TBIL-Cal Comparison Tests (SQ6)

$\sum_{n = 1}^{\infty} b_{n}$ converges but the value cannot be determined.
$\sum_{n = 1}^{\infty} b_{n}$ converges to $3 \cdot 5 = 15 .$
$\sum_{n = 1}^{\infty} b_{n}$ converges to some value other than 15.
$\sum_{n = 1}^{\infty} b_{n}$ diverges.
It cannot be determined whether $\sum_{n = 1}^{\infty} b_{n}$ converges or diverges.

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(b)

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\sum_{n = 1}^{\infty} a_{n}

diverges, what can be said about

\sum_{n = 1}^{\infty} b_{n} ?

$\sum_{n = 1}^{\infty} b_{n}$ converges but the value cannot be determined.
$\sum_{n = 1}^{\infty} b_{n}$ converges and the value can be determined.
$\sum_{n = 1}^{\infty} b_{n}$ diverges.
It cannot be determined whether $\sum_{n = 1}^{\infty} b_{n}$ converges or diverges.

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Activity 8.6.3.

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Using Fact 8.4.2, we know the geometric series

\sum_{n = 0}^{\infty} \frac{1}{2^{n}} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^{n}} + \dots = \frac{1}{1 - \frac{1}{2}} = 2.

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(a)

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What can we say about the series

3 + \frac{3}{2} + \frac{3}{4} + \frac{3}{8} + \dots + \frac{3}{2^{n}} + \dots ?

$3 + \frac{3}{2} + \frac{3}{4} + \frac{3}{8} + \dots + \frac{3}{2^{n}} + \dots$ converges to $3 \cdot 2 = 6 .$
$3 + \frac{3}{2} + \frac{3}{4} + \frac{3}{8} + \dots + \frac{3}{2^{n}} + \dots$ converges to some value other than 6.
$3 + \frac{3}{2} + \frac{3}{4} + \frac{3}{8} + \dots + \frac{3}{2^{n}} + \dots$ diverges.

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(b)

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What do you think we can say about the series

\frac{3.1}{2} + \frac{3.01}{4} + \frac{3.001}{8} + \dots + \frac{3 + (0.1)^{n}}{2^{n}} + \dots ?

$3 + \frac{3.1}{2} + \frac{3.01}{4} + \frac{3.001}{8} + \dots + \frac{3 + (0.1)^{n}}{2^{n}} + \dots$ converges to $3 \cdot 2 = 6 .$
$3 + \frac{3.1}{2} + \frac{3.01}{4} + \frac{3.001}{8} + \dots + \frac{3 + (0.1)^{n}}{2^{n}} + \dots$ converges to some value other than 6.
$3 + \frac{3.1}{2} + \frac{3.01}{4} + \frac{3.001}{8} + \dots + \frac{3 + (0.1)^{n}}{2^{n}} + \dots$ diverges.

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Activity 8.6.4.

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From Fact 8.4.2, we know

1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n} + \dots

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diverges.

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(a)

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What can we say about the series

5 + \frac{5}{2} + \frac{5}{3} + \frac{5}{4} + \dots + \frac{5}{n} + \dots ?

$5 + \frac{5}{2} + \frac{5}{3} + \frac{5}{4} + \dots + \frac{5}{n} + \dots$ converges to a known value we can compute.
$5 + \frac{5}{2} + \frac{5}{3} + \frac{5}{4} + \dots + \frac{5}{n} + \dots$ converges to some unknown value.
$5 + \frac{5}{2} + \frac{5}{3} + \frac{5}{4} + \dots + \frac{5}{n} + \dots$ diverges.

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(b)

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What do you think we can say about the series

4.9 + \frac{4.99}{2} + \frac{4.999}{3} + \frac{4.9999}{4} + \dots + \frac{5 - (0.1)^{n}}{n} + \dots ?

$4.9 + \frac{4.99}{2} + \frac{4.999}{3} + \frac{4.9999}{4} + \dots + \frac{5 - (0.1)^{n}}{n} + \dots$ converges to a known value we can compute.
$4.9 + \frac{4.99}{2} + \frac{4.999}{3} + \frac{4.9999}{4} + \dots + \frac{5 - (0.1)^{n}}{n} + \dots$ converges to some unknown value.
$4.9 + \frac{4.99}{2} + \frac{4.999}{3} + \frac{4.9999}{4} + \dots + \frac{5 - (0.1)^{n}}{n} + \dots$ diverges.

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Fact 8.6.5. The Limit Comparison Test.

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Let

\sum a_{n}

and

\sum b_{n}

be series with positive terms. If

lim_{n \to \infty} \frac{b_{n}}{a_{n}} = c

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for some positive (finite) constant

c,

then

\sum a_{n}

and

\sum b_{n}

either both converge or both diverge.

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Activity 8.6.6.

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Recall that

\sum_{n = 1}^{\infty} \frac{1}{2^{n}}

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converges.

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(a)

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Let

b_{n} = \frac{1}{n} .

