3.1 Sequences

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We determine if a sequence converges or diverges.

Sequence A sequence, denoted $\displaystyle \{a_n\}$ or $\displaystyle \{a_n\}_{1}^\infty$ , is a function whose domain is the natural numbers: $1, 2, 3, \dots$ , i.e, $a_n = f(n)$

We think of a sequence as an ordered list of numbers, $a_1, a_2, a_3, \dots$ . We can also start our sequence with $a_0$ instead of $a_1$

Some examples of sequences are

$\begin{align*} i) \;\; & \{a_n\}_{1}^\infty = \{n^2\} = 1^2, 2^2, 3^2, 4^2, \dots = 1, 4, 9, 16, \dots\\ ii) \;\; & \{a_n\}_{1}^\infty = \{(-1)^n\} = (-1)^1, (-1)^2, (-1)^3, \dots = -1, 1, -1, 1, \dots\\ iii) \;\; & \{a_n\}_{1}^\infty = \left\{\frac{1}{2^n}\right\} = \frac{1}{2^1}, \frac{1}{2^2}, \frac{1}{2^3}, \frac{1}{2^4}, \dots = \frac12, \frac14, \frac18, \frac{1}{16}, \dots\\ iv) \;\; & \{a_n\}_{1}^\infty = \left\{\frac{n}{n+1}\right\} = \frac{1}{1+1}, \frac{2}{2+1}, \frac{3}{3+1}, \frac{4}{4+1}, \dots = \frac12, \frac23, \frac34, \frac45, \dots \end{align*}$

The question we will have about a given sequence is if it has a limit.

A sequence $\{a_n\}$ has a limit $L$ , written $\lim _{n \to \infty } a_n \quad \mbox {or} \quad a_n \to L \; \mbox {as} \; n \to \infty$ if the terms of the sequence get arbitrarily close to $L$ as $n$ increases without bound.
If a sequence has a (finite) limit, we say that the sequence converges. Otherwise, we say that it diverges.

Does the sequence $\displaystyle \{a_n\} = \left \{\frac {1}{2^n}\right \}$ converge or diverge?
We compute $\lim _{n \to \infty } \frac {1}{2^n}$ Since $2^n \to \infty$ as $n \to \infty$ , we can conclude that $\lim _{n \to \infty } \frac {1}{2^n} = 0$ and so the sequence converges to $0$ . The diagram below shows the terms of the sequence approaching $0$ .

The sequence in the previous problem could be written as $\displaystyle \{a_n\} = \left \{\left (\frac 12\right )^n\right \}$ .
In general, a sequence of the form $\{a_n\} = \{r^n\}$ converges if $0<r<1$ and diverges if $r>1$ .

Does the sequence $\displaystyle \{a_n\} = \left \{(-1)^n\right \}$ converge or diverge?
We compute $\lim _{n \to \infty } (-1)^n$ Since $(-1)^n$ oscillates between $-1$ and $1$ we can conclude that $\lim _{n \to \infty } (-1)^n = \text {DNE}$ and so the sequence diverges. The diagram below shows the oscillation of the terms of the sequence.

Does the sequence $\displaystyle \{a_n\} = \left \{n^2\right \}$ converge or diverge?
Since $\lim _{n \to \infty } n^2 = \infty$ we conclude that the sequence diverges. The diagram below shows that the terms of the sequence grow without bound.

Does the sequence $\displaystyle \{a_n\} = \left \{\frac {n}{n+1}\right \}$ converge or diverge?
We compute $\begin{align*} \lim_{n \to \infty} \frac{n}{n+1} &=\lim_{n \to \infty} \frac{n}{n\left(1+\frac{1}{n}\right)} \\ &=\lim_{n \to \infty} \frac{1}{1+\frac{1}{n}}\\ &=\frac{1}{1+0}\\ &= 1 \end{align*}$ Hence, the sequence converges to $1$ . The diagram below shows the terms of the sequence approaching $1$ .

Does the sequence $\displaystyle \{a_n\} = \left \{\frac {n^2}{e^n}\right \}$ converge or diverge?
To compute the limit $\lim _{n \to \infty } \frac {n^2}{e^n}$ we use the fact that if a function, $f(x)$ , has a limit, then the associated sequence $a_n = f(n)$ has the same limit. Using L’Hopital’s rule twice, we have $\begin{align*} \lim_{x \to \infty} \frac{x^2}{e^x} &= \frac{\infty}{\infty} = \lim_{x \to \infty} \frac{2x}{e^x} \;\text{(L'Hopital)}\\ &=\frac{\infty}{\infty} = \lim_{x \to \infty} \frac{2}{e^x} \;\;\text{(L'Hopital again)}\\ &=0 \end{align*}$ Hence, $\lim _{n \to \infty } \frac {n^2}{e^n} =0$ as well, and the sequence converges to $0$ .

A sequence $\{a_n\}$ is called increasing if $a_{n+1} > a_n$ for all $n$ . Similarly, $\{a_n\}$ is called decreasing if $a_{n+1} < a_n$ for all $n$ .

