Σ arⁿ converges to a/(1 − r) exactly when |r| < 1. 70 problems with worked solutions; in 70 of them every
equation is proved by a computer algebra system.
19 worked problems
Try each one before opening the solution. Every line is labeled with what checked it. How we verify
Does \( \displaystyle \sum_{n=0}^{\infty} - \left(- \frac{3}{5}\right)^{n} \) converge? If so, find its sum.
\[ -1 \]
The first term, a.Reviewed
\[ - \frac{3}{5} \]
The common ratio, r.Reviewed
|r| = 3/5 < 1, so the series converges to a/(1 − r).
Reviewed
\[ - \frac{5}{8} \]
a/(1 − r).Reviewed
Answer \( - \frac{5}{8} \)
Bars show the terms -(-3/5)^n for n = 0 to 23; dots show the running totals. The totals level off at -5/8.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=0}^{\infty} - 2 \left(- \frac{1}{3}\right)^{n} \) converge? If so, find its sum.
\[ -2 \]
The first term, a.Reviewed
\[ - \frac{1}{3} \]
The common ratio, r.Reviewed
|r| = 1/3 < 1, so the series converges to a/(1 − r).
Reviewed
\[ - \frac{3}{2} \]
a/(1 − r).Reviewed
Answer \( - \frac{3}{2} \)
Bars show the terms -2(-1/3)^n for n = 0 to 23; dots show the running totals. The totals level off at -3/2.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=0}^{\infty} \left(- \frac{3}{5}\right)^{n} \) converge? If so, find its sum.
\[ 1 \]
The first term, a.Reviewed
\[ - \frac{3}{5} \]
The common ratio, r.Reviewed
|r| = 3/5 < 1, so the series converges to a/(1 − r).
Reviewed
\[ \frac{5}{8} \]
a/(1 − r).Reviewed
Answer \( \frac{5}{8} \)
Bars show the terms (-3/5)^n for n = 0 to 23; dots show the running totals. The totals level off at 5/8.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=1}^{\infty} 4 \left(- \frac{1}{3}\right)^{n - 1} \) converge? If so, find its sum.
\[ 4 \]
The first term, a.Reviewed
\[ - \frac{1}{3} \]
The common ratio, r.Reviewed
|r| = 1/3 < 1, so the series converges to a/(1 − r).
Reviewed
\[ 3 \]
a/(1 − r).Reviewed
Answer \( 3 \)
Bars show the terms 4(-1/3)^(n - 1) for n = 1 to 24; dots show the running totals. The totals level off at 3.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=0}^{\infty} 2 \left(\frac{4}{5}\right)^{n} \) converge? If so, find its sum.
\[ 2 \]
The first term, a.Reviewed
\[ \frac{4}{5} \]
The common ratio, r.Reviewed
|r| = 4/5 < 1, so the series converges to a/(1 − r).
Reviewed
\[ 10 \]
a/(1 − r).Reviewed
Answer \( 10 \)
Bars show the terms 2(4/5)^n for n = 0 to 23; dots show the running totals. The totals level off at 10.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=0}^{\infty} \left(\frac{1}{5}\right)^{n} \) converge? If so, find its sum.
\[ 1 \]
The first term, a.Reviewed
\[ 1 \cdot \frac{1}{5} = \frac{1}{5} \]
The common ratio, r.Reviewed
|r| = 1/5 < 1, so the series converges to a/(1 − r).
Reviewed
\[ \frac{5}{4} \]
a/(1 − r).Reviewed
Answer \( \frac{5}{4} \)
Bars show the terms (1/5)^n for n = 0 to 23; dots show the running totals. The totals level off at 5/4.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=0}^{\infty} - 5^{- n} \) converge? If so, find its sum.
\[ -1 \]
The first term, a.Reviewed
\[ 1 \cdot \frac{1}{5} = \frac{1}{5} \]
The common ratio, r.Reviewed
|r| = 1/5 < 1, so the series converges to a/(1 − r).
Reviewed
\[ - \frac{5}{4} \]
a/(1 − r).Reviewed
Answer \( - \frac{5}{4} \)
Bars show the terms -1/5^n for n = 0 to 23; dots show the running totals. The totals level off at -5/4.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=0}^{\infty} 5 \left(- \frac{2}{3}\right)^{n} \) converge? If so, find its sum.
\[ 5 \]
The first term, a.Reviewed
\[ - \frac{2}{3} \]
The common ratio, r.Reviewed
|r| = 2/3 < 1, so the series converges to a/(1 − r).
Reviewed
\[ 3 \]
a/(1 − r).Reviewed
Answer \( 3 \)
Bars show the terms 5(-2/3)^n for n = 0 to 23; dots show the running totals. The totals level off at 3.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=0}^{\infty} 3 \cdot 5^{- n} \) converge? If so, find its sum.
\[ 3 \]
The first term, a.Reviewed
\[ 1 \cdot \frac{1}{5} = \frac{1}{5} \]
The common ratio, r.Reviewed
|r| = 1/5 < 1, so the series converges to a/(1 − r).
Reviewed
\[ \frac{15}{4} \]
a/(1 − r).Reviewed
Answer \( \frac{15}{4} \)
Bars show the terms 3/5^n for n = 0 to 23; dots show the running totals. The totals level off at 15/4.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=0}^{\infty} - 3 \cdot 2^{- n} \) converge? If so, find its sum.
\[ -3 \]
The first term, a.Reviewed
\[ 1 \cdot \frac{1}{2} = \frac{1}{2} \]
The common ratio, r.Reviewed
|r| = 1/2 < 1, so the series converges to a/(1 − r).
