When does an infinite geometric series converge?

When the first term is 0, or when the ratio r satisfies -1 < r < 1. For r = 1 or r = -1 (with a nonzero first term) it diverges.

What is $\lim r^n$ when $r = -1$?

The terms alternate 1, -1, 1, -1, \ldots, so there is no limit.

Math III

Limits of sequences and infinite series

Q: What is $\lim r^n$ when $r = -1$?

The terms alternate 1, -1, 1, -1, \ldots, so there is no limit.

Find what value a sequence approaches as $n$ grows without bound. The key tools are handling indeterminate forms and the sum of an infinite geometric series.

Basic, Standard: Grade 12 · Term 1 / Advanced: Grade 12 · Exam prep

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The forms ∞/∞ and ∞ − ∞

For a limit of the form $\frac{\infty}{\infty}$ , divide the numerator and denominator by the highest power in the denominator and use $\frac{1}{n} \to 0$ .

\lim_{n \to \infty} \frac{3n^2 - n}{2n^2 + 1} = \lim_{n \to \infty} \frac{3 - \frac{1}{n}}{2 + \frac{1}{n^2}} = \frac{3}{2}

For $\infty - \infty$ forms such as $\sqrt{n^2 + 4n} - n$ , rationalize:

\sqrt{n^2 + 4n} - n = \frac{4n}{\sqrt{n^2 + 4n} + n} = \frac{4}{\sqrt{1 + \frac{4}{n}} + 1} \to 2

Caution

$\infty - \infty$ need not be $0$ , and $\frac{\infty}{\infty}$ need not be $1$ . Always transform the expression before taking the limit.

Limits of $r^n$ and geometric series

Formulas

\lim_{n \to \infty} r^n = \begin{cases} \infty & (r > 1) \\ 1 & (r = 1) \\ 0 & (|r| < 1) \end{cases} \qquad (\text{it oscillates for } r \leqq -1)

The infinite geometric series $\displaystyle\sum_{n=1}^{\infty} ar^{n-1}$ converges when $a = 0$ or $|r| < 1$ ; for $a \neq 0$ and $|r| < 1$ its sum is

\frac{a}{1 - r}

For expressions mixing powers, such as $\frac{3^{n+1} + 2^n}{3^n - 2^n}$ , divide by the dominant power $3^n$ so that $\left(\frac{2}{3}\right)^n \to 0$ .

When asked for the values of $x$ that make a series converge, remember the case where the first term is $0$ : then every term is $0$ and the series converges.

Telescoping sums

For series that are not geometric, find the partial sum $S_n$ first and then let $n \to \infty$ . Partial fractions make the middle terms cancel:

\sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{k+1}\right) = 1 - \frac{1}{n+1} \to 1

For "arithmetic times geometric" series such as $\displaystyle\sum_{n=1}^{\infty} nr^n$ , compute $S_n - rS_n$ to reduce to a geometric sum.

The squeeze theorem

Squeeze theorem

If $a_n \leqq c_n \leqq b_n$ and $\displaystyle\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n = \alpha$ , then $\displaystyle\lim_{n \to \infty} c_n = \alpha$ .

Example: $-1 \leqq \sin n \leqq 1$ gives $-\frac{1}{n} \leqq \frac{\sin n}{n} \leqq \frac{1}{n}$ , so $\frac{\sin n}{n} \to 0$ . Taking $n$ -th roots of $3^n < 2^n + 3^n < 2 \cdot 3^n$ shows $\sqrt[n]{2^n + 3^n} \to 3$ .

Q1BasicGr 12 · Term 1Limits of sequences and infinite series

Find the limit.

\lim_{n \to \infty} \frac{5^{n} + 2^{n}}{-2 \cdot 5^{n} + 2^{n}}

Answer

-\frac{1}{2}

Solution

Dividing the numerator and denominator by $5^n$ ,
$\lim_{n \to \infty} \frac{1 + \left(\frac{2}{5}\right)^{n}}{-2 + \left(\frac{2}{5}\right)^{n}}$
Since $\left|\frac{2}{5}\right| < 1$ , $\left(\frac{2}{5}\right)^{n} \to 0$ . Hence
$\lim_{n \to \infty} \frac{5^{n} + 2^{n}}{-2 \cdot 5^{n} + 2^{n}} = -\frac{1}{2}$

Q2StandardGr 12 · Term 1Limits of sequences and infinite series

Find the limit.

\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + \cdots + n^2}{2n^{3} + 4}

Answer

\frac{1}{6}

Solution

By the sum formula,
$1^2 + 2^2 + \cdots + n^2 = \frac{1}{6}n(n+1)(2n+1)$
Dividing by $n^3$ and taking the limit,
$\lim_{n \to \infty} \frac{\frac{1}{6}\left(1 + \frac{1}{n}\right)\left(2 + \frac{1}{n}\right)}{2 + \frac{4}{n^3}} = \frac{2}{12} = \frac{1}{6}$

Q3AdvancedExam prepLimits of sequences and infinite series

Find the sum of the infinite series.

\sum_{n=1}^{\infty} \frac{n}{4^n}

Answer

\frac{4}{9}

Solution

Let $r = \frac{1}{4}$ and $S_n = \sum_{k=1}^n kr^k$ . Then
$(1 - r)S_n = r + r^2 + \cdots + r^n - nr^{n+1} = \frac{r(1 - r^n)}{1 - r} - nr^{n+1}$
Since $|r| < 1$ , we have $r^n \to 0$ and $nr^{n+1} \to 0$ , so
$\lim_{n \to \infty} S_n = \frac{r}{(1 - r)^2} = \frac{\frac{1}{4}}{\left(1 - \frac{1}{4}\right)^2} = \frac{4}{9}$

Q1BasicGr 12 · Term 1Limits of sequences and infinite series

Write the repeating decimal $0.8\overline{1}$ as a fraction.

Q2StandardGr 12 · Term 1Limits of sequences and infinite series

Find the sum of the infinite series.

\sum_{n=1}^{\infty} \frac{3^{n} + 4^{n}}{5^n}

Q3AdvancedExam prepLimits of sequences and infinite series

Consider the infinite series $\displaystyle \sum_{n=1}^{\infty} x(3 - x)^{n-1}$ .

(1)Find the values of $x$ for which the series converges.
(2)Find the sum when $x = \frac{5}{2}$ .

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