Telescoping Series, p-Series, and Integral Test: Understanding Sₙ and Convergence

To find S_3, we calculate the first three terms and sum them up. For n=1, a₁ = 2 + (-1)^1 / 2 = 2 - 0.5 = 1.5. For n=2, a_2 = 2 + (-1)² / 2 = 2 + 0.5 = 2.5. For n=3, a_3 = 2 + (-1)³ / 2 = 2 - 0.5 = 1.5. Adding these together, S_3 = 1.5 + 2.5 + 1.5. First, 1.5 + 2.5 equals 4.0. Then, 4.0 + 1.5 equals 5.5.

The partial sum S_2 consists of the first term plus the second term. The first term a₁ is given as 5. The second term a_2 is found by multiplying the first term by the common ratio r. So, a_2 = 5 * 0.2 = 1. Now we add the first two terms: S_2 = a₁ + a_2 = 5 + 1 = 6.

The definition of the convergence of an infinite series is based entirely on the behavior of its partial sums. We say a series converges if and only if the sequence of partial sums {Sₙ} converges to a specific, finite limit. While it is true that for a series to converge, the limit of the terms aₙ must be 0 (Option D), that is a necessary condition but not a sufficient one (e.g., the harmonic series). Option B is the definition of convergence itself.

The Alternating Series Estimation Theorem states that the error involved in using the n-th partial sum Sₙ to approximate the total sum S is bounded by the absolute value of the first unused term, which is the (n+1)-th term. Here, we are using S₁0, so the error is bounded by the magnitude of the 11th term. The general term magnitude is 1/n³. Substituting n=11 into this expression gives 1/11³. Therefore, the maximum error is 1/11³.

The partial sum S₁00 represents the sum of the first 100 terms (a₁ + ... + a_99 + a₁00). The partial sum S_99 represents the sum of the first 99 terms (a₁ + ... + a_99). To isolate the 100th term, we subtract the sum of the first 99 terms from the sum of the first 100 terms. Therefore, a₁00 = S₁00 - S_99.

We examine the expression Sₙ = 2 - 1/n. Since n starts at 1 and increases to infinity, the fraction 1/n starts at 1 and decreases towards 0. The value of Sₙ is smallest when the term being subtracted (1/n) is largest. The largest value of 1/n occurs at n=1, where 1/n = 1. Thus, the smallest value for Sₙ is S₁ = 2 - 1 = 1. As n increases, 1/n gets smaller, so we subtract less, and Sₙ gets closer to 2. Therefore, the lower bound is 1.

The first partial sum S₁ is equal to the first term a₁, because S₁ sums only the first term. We can find S₁ by plugging n=1 into the given formula. S₁ = 4 - 4(⅓)^1 = 4 - 4/3. To subtract these, we find a common denominator: 4 is 12/3. So, 12/3 - 4/3 = 8/3. Therefore, a₁ is 8/3.

The operation of summation is linear. This means that the partial sum of a linear combination of sequences is the linear combination of their partial sums. Mathematically, sum from k=1 to n of (aₖ - 2bₖ) is equal to (sum from k=1 to n of aₖ) - 2*(sum from k=1 to n of bₖ). Substituting the given partial sum formulas, we get aₙ - 2Bₙ = n² - 2(n) = n² - 2n.

We can compare the partial sum to the integral of the function f(x) = 1/x. By viewing the sum as a left Riemann sum of the decreasing function 1/x from x=1 to x=n+1, the area of the rectangles (the sum) is greater than the area under the curve. The integral of 1/x from 1 to n+1 is ln(n+1) - ln(1) = ln(n+1). Therefore, the partial sum Sₙ is strictly greater than ln(n+1).

This is a direct application of the Monotone Convergence Theorem for sequences. Since the terms aₙ are positive, the sequence of partial sums Sₙ is strictly increasing (Sₙ+1 = Sₙ + aₙ+1 > Sₙ). A sequence that is both increasing and bounded above must converge to a limit. Therefore, the series converges.

This is a finite geometric series with first term a=1, common ratio r=x, and the number of terms is n+1 (counting from power 0 to power n). The standard formula for the sum of a geometric progression is a(1 - r^k) / (1 - r), where k is the number of terms. Substituting our values, we get 1 * (1 - xⁿ⁺¹) / (1 - x). Note that option A is the sum for n terms (powers 0 to n-1), but the question specifies the sum up to xⁿ.

The remainder Rₙ of a convergent series is defined as the difference between the actual infinite sum S and the n-th partial sum Sₙ. It represents the "tail" of the series that has not yet been summed (i.e., the sum from k=n+1 to infinity). Therefore, Rₙ = S - Sₙ.

This partial sum is a Riemann sum. Specifically, it is the right-endpoint approximation for the area under a curve on the interval [0, 1]. The width of each subinterval is delta_x = 1/n, and the height of the function at the k-th interval is (k/n)². This corresponds to the function f(x) = x² evaluated at x = k/n. As n approaches infinity, the Riemann sum converges to the definite integral of x² from 0 to 1.