Reviewed
\[ -6 \]
a/(1 − r).Reviewed
Answer \( -6 \)
Bars show the terms -3/2^n for n = 0 to 23; dots show the running totals. The totals level off at -6.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=1}^{\infty} 5 \cdot 3^{1 - n} \) converge? If so, find its sum.
\[ 5 \]
The first term, a.Reviewed
\[ 1 \cdot \frac{1}{3} = \frac{1}{3} \]
The common ratio, r.Reviewed
|r| = 1/3 < 1, so the series converges to a/(1 − r).
Reviewed
\[ \frac{15}{2} \]
a/(1 − r).Reviewed
Answer \( \frac{15}{2} \)
Bars show the terms 5·3^(1 - n) for n = 1 to 24; dots show the running totals. The totals level off at 15/2.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=1}^{\infty} 5 \left(\frac{2}{5}\right)^{n - 1} \) converge? If so, find its sum.
\[ 5 \]
The first term, a.Reviewed
\[ \frac{2}{5} \]
The common ratio, r.Reviewed
|r| = 2/5 < 1, so the series converges to a/(1 − r).
Reviewed
\[ \frac{25}{3} \]
a/(1 − r).Reviewed
Answer \( \frac{25}{3} \)
Bars show the terms 5(2/5)^(n - 1) for n = 1 to 24; dots show the running totals. The totals level off at 25/3.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=1}^{\infty} - 5 \cdot 3^{1 - n} \) converge? If so, find its sum.
\[ -5 \]
The first term, a.Reviewed
\[ 1 \cdot \frac{1}{3} = \frac{1}{3} \]
The common ratio, r.Reviewed
|r| = 1/3 < 1, so the series converges to a/(1 − r).
Reviewed
\[ - \frac{15}{2} \]
a/(1 − r).Reviewed
Answer \( - \frac{15}{2} \)
Bars show the terms -5·3^(1 - n) for n = 1 to 24; dots show the running totals. The totals level off at -15/2.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=1}^{\infty} - 2 \left(- \frac{1}{2}\right)^{n - 1} \) converge? If so, find its sum.
\[ -2 \]
The first term, a.Reviewed
\[ - \frac{1}{2} \]
The common ratio, r.Reviewed
|r| = 1/2 < 1, so the series converges to a/(1 − r).
Reviewed
\[ - \frac{4}{3} \]
a/(1 − r).Reviewed
Answer \( - \frac{4}{3} \)
Bars show the terms -2(-1/2)^(n - 1) for n = 1 to 24; dots show the running totals. The totals level off at -4/3.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=1}^{\infty} \left(- \frac{2}{3}\right)^{n - 1} \) converge? If so, find its sum.
\[ 1 \]
The first term, a.Reviewed
\[ - \frac{2}{3} \]
The common ratio, r.Reviewed
|r| = 2/3 < 1, so the series converges to a/(1 − r).
Reviewed
\[ \frac{3}{5} \]
a/(1 − r).Reviewed
Answer \( \frac{3}{5} \)
Bars show the terms (-2/3)^(n - 1) for n = 1 to 24; dots show the running totals. The totals level off at 3/5.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=1}^{\infty} - 2^{1 - n} \) converge? If so, find its sum.
\[ -1 \]
The first term, a.Reviewed
\[ 1 \cdot \frac{1}{2} = \frac{1}{2} \]
The common ratio, r.Reviewed
|r| = 1/2 < 1, so the series converges to a/(1 − r).
Reviewed
\[ -2 \]
a/(1 − r).Reviewed
Answer \( -2 \)
Bars show the terms -2^(1 - n) for n = 1 to 24; dots show the running totals. The totals level off at -2.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=1}^{\infty} - 2 \left(\frac{2}{3}\right)^{n - 1} \) converge? If so, find its sum.
\[ -2 \]
The first term, a.Reviewed
\[ \frac{2}{3} \]
The common ratio, r.Reviewed
|r| = 2/3 < 1, so the series converges to a/(1 − r).
Reviewed
\[ -6 \]
a/(1 − r).Reviewed
Answer \( -6 \)
Bars show the terms -2(2/3)^(n - 1) for n = 1 to 24; dots show the running totals. The totals level off at -6.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=1}^{\infty} - \left(\frac{2}{5}\right)^{n - 1} \) converge? If so, find its sum.
\[ -1 \]
The first term, a.Reviewed
\[ \frac{2}{5} \]
The common ratio, r.Reviewed
|r| = 2/5 < 1, so the series converges to a/(1 − r).
Reviewed
\[ - \frac{5}{3} \]
a/(1 − r).Reviewed
Answer \( - \frac{5}{3} \)
Bars show the terms -(2/5)^(n - 1) for n = 1 to 24; dots show the running totals. The totals level off at -5/3.
✓ Nihil obstat Lines: 4 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
Does \( \displaystyle \sum_{n=1}^{\infty} - 3 \left(-1\right)^{n - 1} \) converge? If so, find its sum.
\[ -3 \]
The first term, a.Reviewed
\[ -1 \]
The common ratio, r.Reviewed
|r| = 1 ≥ 1, so the series diverges.
Reviewed
Answer \( \text{diverges} \)
Bars show the terms -3(-1)^(n - 1) for n = 1 to 24; dots show the running totals. The totals never settle: the series diverges.
✓ Nihil obstat Lines: 3 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