Limits and Infinite Geometric Series

Intuitive Notion of the Limit of a Sequence

Consider the sequence $(a_n)$ defined by $a_n = \dfrac{n}{n+1}$ . Its first few terms are:

\frac{1}{2},\; \frac{2}{3},\; \frac{3}{4},\; \frac{4}{5},\; \frac{5}{6},\; \frac{6}{7},\; \frac{7}{8},\; \frac{8}{9},\; \ldots

If we plot the terms of this sequence on the number line, the points get closer and closer to the point with coordinate 1.

📐Definition — Limit of a Sequence

We say that as $n$ increases, the terms of the sequence $(a_n)$ approach the number 1. In other words, as $n$ increases, the difference $|a_n - 1|$ becomes smaller and smaller. For example, $|a_n - 1| < 0.1$ when $n \geqslant 10$ , $|a_n - 1| < 0.0001$ when $n \geqslant 10\,000$ , etc.

In general, starting from some number $n$ , the difference $|a_n - 1|$ can become smaller than any given positive number.

In this case, the number 1 is called the limit of the sequence $a_n$ , and we write:

\lim_{n \to \infty} a_n = 1 \quad \text{or} \quad \frac{n}{n+1} \to 1 \text{ as } n \to \infty.

📐Definition — Convergent Sequence

A sequence that has a limit is called convergent. For example, the sequence with $a_n = \dfrac{1}{2^n}$ is convergent: as $n$ increases, its terms approach 0, i.e., $\displaystyle\lim_{n \to \infty} \frac{1}{2^n} = 0$ .

Not every sequence is convergent. For example, the sequence of natural numbers has no limit. Nor does the sequence $(a_n)$ where $a_n = (-1)^n$ .

Sum of an Infinite Geometric Series

So far we have considered sums with a finite number of terms. However, some problems require us to consider sums of infinitely many terms.

Consider a square with side 1. Divide it into 2 equal parts and shade one part. Then divide the unshaded part into 2 equal parts, shade one of them, and continue.

After the $n$ -th step, the shaded area equals:

S_n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^n}.

Using the formula for the sum of a geometric progression:

S_n = \frac{\frac{1}{2}\left(\left(\frac{1}{2}\right)^n - 1\right)}{\frac{1}{2} - 1} = 1 - \left(\frac{1}{2}\right)^n = 1 - \frac{1}{2^n}.

Since $\displaystyle\lim_{n \to \infty} \frac{1}{2^n} = 0$ , we get $\displaystyle\lim_{n \to \infty} S_n = 1$ .

Formula for the Sum of an Infinite Geometric Series

Consider an arbitrary infinite geometric progression $b_1, b_2, b_3, \ldots, b_n, \ldots$ with $|q| < 1$ .

The sum of its first $n$ terms is $S_n = \dfrac{b_1(q^n - 1)}{q - 1}$ . We can rewrite:

S_n = \frac{b_1(q^n - 1)}{q - 1} = \frac{b_1 - b_1 q^n}{1 - q} = \frac{b_1}{1-q} - \frac{b_1}{1-q} \cdot q^n.

Since $|q| < 1$ , we have $\displaystyle\lim_{n \to \infty} q^n = 0$ , so $\displaystyle\lim_{n \to \infty} \frac{b_1}{1-q} \cdot q^n = 0$ . Therefore:

\lim_{n \to \infty} S_n = \frac{b_1}{1-q}.

⚡Theorem — Sum of an Infinite Geometric Series

The number $\dfrac{b_1}{1-q}$ is called the sum of an infinite geometric series $(b_n)$ with $|q| < 1$ :

b_1 + b_2 + b_3 + \ldots + b_n + \ldots = \frac{b_1}{1-q}.

Denoting this sum by $S$ :

S = \frac{b_1}{1 - q}

ℹNote — When Does the Sum Exist?

The sum of an infinite geometric series exists only when $|q| < 1$ . If $|q| \geqslant 1$ , the partial sums $S_n$ do not approach any number as $n$ increases, and the sum does not exist.

Converting Repeating Decimals to Fractions

The formula for the sum of an infinite geometric series can be used to convert repeating decimals to fractions.

✎Example — Converting the Repeating Decimal 0.(45) to a Fraction

Problem. Express the repeating decimal $0.(45)$ as a fraction.

Solution. Write the number as a sum:

0.(45) = 0.454545\ldots = 0.45 + 0.0045 + 0.000045 + \ldots

The terms $0.45, 0.0045, 0.000045, \ldots$ form an infinite geometric series with $b_1 = 0.45$ and $q = 0.01$ . Since $|q| < 1$ :

S = \frac{0.45}{1 - 0.01} = \frac{0.45}{0.99} = \frac{45}{99} = \frac{5}{11}.

Therefore $0.(45) = \dfrac{5}{11}$ .

Answer: $\dfrac{5}{11}$ .

✎Example — Converting 0.2(54) to a Fraction

Problem. Express the repeating decimal $0.2(54)$ as a fraction.

Solution. We have:

0.2(54) = 0.2545454\ldots = 0.2 + 0.054 + 0.00054 + 0.0000054 + \ldots

The repeating part $0.0545454\ldots$ is the sum of an infinite geometric series with $b_1 = 0.054$ and $q = 0.01$ :

0.0545454\ldots = \frac{0.054}{1 - 0.01} = \frac{0.054}{0.99} = \frac{54}{990} = \frac{3}{55}.

Therefore $0.2(54) = 0.2 + 0.0(54) = \dfrac{1}{5} + \dfrac{3}{55} = \dfrac{11}{55} + \dfrac{3}{55} = \dfrac{14}{55}$ .

Answer: $\dfrac{14}{55}$ .

✎Example — Finding the Common Ratio of an Infinite GP

Problem. Find the common ratio of an infinite geometric progression ( $|q| < 1$ ) in which each term is 4 times the sum of all subsequent terms.

Solution. Let $a_n$ be an arbitrary term and $a_{n+1}$ the next term. The infinite GP starting from $a_{n+1}$ with ratio $q$ ( $|q| < 1$ ) has sum $\dfrac{a_{n+1}}{1-q}$ . By the condition, $a_n = \dfrac{4a_{n+1}}{1-q}$ , i.e., $a_n = \dfrac{4a_n q}{1-q}$ .

Since no term of a GP is zero: $1 = \dfrac{4q}{1-q}$ . Solving gives $q = \dfrac{1}{5}$ .

Answer: $\dfrac{1}{5}$ .

Exercises

✏Exercise — Infinite Geometric Series Problems

35.1. Find the sum of the infinite geometric progression $(b_n)$ with common ratio $q$ if:

$b_1 = 24$ , $q = \dfrac{3}{4}$
$b_1 = -84$ , $q = -\dfrac{1}{3}$

35.5. Express the repeating decimal as a fraction:

$0.1111\ldots$
$0.(5)$
$0.(24)$
$0.416416416\ldots$

35.7. Find the sum of the infinite geometric progression:

$\sqrt{2},\; -1,\; \dfrac{1}{\sqrt{2}},\; \ldots$
$3\sqrt{3},\; 3,\; \sqrt{3},\; \ldots$