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Summing a series by differentiating or integrating practice problems

Sum Σ n xⁿ or Σ xⁿ/n by differentiating or integrating the geometric series. 20 problems with worked solutions; in 20 of them every equation is proved by a computer algebra system.

4 worked problems

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

Problem 7.615

hard
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \left(- \frac{1}{2}\right)^{n} n \).
  1. Differentiate Σ xⁿ = 1/(1 − x) term by term and multiply by x: Σ n xⁿ = x/(1 − x)² for |x| < 1.
    Reviewed
  2. \[ \frac{d}{d x} \frac{1}{1 - x} = \frac{1}{\left(1 - x\right)^{2}} \]
    d/dx of 1/(1 − x).✓ Proved
  3. \[ \left. x \frac{d}{d x} 1 \frac{1}{1 - x} \right|_{\substack{ x=- \frac{1}{2} }} = - \frac{2}{9} \]
    Put x = -1/2.✓ Proved
  4. \[ \sum_{n=1}^{\infty} \left(- \frac{1}{2}\right)^{n} n = - \frac{2}{9} \]
    Summed directly, the series agrees.✓ Proved
Answer \( - \frac{2}{9} \)

✓ Nihil obstat Lines: 3 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 7.617

hard
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(- \frac{1}{2}\right)^{n}}{n} \).
  1. Integrate Σ xⁿ⁻¹ = 1/(1 − x) from 0 to x term by term: Σ xⁿ/n = −ln(1 − x) for |x| < 1.
    Reviewed
  2. \[ \frac{d}{d x} \left(- \ln{\left(1 - x \right)}\right) = \frac{1}{1 - x} \]
    −ln(1 − x) is the antiderivative of 1/(1 − x) that is 0 at x = 0.✓ Proved
  3. \[ \left. - \ln{\left(1 - x \right)} \right|_{\substack{ x=- \frac{1}{2} }} = - \ln{\left(3 \right)} + \ln{\left(2 \right)} \]
    Put x = -1/2.✓ Proved
  4. \[ \sum_{n=1}^{\infty} \frac{\left(- \frac{1}{2}\right)^{n}}{n} = - \ln{\left(3 \right)} + \ln{\left(2 \right)} \]
    Summed directly, the series agrees.✓ Proved
Answer \( - \ln{\left(3 \right)} + \ln{\left(2 \right)} \)

✓ Nihil obstat Lines: 3 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 7.618

hard
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} 3^{- n} n \).
  1. Differentiate Σ xⁿ = 1/(1 − x) term by term and multiply by x: Σ n xⁿ = x/(1 − x)² for |x| < 1.
    Reviewed
  2. \[ \frac{d}{d x} \frac{1}{1 - x} = \frac{1}{\left(1 - x\right)^{2}} \]
    d/dx of 1/(1 − x).✓ Proved
  3. \[ \left. x \frac{d}{d x} 1 \frac{1}{1 - x} \right|_{\substack{ x=1 \cdot \frac{1}{3} }} = \frac{3}{4} \]
    Put x = 1/3.✓ Proved
  4. \[ \sum_{n=1}^{\infty} 3^{- n} n = \frac{3}{4} \]
    Summed directly, the series agrees.✓ Proved
Answer \( \frac{3}{4} \)

✓ Nihil obstat Lines: 3 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 7.622

hard
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \frac{3^{- n}}{n} \).
  1. Integrate Σ xⁿ⁻¹ = 1/(1 − x) from 0 to x term by term: Σ xⁿ/n = −ln(1 − x) for |x| < 1.
    Reviewed
  2. \[ \frac{d}{d x} \left(- \ln{\left(1 - x \right)}\right) = \frac{1}{1 - x} \]
    −ln(1 − x) is the antiderivative of 1/(1 − x) that is 0 at x = 0.✓ Proved
  3. \[ \left. - \ln{\left(1 - x \right)} \right|_{\substack{ x=1 \cdot \frac{1}{3} }} = - \ln{\left(2 \right)} + \ln{\left(3 \right)} \]
    Put x = 1/3.✓ Proved
  4. \[ \sum_{n=1}^{\infty} \frac{3^{- n}}{n} = - \ln{\left(2 \right)} + \ln{\left(3 \right)} \]
    Summed directly, the series agrees.✓ Proved
Answer \( - \ln{\left(2 \right)} + \ln{\left(3 \right)} \)

✓ Nihil obstat Lines: 3 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 summing a series by differentiating or integrating practice

Find the sum of \( \displaystyle \sum_{n=1}^{\infty} 10^{- n} n \).
Problem 7.614hard✓ Every equation proved
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \left(- \frac{1}{5}\right)^{n} n \).
Problem 7.616hard✓ Nihil obstat
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} 2^{- n} n \).
Problem 7.619hard✓ Nihil obstat
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \left(- \frac{1}{4}\right)^{n} n \).
Problem 7.620hard✓ Nihil obstat
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(- \frac{1}{4}\right)^{n}}{n} \).
Problem 7.621hard✓ Nihil obstat
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \frac{5^{- n}}{n} \).
Problem 7.623hard✓ Nihil obstat
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \left(- \frac{1}{3}\right)^{n} n \).
Problem 7.624hard✓ Every equation proved
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(- \frac{1}{10}\right)^{n}}{n} \).
Problem 7.625hard✓ Nihil obstat
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(- \frac{1}{3}\right)^{n}}{n} \).
Problem 7.626hard✓ Nihil obstat
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} 5^{- n} n \).
Problem 7.627hard✓ Nihil obstat
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \frac{4^{- n}}{n} \).
Problem 7.628hard✓ Nihil obstat
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} 4^{- n} n \).
Problem 7.684hard✓ Nihil obstat
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \frac{\left(- \frac{1}{5}\right)^{n}}{n} \).
Problem 7.685hard✓ Nihil obstat
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \frac{2^{- n}}{n} \).
Problem 7.686hard✓ Nihil obstat
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \frac{10^{- n}}{n} \).
Problem 7.687hard✓ Nihil obstat
Find the sum of \( \displaystyle \sum_{n=1}^{\infty} \left(- \frac{1}{10}\right)^{n} n \).
Problem 7.720hard✓ Nihil obstat