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Partial sums of a series practice problems

The first partial sums of a series, and what they say about convergence. 12 problems with worked solutions; in 12 of them every equation is proved by a computer algebra system.

8 worked problems

Try each one before opening the solution. Every line is labeled with what checked it. How we verify

Problem 7.557

easy
Find the first four partial sums \( \displaystyle S_1, \dots, S_4 \) of \( \displaystyle \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n} \).
  1. S_k adds the first k terms: a₁ = 1/2, a₂ = 1/4, and so on.
    Reviewed
  2. \[ \left. \left(1 \cdot \frac{1}{2}\right)^{n} \right|_{\substack{ n=1 }} = \frac{1}{2} \]
    S1.✓ Proved
  3. \[ 1 \cdot \frac{1}{4} + 1 \cdot \frac{1}{2} = \frac{3}{4} \]
    S2.✓ Proved
  4. \[ 1 \cdot \frac{1}{8} + 1 \cdot \frac{1}{4} + 1 \cdot \frac{1}{2} = \frac{7}{8} \]
    S3.✓ Proved
  5. \[ 1 \cdot \frac{1}{16} + 1 \cdot \frac{1}{8} + 1 \cdot \frac{1}{4} + 1 \cdot \frac{1}{2} = \frac{15}{16} \]
    S4.✓ Proved
Answer \( S_1 = \frac{1}{2},\ S_2 = \frac{3}{4},\ S_3 = \frac{7}{8},\ S_4 = \frac{15}{16} \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 7.558

easy
Find the first four partial sums \( \displaystyle S_1, \dots, S_4 \) of \( \displaystyle \sum_{n=1}^{\infty} 5 \cdot 3^{- n} \).
  1. S_k adds the first k terms: a₁ = 5/3, a₂ = 5/9, and so on.
    Reviewed
  2. \[ \left. \frac{5}{3^{n}} \right|_{\substack{ n=1 }} = \frac{5}{3} \]
    S1.✓ Proved
  3. \[ \frac{5}{9} + \frac{5}{3} = \frac{20}{9} \]
    S2.✓ Proved
  4. \[ \frac{5}{27} + \frac{5}{9} + \frac{5}{3} = \frac{65}{27} \]
    S3.✓ Proved
  5. \[ \frac{5}{81} + \frac{5}{27} + \frac{5}{9} + \frac{5}{3} = \frac{200}{81} \]
    S4.✓ Proved
Answer \( S_1 = \frac{5}{3},\ S_2 = \frac{20}{9},\ S_3 = \frac{65}{27},\ S_4 = \frac{200}{81} \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 7.559

easy
Find the first four partial sums \( \displaystyle S_1, \dots, S_4 \) of \( \displaystyle \sum_{n=1}^{\infty} 2 n - 1 \).
  1. S_k adds the first k terms: a₁ = 1, a₂ = 3, and so on.
    Reviewed
  2. \[ \left. 2 n - 1 \right|_{\substack{ n=1 }} = 1 \]
    S1.✓ Proved
  3. \[ 1 + 3 = 4 \]
    S2.✓ Proved
  4. \[ 1 + 3 + 5 = 9 \]
    S3.✓ Proved
  5. \[ 1 + 3 + 5 + 7 = 16 \]
    S4.✓ Proved
Answer \( S_1 = 1,\ S_2 = 4,\ S_3 = 9,\ S_4 = 16 \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 7.561

easy
Find the first four partial sums \( \displaystyle S_1, \dots, S_4 \) of \( \displaystyle \sum_{n=1}^{\infty} \frac{1}{n \left(n + 1\right)} \).
  1. S_k adds the first k terms: a₁ = 1/2, a₂ = 1/6, and so on.
    Reviewed
  2. \[ \left. 1 \frac{1}{n \left(n + 1\right)} \right|_{\substack{ n=1 }} = \frac{1}{2} \]
    S1.✓ Proved
  3. \[ 1 \cdot \frac{1}{6} + 1 \cdot \frac{1}{2} = \frac{2}{3} \]
    S2.✓ Proved
  4. \[ 1 \cdot \frac{1}{12} + 1 \cdot \frac{1}{6} + 1 \cdot \frac{1}{2} = \frac{3}{4} \]
    S3.✓ Proved
  5. \[ 1 \cdot \frac{1}{20} + 1 \cdot \frac{1}{12} + 1 \cdot \frac{1}{6} + 1 \cdot \frac{1}{2} = \frac{4}{5} \]
    S4.✓ Proved
Answer \( S_1 = \frac{1}{2},\ S_2 = \frac{2}{3},\ S_3 = \frac{3}{4},\ S_4 = \frac{4}{5} \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 7.562

