Using Partial Sums to Approximate Infinite Series

This is a geometric series with a₁=1 and r=1/2. The nth partial sum is Sₙ = (1 - (½)ⁿ)/(1 - 1/2) = (1 - (½)ⁿ)/(½) = 2(1 - (½)ⁿ). For n=6, (½)⁶ = 1/64, so S₆ = 2(1 - 1/64) = 2(63/64) = 63/32.

The relationship between a specific term aₙ and the partial sums is given by aₙ = Sₙ - Sn-1. To find the fourth term a₄, we need to calculate S₄ and S_3. First, we calculate S₄ by substituting n=4 into the formula n² + 3n, which gives 4² + 3(4) = 16 + 12 = 28. Next, we calculate S_3 by substituting n=3 into the formula, which gives 3² + 3(3) = 9 + 9 = 18. Finally, we subtract S_3 from S₄ to find a₄, so 28 - 18 equals 10.

Option B is false because there are divergent series whose partial sums do not go to infinity, such as the series 1 - 1 + 1 - 1 + ... which oscillates between 0 and 1. Divergent means the partial sums do not approach a finite limit, but they may oscillate or behave erratically without necessarily going to infinity.

The error bound for an alternating series is the absolute value of the next term. So we need to find the smallest n such that |a_{n+1}| < 0.1. For this series, a_{n+1} = 1/(n+1). So we need 1/(n+1) < 0.1, which means n+1 > 10, so n > 9. Thus, the minimum n is 10. That means if we use the partial sum S₁₀, the error is at most 1/11 ≈ 0.0909 < 0.1. So the answer is 10.

This is a telescoping series involving square roots. We calculate the first three terms.

For n=1: 1/1 - 1/√(2) = 1 - 1/√(2).

For n=2: 1/√(2) - 1/√(3).

For n=3: 1/√(3) - 1/√(4) = 1/√(3) - 1/2.

When we add these terms to find S_3, the -1/√(2) cancels with +1/√(2), and the -1/√(3) cancels with +1/√(3). We are left with the first part of the first term and the last part of the third term: 1 - 1/2 = 0.5.

The terms are aₙ = (-1)ⁿ⁻¹/(2n-1). S₅ uses terms for n=1 to 5. The next term is for n=6: a₆ = (-1)⁵/(2*6-1) = (-1)/(11) = -1/11. The absolute value is 1/11. So the maximum error is 1/11.

The value of the infinite series is the limit of Sₙ as n→∞. lim (n→∞) [4 - 2/(n+1)] = 4 - 0 = 4.

We use the property of logarithms that ln(a/b) = ln(a) - ln(b). This allows us to rewrite the general term as ln(k) - ln(k+1). This transforms the series into a telescoping sum. Writing out the terms, we get: (ln(1) - ln(2)) + (ln(2) - ln(3)) + (ln(3) - ln(4)) + ... + (ln(n) - ln(n+1)). The -ln(2) cancels with +ln(2), -ln(3) cancels with +ln(3), and so on. We are left with the first term ln(1) and the last term -ln(n+1). Since ln(1) is 0, the final expression for the partial sum is simply -ln(n+1).

First, we factor the denominator n² + 3n + 2 into (n+1)(n+2). We then use partial fractions to write 1/((n+1)(n+2)) as A/(n+1) + B/(n+2). Solving for A and B, we find that A=1 and B=-1, so the term is 1/(n+1) - 1/(n+2). This creates a telescoping series. We write out the first few terms: (1/2 -⅓) + (1/3 - 1/4) + (1/4 - 1/5) + ... + (1/(n+1) - 1/(n+2)). We observe that -1/3 cancels with +1/3, -1/4 cancels with +1/4, and so on. The only terms that survive are the very first positive term, 1/2, and the very last negative term, -1/(n+2). Thus, Sₙ = 1/2 - 1/(n+2).

The nth term aₙ = Sₙ - Sₙ₋₁. So a₁₀₀ = S₁₀₀ - S₉₉. Compute S₁₀₀ = 3×100/(100+1) = 300/101. Compute S₉₉ = 3×99/(99+1) = 297/100. Then a₁₀₀ = 300/101 - 297/100 = (300×100 - 297×101)/(101×100) = (30000 - 29997)/(10100) = 3/10100.

The sum of the first 10 terms is S₁₀. Substitute n=10 into the formula: S₁₀ = (10² + 3×10)/(10² + 1) = (100 + 30)/(100 + 1) = 130/101, an exact value. Although the sum is approximately  ≈1.287, Option D should be understood as a decimal approximation and not the exact sum.

For an infinite series to converge, the sequence of its partial sums Sₙ must approach a single, finite limit. If the partial sums oscillate between two different values, they do not approach a single limit. Therefore, by definition, the series diverges.

We have S₂ = a₁ + a₂ = 8, and S₃ = a₁ + a₂ + a₃ = 12. Subtracting, S₃ - S₂ = a₃ = 12 - 8 = 4.