Exercises: Sequences

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Exercises relating to sequences.

The sequence $a_n = 1/n^3$ has limit $L = 0$ . Suppose $\epsilon = 1/64$ ; find a threshold $N$ such that $|a_n -L| < \epsilon$ is guaranteed to hold for all $n > N$ (take your value of $N$ as small as possible). $N = \answer {4}.$

The sequence $a_n = (2n^2 + (-1)^n)/n^2$ has limit $L = 2$ . Suppose $\epsilon = 1/100$ ; find a threshold $N$ such that $|a_n -L| < \epsilon$ is guaranteed to hold for all $n > N$ (take your value of $N$ as small as possible). $N = \answer {10}.$

Determine the $n^\text {th}$ term of the given sequence. $a_1 = 4$ , $a_2 = 7$ , $a_3 = 10$ , $a_4 = 13$ , $a_5 = 16$ , $\ldots$ $a_n = \answer {3n+1}.$

Determine the $n^\text {th}$ term of the given sequence. $a_1 = 3$ , $a_2 = -5/2$ , $a_3 = 7/4$ , $a_4 = -9/8$ , $a_5 = 11/16$ , $\ldots$ $a_n = \answer {\frac {(-1)^{n-1} (2n+1)}{2^{n-1}}}.$

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $a_n = \left \{(-1)^n\frac {n}{n+1}\right \}$ The sequence to $\answer {N/A}$ (enter N/A if the sequence does not converge to a finite answer).

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $a_n = \frac {4n^2-n+5}{3n^2+1}$ The sequence to $\answer {4/3}$ (enter N/A if the sequence does not converge to a finite answer).

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $a_n = \frac {4^n}{5^n}$ The sequence to $\answer {0}$ (enter N/A if the sequence does not converge to a finite answer).

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $a_n = \left (1 - \frac {3}{n} \right )^{-n}$ The sequence to $\answer {e^3}$ (enter N/A if the sequence does not converge to a finite answer).

Take the reciprocal and compare to your reference list of commonly-occurring limits.

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $a_n = \left (1 - \frac {3}{n}\right )^{-n^2}$ The sequence to $\answer {N/A}$ (enter N/A if the sequence does not converge to a finite answer).

If a sequence $b_n$ tends to $e^3$ , what will the sequence $(b_n)^n$ do?

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $a_n = \frac {(1.1)^n}{n}$ The sequence to $\answer {N/A}$ (enter N/A if the sequence does not converge to a finite answer).

What are the relative orders of growth of numerator versus denominator?

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $a_n = \frac {(0.9)^n}{n}$ The sequence to $\answer {0}$ (enter N/A if the sequence does not converge to a finite answer).

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $a_n = n^{1000000} (0.9)^n$ The sequence to $\answer {0}$ (enter N/A if the sequence does not converge to a finite answer).

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $a_n = \frac {\ln n}{n}$ The sequence to $\answer {0}$ (enter N/A if the sequence does not converge to a finite answer).

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $a_n = \frac {\ln n}{n^{0.00001}}$ The sequence to $\answer {0}$ (enter N/A if the sequence does not converge to a finite answer).

Let $b_n$ be the sequence given by $b_1 = 0 \ \text { and } \ b_{n+1} = \frac {2 + b_n}{3} \ \text { for } n \geq 1$ converges. Compute its limit. $\lim _{n \rightarrow \infty } b_n = \answer {1}.$

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $a_1 = 1 \ \text { and } \ a_{n+1} = a_n + \frac {1}{a_n} \ \text { for } n \geq 1$ The sequence to $\answer {N/A}$ (enter N/A if the sequence does not converge to a finite answer).

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $a_1 = 1 \ \text { and } \ a_{n+1} = a_n + \frac {1}{4} (4 - a_n^2) \ \text { for } n \geq 1$ The sequence to $\answer {2}$ (enter N/A if the sequence does not converge to a finite answer).