easy
Find the first four partial sums \( \displaystyle S_1, \dots, S_4 \) of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(-1\right)^{n + 1}}{n} \).
  1. S_k adds the first k terms: a₁ = 1, a₂ = -1/2, and so on.
    Reviewed
  2. \[ \left. \frac{\left(-1\right)^{n + 1}}{n} \right|_{\substack{ n=1 }} = 1 \]
    S1.✓ Proved
  3. \[ 1 - \frac{1}{2} = \frac{1}{2} \]
    S2.✓ Proved
  4. \[ - \frac{1}{2} + 1 \cdot \frac{1}{3} + 1 = \frac{5}{6} \]
    S3.✓ Proved
  5. \[ - \frac{1}{2} - 1 \cdot \frac{1}{4} + 1 \cdot \frac{1}{3} + 1 = \frac{7}{12} \]
    S4.✓ Proved
Answer \( S_1 = 1,\ S_2 = \frac{1}{2},\ S_3 = \frac{5}{6},\ S_4 = \frac{7}{12} \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 7.563

easy
Find the first four partial sums \( \displaystyle S_1, \dots, S_4 \) of \( \displaystyle \sum_{n=1}^{\infty} n^{2} \).
  1. S_k adds the first k terms: a₁ = 1, a₂ = 4, and so on.
    Reviewed
  2. \[ \left. n^{2} \right|_{\substack{ n=1 }} = 1 \]
    S1.✓ Proved
  3. \[ 1 + 4 = 5 \]
    S2.✓ Proved
  4. \[ 1 + 4 + 9 = 14 \]
    S3.✓ Proved
  5. \[ 1 + 4 + 9 + 16 = 30 \]
    S4.✓ Proved
Answer \( S_1 = 1,\ S_2 = 5,\ S_3 = 14,\ S_4 = 30 \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 7.565

easy
Find the first four partial sums \( \displaystyle S_1, \dots, S_4 \) of \( \displaystyle \sum_{n=1}^{\infty} n \).
  1. S_k adds the first k terms: a₁ = 1, a₂ = 2, and so on.
    Reviewed
  2. \[ \left. n \right|_{\substack{ n=1 }} = 1 \]
    S1.✓ Proved
  3. \[ 1 + 2 = 3 \]
    S2.✓ Proved
  4. \[ 1 + 2 + 3 = 6 \]
    S3.✓ Proved
  5. \[ 1 + 2 + 3 + 4 = 10 \]
    S4.✓ Proved
Answer \( S_1 = 1,\ S_2 = 3,\ S_3 = 6,\ S_4 = 10 \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 7.566

easy
Find the first four partial sums \( \displaystyle S_1, \dots, S_4 \) of \( \displaystyle \sum_{n=1}^{\infty} \frac{1}{n} \).
  1. S_k adds the first k terms: a₁ = 1, a₂ = 1/2, and so on.
    Reviewed
  2. \[ \left. 1 \frac{1}{n} \right|_{\substack{ n=1 }} = 1 \]
    S1.✓ Proved
  3. \[ 1 \cdot \frac{1}{2} + 1 = \frac{3}{2} \]
    S2.✓ Proved
  4. \[ 1 \cdot \frac{1}{3} + 1 \cdot \frac{1}{2} + 1 = \frac{11}{6} \]
    S3.✓ Proved
  5. \[ 1 \cdot \frac{1}{4} + 1 \cdot \frac{1}{3} + 1 \cdot \frac{1}{2} + 1 = \frac{25}{12} \]
    S4.✓ Proved
Answer \( S_1 = 1,\ S_2 = \frac{3}{2},\ S_3 = \frac{11}{6},\ S_4 = \frac{25}{12} \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

More partial sums of a series practice

Find the first four partial sums \( \displaystyle S_1, \dots, S_4 \) of \( \displaystyle \sum_{n=1}^{\infty} \left(-1\right)^{n} \).
Problem 7.560easy✓ Every equation proved
Find the first four partial sums \( \displaystyle S_1, \dots, S_4 \) of \( \displaystyle \sum_{n=1}^{\infty} 3 \cdot 3^{- n} \).
Problem 7.564easy✓ Nihil obstat
Find the first four partial sums \( \displaystyle S_1, \dots, S_4 \) of \( \displaystyle \sum_{n=1}^{\infty} 4 \cdot 3^{- n} \).
Problem 7.567easy✓ Nihil obstat
Find the first four partial sums \( \displaystyle S_1, \dots, S_4 \) of \( \displaystyle \sum_{n=1}^{\infty} 2 \cdot 3^{- n} \).
Problem 7.568easy✓ Nihil obstat