Show that the sequence $\left \{\frac {1}{n}\right \}$ is decreasing.
We will use the fact that if $0<a<b$ then $\frac {1}{a} > \frac {1}{b}$ . Since $n+1 > n \quad \text {for} \; n = 1, 2, 3, \ldots$ Taking reciprocals gives $\frac {1}{n+1} < \frac {1}{n} \quad \text {for} \; n = 1, 2, 3, \ldots$ and hence the sequence $\left \{\frac {1}{n}\right \}$ is decreasing.

If the function $f(x)$ associated with the sequence $\{a_n\} = \{f(n)\}$ is an increasing (or decreasing) function, then the sequence is increasing (or decreasing).

Show that the sequence $\left \{\frac {n+1}{n^2 + 1}\right \}$ is decreasing.
We will show that the associated function $f(x) = \frac {x+1}{x^2 + 1}$ is an increasing function. To do this recall that if $f'(x)< 0$ for $x>a$ , then $f(x)$ is decreasing on the interval $(a,\infty )$ . The quotient rule gives $f'(x) = \frac {\left (x^2+1\right ) - 2x(x+1)}{(x^2+1)^2} = \frac {-x^2-2x+1}{(x^2+1)^2}$ which is negative if $x>1/2$ . Thus $f(x)$ is decreasing on the interval $(1/2, \infty )$ and $\left \{\frac {n+1}{n^2 + 1} \right \}$ is a decreasing sequence.

Problems

You can type DNE or infinity into the answer box as needed.

(problem 1) Find the limit of the sequence $\displaystyle \left \{\frac {1}{3^n}\right \}$ (if it exists), and then state whether the sequence converges or diverges.
$\displaystyle \lim _{n\to \infty } \frac {1}{3^n} = \;\;\answer {0}$ The sequence convergesdiverges

(problem 2) Find the limit of the sequence $\displaystyle \left \{\frac {n+1}{n^3}\right \}$ (if it exists), and then state whether the sequence converges or diverges.
$\displaystyle \lim _{n\to \infty } \frac {n+1}{n^3} = \;\;\answer {0}$ The sequence convergesdiverges

(problem 3) Find the limit of the sequence $\displaystyle \left \{\sqrt [n]{2}\right \}$ (if it exists), and then state whether the sequence converges or diverges.
$\displaystyle \lim _{n\to \infty } \sqrt [n]2 = \;\;\answer {1}$ The sequence convergesdiverges

(problem 4) Find the limit of the sequence $\displaystyle \left \{\frac {2^{2n}}{3^n}\right \}$ (if it exists), and then state whether the sequence converges or diverges.
$\displaystyle \lim _{n\to \infty } \frac {2^{2n}}{3^n} = \;\;\answer {\infty }$ The sequence convergesdiverges

(problem 5) Find the limit of the sequence $\displaystyle \left \{\frac {\ln n}{n}\right \}$ (if it exists), and then state whether the sequence converges or diverges.
$\displaystyle \lim _{n\to \infty } \frac {\ln n}{n} = \;\;\answer {0}$ The sequence convergesdiverges

(problem 6) Find the limit of the sequence $\displaystyle \left \{\frac {n^2}{5n+4}\right \}$ (if it exists), and then state whether the sequence converges or diverges.
$\displaystyle \lim _{n\to \infty } \frac {n^2}{5n+4} = \;\;\answer {\infty }$ The sequence convergesdiverges

(problem 7) Find the limit of the sequence $\displaystyle \left \{\frac {(-1)^n}{n}\right \}$ (if it exists), and then state whether the sequence converges or diverges.
$\displaystyle \lim _{n\to \infty } \frac {(-1)^n}{n} = \;\;\answer {0}$ The sequence convergesdiverges

(problem 8) Find the limit of the sequence $\displaystyle \left \{\cos (n\pi )\right \}$ (if it exists), and then state whether the sequence converges or diverges.
$\displaystyle \lim _{n\to \infty } \cos (n\pi ) = \;\;\answer {DNE}$ The sequence convergesdiverges

(problem 9) Find the limit of the sequence $\displaystyle \left \{\frac {n}{2^n}\right \}$ (if it exists), and then state whether the sequence converges or diverges.
$\displaystyle \lim _{n\to \infty } \frac {n}{2^n} = \;\;\answer {0}$ The sequence convergesdiverges

(problem 10) Find the limit of the sequence $\displaystyle \left \{\frac {2n+1}{3n+2}\right \}$ (if it exists), and then state whether the sequence converges or diverges.
$\displaystyle \lim _{n\to \infty } \frac {2n+1}{3n+2} = \;\;\answer {2/3}$ The sequence convergesdiverges

(problem 11) Determine whether the sequence is increasing, decreasing or neither.
a) $\displaystyle \left \{\frac {1}{\sqrt n}\right \}$ increasing decreasing neither
b) $\displaystyle \left \{\frac {n}{n^2 +1}\right \}$ increasing decreasing neither
c) $\displaystyle \left \{\ln n\right \}$ increasing decreasing neither
d) $\displaystyle \left \{\frac {\cos (n\pi )}{n^2}\right \}$ increasing decreasing neither
e) $\displaystyle \left \{\frac {n^2 + 1}{n^3 + 1}\right \}$ increasing decreasing neither