Compute

lim_{n \to \infty} \frac{\frac{1}{n}}{\frac{1}{2^{n}}} .

$- \infty .$
$0 .$
$\frac{1}{2} .$
$1 .$
$\infty .$

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(b)

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Does

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

converge or diverge?

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(c)

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Let

b_{n} = \frac{1}{n^{2}} .

Compute

lim_{n \to \infty} \frac{\frac{1}{n^{2}}}{\frac{1}{2^{n}}} .

$\infty .$
$\ln (2) .$
$1 .$
$\frac{1}{2} .$
$- \infty .$

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(d)

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Does

\sum_{n = 1}^{\infty} \frac{1}{n^{2}}

converge or diverge?

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(e)

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Let

\sum a_{n}

and

\sum b_{n}

be series with positive terms. If

lim_{n \to \infty} \frac{b_{n}}{a_{n}}

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diverges, can we conclude that

\sum b_{n}

converges or diverges?

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Activity 8.6.7.

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We wish to determine if

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

converges or diverges using Fact 8.6.5.

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(a)

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Compute

lim_{n \to \infty} \frac{\frac{1}{4^{n} - 1}}{\frac{1}{4^{n}}} .

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(b)

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Does the geometric series

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

converge or diverge by Fact 8.4.2?

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(c)

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Does

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

converge or diverge?

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Activity 8.6.8.

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We wish to determine if

\sum_{n = 2}^{\infty} \frac{2}{\sqrt{n + 3}}

converges or diverges using Fact 8.6.5.

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(a)

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To which of the following should we compare

{a_{n}} = {\frac{2}{\sqrt{n + 3}}} ?

${\frac{1}{n}} .$
${\frac{1}{\sqrt{n}}} .$
${\frac{1}{n^{2}}} .$
${\frac{1}{2^{n}}} .$

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(b)

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Compute

lim_{n \to \infty} \frac{b_{n}}{a_{n}} .

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(c)

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Compute

lim_{n \to \infty} \frac{a_{n}}{b_{n}} .

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(d)

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What is true about

lim_{n \to \infty} \frac{b_{n}}{a_{n}}

and

lim_{n \to \infty} \frac{a_{n}}{b_{n}} ?

Their values are reciprocals.
Their values negative reciprocals.
They are both positive finite constants.
Only one value is a finite positive constant.
One value is $0$ and the other value is infinite.

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(e)

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Does the series

\sum_{n = 2}^{\infty} \frac{1}{\sqrt{n}}

converge or diverge?

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(f)

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Using your chosen sequence and Fact 8.6.5, does

\sum_{n = 2}^{\infty} \frac{2}{\sqrt{n + 3}}

converge or diverge?

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Activity 8.6.9.

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We wish to determine if

\sum_{n = 1}^{\infty} \frac{3}{n^{2} + 8 n + 5}

converges or diverges using Fact 8.6.5.

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(a)

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To which of the following should we compare

{x_{n}} = {\frac{3}{n^{2} + 8 n + 5}} ?

${\frac{1}{n}} .$
${\frac{1}{\sqrt{n}}} .$
${\frac{1}{n^{2}}} .$
${\frac{1}{2^{n}}} .$

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(b)

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Using your chosen sequence and Fact 8.6.5, does

\frac{3}{n^{2} + 8 n + 5}

converge or diverge?

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Activity 8.6.10.

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Use Fact 8.6.5 to determine if the series

\sum_{n = 5}^{\infty} \frac{2}{4^{n}}

converges or diverges.

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Activity 8.6.11.

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Consider sequences

{a_{n}}, {b_{n}}

where

a_{n} \geq b_{n} \geq 0 .

described in detail following the image — Figure 181. Plots of ${a_{n}}, {b_{n}}$

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(a)

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Suppose that

\sum_{n = 0}^{\infty} a_{n}

converges. What could be said about

{b_{n}} ?

$\sum_{n = 0}^{\infty} b_{n}$ converges.
$\sum_{n = 0}^{\infty} b_{n}$ diverges.
Whether or not $\sum_{n = 0}^{\infty} b_{n}$ converges or diverges cannot be determined with this information.

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(b)

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Suppose that

\sum_{n = 1}^{\infty} a_{n} = \sum_{n = 1}^{\infty} \frac{1}{n + 1}

which diverges. Which of the following statements are true?

$0 \leq \frac{1}{2 n^{2}} \leq \frac{1}{n + 1}$ for each $n \geq 1$ and $\sum_{n = 1}^{\infty} \frac{1}{2 n^{2}}$ is a convergent $p$ -series where $p = 2 .$
$0 \leq \frac{1}{2 n} \leq \frac{1}{n + 1}$ for each $n \geq 1$ and $\sum_{n = 1}^{\infty} \frac{1}{2 n}$ is a divergent $p$ -series where $p = 1 .$

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(c)

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Suppose that

\sum_{n = 0}^{\infty} a_{n}

was some series that diverges. What could be said about

{b_{n}} ?

