0.5, 1, 1.5, and 2. The Riemann sum is f(0.5)Δx + f(1)Δx + f(1.5)Δx + f(2)Δx = (0.5² + 1² + 1.5² + 2²) × 0.5. This is a partial sum of the function values multiplied by the width.

The partial sum Sₙ is defined as the sum of the first n terms: Sₙ = S_(n-1) + aₙ. We are given that every term aₙ is strictly positive (aₙ > 0). When you take a current sum and add a positive number to it, the new sum must be larger than the previous one. Therefore, S₁ < S_2 < S_3, and so on. This means the sequence of partial sums is strictly increasing.

We use the recursive definition provided. We start with S₁ = 3. To find the second partial sum S_2, we plug n=2 into the formula Sₙ = S_(n-1) + 2n, which gives S_2 = 3 + 2(2) = 7. To find the third partial sum S_3, we plug n=3 into the formula, using the value we just found for S_2. This gives S_3 = 7 + 2(3) = 7 + 6 = 13.

The Alternating Series Estimation Theorem states that the error in using Sₙ to approximate the sum is bounded by the magnitude of the first neglected term, which is aₙ₊₁. Since we are using S₄, the error is bounded by the absolute value of the 5th term, a_5. The formula for the term is 10/n!. Substituting n=5, we get 10/5!. Since 5! = 120, the error bound is 10/120.

This is a telescoping series with a "gap" of 2. The terms are (1 -⅓) + (1/2 - 1/4) + (1/3 - 1/5)... The cancellation does not start immediately. 1 and 1/2 never get cancelled. The -1/3 cancels with +1/3 later. At the very end of the partial sum S₁0, the terms that have not yet been cancelled by a future term are the negative parts of the final two terms: -1/(10+1) and -1/(10+2). Thus, S₁0 consists of the initial fixed parts (1 + 1/2) and the trailing parts (-1/11 - 1/12).