Worked examples · Sheet 1

Sums of geometric series

Theory used: Geometric series Series, convergence, sum

Exercise 1

Compute the sum of the series

n = 0 \sum \infty (\frac{1}{3})^{n} .

q = \frac{1}{3}, ∣ q ∣ < 1

n = 0 \sum \infty (\frac{1}{3})^{n} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}

The series is geometric with ratio $q = \frac{1}{3}$ . Since $∣ q ∣ < 1$ , the series converges and the sum is the first term divided by $1 - q$ . The first term is the one with index $n = 0$ , that is, $1$ , therefore the sum equals $\frac{1}{1 - \frac{1}{3}} = \frac{3}{2}$ .

Exercise 2

Compute the sum of the series

n = 1 \sum \infty \frac{2 ^{n}}{5 ^{n + 1}} .

\frac{2 ^{n}}{5 ^{n + 1}} = \frac{1}{5} (\frac{2}{5})^{n}

n = 1 \sum \infty \frac{2 ^{n}}{5 ^{n + 1}} = \frac{1}{5} n = 1 \sum \infty (\frac{2}{5})^{n} = \frac{1}{5} \cdot \frac{\frac{2}{5}}{1 - \frac{2}{5}} = \frac{1}{5} \cdot \frac{2}{3} = \frac{2}{15}

The general term is not yet in the form $q^{n}$ , therefore it is rewritten: $5^{n + 1} = 5 \cdot 5^{n}$ , hence $\frac{2 ^{n}}{5 ^{n + 1}} = \frac{1}{5} (\frac{2}{5})^{n}$ . The constant $\frac{1}{5}$ comes out of the sum by linearity. What remains is a geometric series with ratio $q = \frac{2}{5}$ , with $∣ q ∣ < 1$ , starting from $n = 1$ : the sum is the first term $\frac{2}{5}$ divided by $1 - q = \frac{3}{5}$ , that is, $\frac{2}{3}$ . Multiplying by the constant $\frac{1}{5}$ gives $\frac{2}{15}$ .

Exercise 3

Compute the sum of the series

n = 2 \sum \infty \frac{( - 1 ) ^{n} 3 ^{n}}{4 ^{n}} .

\frac{( - 1 ) ^{n} 3 ^{n}}{4 ^{n}} = (- \frac{3}{4})^{n}, q = - \frac{3}{4}, ∣ q ∣ < 1

n = 2 \sum \infty (- \frac{3}{4})^{n} = \frac{( - \frac{3}{4} ) ^{2}}{1 - ( - \frac{3}{4} )} = \frac{\frac{9}{16}}{\frac{7}{4}} = \frac{9}{16} \cdot \frac{4}{7} = \frac{9}{28}

The factors $(- 1)^{n}$ , $3^{n}$ and $4^{n}$ have the same exponent, therefore the general term can be gathered into a single power: the ratio is $q = - \frac{3}{4}$ . Since $∣ q ∣ = \frac{3}{4} < 1$ , the series converges even with a negative ratio. The index starts from $n = 2$ , hence the first term is $q^{2} = \frac{9}{16}$ and the sum is the first term divided by $1 - q = 1 + \frac{3}{4} = \frac{7}{4}$ . The result is $\frac{9}{28}$ .