$\sum_{n = 0}^{\infty} b_{n}$ converges.
$\sum_{n = 0}^{\infty} b_{n}$ diverges.
Whether or not $\sum_{n = 0}^{\infty} b_{n}$ converges or diverges cannot be determined with this information.

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(d)

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Suppose that

\sum_{n = 0}^{\infty} b_{n}

diverges. What could be said about

{a_{n}} ?

$\sum_{n = 0}^{\infty} a_{n}$ converges.
$\sum_{n = 0}^{\infty} a_{n}$ diverges.
Whether or not $\sum_{n = 0}^{\infty} a_{n}$ converges or diverges cannot be determined with this information.

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(e)

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Suppose that

\sum_{n = 0}^{\infty} b_{n} = \sum_{n = 0}^{\infty} \frac{1}{3^{n}}

which converges. Which of the following statements are true?

$0 \leq \frac{1}{3^{n}} \leq \frac{1}{2^{n}}$ for each $n$ and $\sum_{n = 0}^{\infty} \frac{1}{2^{n}}$ is a convergent geometric series where $| r | = \frac{1}{2} < 1 .$
$0 \leq \frac{1}{3^{n}} \leq 1$ for each $n$ and $\sum_{n = 0}^{\infty} 1$ diverges by the Divergence Test.

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(f)

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Suppose that

\sum_{n = 0}^{\infty} b_{n}

was some series that converges. What could be said about

{a_{n}} ?

$\sum_{n = 0}^{\infty} a_{n}$ converges.
$\sum_{n = 0}^{\infty} a_{n}$ diverges.
Whether or not $\sum_{n = 0}^{\infty} a_{n}$ converges or diverges cannot be determined with this information.

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Fact 8.6.12.

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Supppose we have sequences

{a_{n}}, {b_{n}}

so that for some

k

we have that

0 \leq b_{n} \leq a_{n}

for each

k \geq n .

Then we have the following results:

If $\sum_{k = n}^{\infty} a_{n}$ converges, then so does $\sum_{k = n}^{\infty} b_{n} .$
If $\sum_{k = n}^{\infty} b_{n}$ diverges, then so does $\sum_{k = n}^{\infty} a_{n} .$

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Activity 8.6.13.

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Suppose that you were handed positive sequences

{a_{n}}, {b_{n}} .

For the first few values

a_{n} \geq b_{n},

but after that what happens is unclear until

n = 100 .

Then for any

n \geq 100

we have that

a_{n} \leq b_{n} .

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(a)

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How might we best utilize Fact 8.6.12 to determine the convergence of

\sum_{n = 0}^{\infty} a_{n}

\sum_{n = 0}^{\infty} b_{n} ?

Since $a_{n}$ is sometimes greater than, and sometimes less than $b_{n},$ there is no way to utilize Fact 8.6.12.
Since initially, we have $b_{n} \leq a_{n},$ we can utilize Fact 8.6.12 by assuming $a_{n} \geq b_{n} .$
Since we can rewrite $\sum_{n = 0}^{\infty} a_{n} = \sum_{n = 0}^{99} a_{n} + \sum_{n = 100}^{\infty} a_{n}$ and $\sum_{n = 0}^{\infty} b_{n} = \sum_{n = 0}^{99} b_{n} + \sum_{n = 100}^{\infty} b_{n}$ and $\sum_{n = 0}^{99} a_{n}, \sum_{n = 0}^{99} b_{n}$ are necessarily finite, we can compare $\sum_{n = 100}^{\infty} a_{n}, \sum_{n = 100}^{\infty} b_{n}$ with Fact 8.6.12.

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Fact 8.6.14. The Direct Comparison Test.

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Let

\sum a_{n}

and

\sum b_{n}

be series with positive terms. If there is a

k

such that

b_{n} \leq a_{n}

for each

n \geq k,

then:

If $\sum a_{n}$ converges, then so does $\sum b_{n} .$
If $\sum b_{n}$ diverges, then so does $\sum a_{n} .$

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Activity 8.6.15.

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Suppose we wish to determine if

\sum_{n = 1}^{\infty} \frac{1}{2 n + 3}

converged using Fact 8.6.14.

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(a)

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Does

\sum_{n = 1}^{\infty} \frac{1}{3 n}

converge or diverge?

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(b)

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For which value

k

\frac{1}{3 n} \leq \frac{1}{2 n + 3}

for each

n \geq k ?

$\frac{1}{3 n} \leq \frac{1}{2 n + 3}$ for each $n \geq k = 0 .$
$\frac{1}{3 n} \leq \frac{1}{2 n + 3}$ for each $n \geq k = 1 .$
$\frac{1}{3 n} \leq \frac{1}{2 n + 3}$ for each $n \geq k = 2 .$
$\frac{1}{3 n} \leq \frac{1}{2 n + 3}$ for each $n \geq k = 3 .$
There is no $k$ for which $\frac{1}{3 n} \leq \frac{1}{2 n + 3}$ for each $n \geq k .$