Let $a_n$ be the sequence given by $a_1 = \frac {1}{4}, \ \text { and } \ a_{n+1} = 2 a_n(1-a_n) \ \text { for } n \geq 1$ converges. Compute its limit. $\lim _{n \rightarrow \infty } a_n = \answer {\frac {1}{2}}.$

If $a_n$ happens to be positive and less than $1/2$ , then $2 (1-a_n) > 1$ , so this forces $2 a_n(1-a_n) > a_n$ (meaning that the term after $a_n$ will be larger than $a_n$ .

The function $2 x(1-x)$ is nonnegative on the interval $[0,1]$ and has a maximum value of $1/2$ attained at $x = 1/2$ . This means that if $a_n$ is anything between $0$ and $1$ , the next term of the sequence will always be between $0$ and $1/2$ .

Sample Quiz Questions

Find the limit of the sequence $\lim _{n \rightarrow \infty } \sqrt {\frac {2n-2}{2n^2-4n+3}}.$ Justify your response. (Hints will not be revealed until after you choose a response.)

Because the square root function is continuous, you can pass the limit through it and compute $\sqrt {\lim _{n \rightarrow \infty } \frac {2n-2}{2n^2-4n+3}}.$

Reduce numerator and denominator to the dominant terms (in the regime $n \rightarrow \infty$ ).

$\begin {aligned} \lim _{n \rightarrow \infty } \sqrt {\frac {2n-2}{2n^2-4n+3}} & = \lim _{n \rightarrow \infty } \sqrt {\frac {2n^{-1}-2n^{-2}}{2-4n^{-1}+3n^{-2}}} \\ & = \sqrt {\frac {\lim _{n \rightarrow \infty } 2n^{-1}-2n^{-2}}{\lim _{n \rightarrow \infty } 2-4n^{-1}+3n^{-2}}} \\ & = \sqrt {\frac {0}{2}} = 0. \end {aligned}$

Determine whether the limit below exists. If it exists, find its value. $\lim _{n \rightarrow \infty } \frac {(-1)^{n+1}n^2 - 2^{-n-1}}{(-1)^{n}n^2 + 4^{-n}}.$ Justify your response. (Hints will not be revealed until after you choose a response.)

Comparing the orders of growth of the terms in the numerator, the first term dominates because $|-1|>|1/2|$ .

Likewise the first term dominates in the denominator because $|-1|>|1/4|$ .

Neglecting non-dominant terms leads to the limit $\lim _{n \rightarrow \infty } \frac {(-1)^{n+1}n^2}{(-1)^{n}n^2}$ which simply equals $-1$ .

Determine whether the limit below exists. If it exists, find its value. $\lim _{n \rightarrow \infty } \left (\frac {4n - 3}{4n + 1}\right )^{n}.$ Justify your response. (Hints will not be revealed until after you choose a response.)

First observe that $\frac {4n - 3}{4n + 1} = 1 - \frac {4}{4n + 1} \rightarrow 1$ as $n \rightarrow \infty$ .

Next, in light of the known limit $(1+x/k)^{k} \rightarrow e^x$ as $k \rightarrow \infty$ , manipulate exponents to see that $\left (1 - \frac {4}{4n + 1}\right )^n = \left ( \left (1 - \frac {4}{4n + 1}\right )^{4n + 1} \right )^{ 1/4} \left (1 - \frac {4}{4n + 1}\right )^{-1/4}.$

As $n \rightarrow \infty$ , the first term on the right-hand side tends to $e^{-1}$ and the second term tends to $1$ . Thus the original sequence tends to $e^{-1}$ as well.

Determine whether the limit below exists. If it exists, find its value. $\lim _{n \rightarrow \infty } \left (\frac {n + 3}{2n - 2}\right )^{n^2}.$ Justify your response. (Hints will not be revealed until after you choose a response.)

First observe that $\frac {n + 3}{2n - 2} = \frac {1}{2} + \frac {2}{n - 1} \rightarrow \frac {1}{2}$ as $n \rightarrow \infty$ .

Since the limit is positive and less than one, raising this expression to increasingly large powers generates a sequence which converges rapidly to zero.

Sample Exam Questions

Determine whether the sequence $\displaystyle a_n = (-1)^{n-1} \frac {n^2}{1 + n^2 + n^3}$ converges or diverges. If it converges, find its limit.

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