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Calculus 2 : Types of Series

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Example Questions

Example Question #1 : Sequences

Calculate the sum of the following infinite geometric series:

$\sum_{k=0}^{\infty}17\left(\frac{1}{6}\right)^k$

Possible Answers:

$\frac{102}{5}$

$\frac{104}{5}$

$\frac{103}{5}$

$\frac{101}{5}$

Correct answer:

$\frac{102}{5}$

Explanation:

This is an infinite geometric series.

The sum of an infinite geometric series can be calculated with the following formula,

$S = \frac{a_{1}}{1-r}$ , where $a_{1}$ is the first value of the summation, and r is the common ratio.

Solution:

Value of $a_{1}$ can be found by setting k = 0

$a_{1}=17$

r is the value contained in the exponent

$r=\frac{1}{6}$

$S=\frac{a_{1}}{1-r}$

$S=\frac{17}{1-\frac{1}{6}} = \frac{17*6}{5}$

$S=\frac{102}{5}$

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Example Question #31 : Types Of Series

Calculate the sum of the following infinite geometric series:

$\sum_{k=0}^{\infty}3\left(\frac{3}{7}\right)^k$

Possible Answers:

$\frac{27}{4}$

$\frac{21}{4}$

$\frac{25}{4}$

$\frac{23}{4}$

Correct answer:

$\frac{21}{4}$

Explanation:

This is an infinite geometric series.

The sum of an infinite geometric series can be calculated with the following formula,

$S = \frac{a_{1}}{1-r}$ , where $a_{1}$ is the first value of the summation, and r is the common ratio.

Solution:

Value of $a_{1}$ can be found by setting k = 0

$a_{1}=3$

r is the value contained in the exponent

$r=\frac{3}{7}$

$S=\frac{a_{1}}{1-r}$

$S=\frac{3}{1-\frac{3}{7}} = \frac{3}{\frac{4}{7}}$

$S=\frac{21}{4}$

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Example Question #32 : Types Of Series

Calculate the sum of the following infinite geometric series:

$\sum_{k=0}^{\infty}8\left(\frac{4}{9}\right)^k$

Possible Answers:

$\frac{72}{5}$

$\frac{74}{5}$

$\frac{71}{5}$

$\frac{73}{5}$

Correct answer:

$\frac{72}{5}$

Explanation:

This is an infinite geometric series.

The sum of an infinite geometric series can be calculated with the following formula,

$S = \frac{a_{1}}{1-r}$ , where $a_{1}$ is the first value of the summation, and r is the common ratio.

Solution:

Value of $a_{1}$ can be found by setting k = 0

$a_{1}=8$

r is the value contained in the exponent

$r=\frac{4}{9}$

$S=\frac{a_{1}}{1-r}$

$S=\frac{8}{1-\frac{4}{9}} = \frac{8}{\frac{5}{9}} = \frac{8*9}{5}$

$S=\frac{72}{5}$

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Example Question #31 : Types Of Series

Evaluate:

$\sum_{x= 0}^{\infty } \frac{1}{n^2+8n+15}$

Possible Answers:

$\frac{7}{24}$

$\frac{8}{15}$

$\frac{9}{40}$

The series does not converge.

$\frac{2}{5}$

Correct answer:

$\frac{7}{24}$

Explanation:

The method of partial fractions can be used to rewrite this expression:

$\frac{1}{n^2+8n+15} = \frac{1}{(n+3)(n+5)} = \frac{A}{n + 3} + \frac{B }{n+ 5}$

$\frac{1}{(n+3)(n+5)}$

$= \frac{A \left ( n+ 5 \right )}{(n+3)(n+5)} + \frac{B \left ( n + 3 \right )}{(n+3)(n+5)}$

$= \frac{A \left ( n+ 5 \right )+B \left ( n + 3 \right )}{(n+3)(n+5)}$

$= \frac{\left (A+B \right )n +5A+3B}{(n+3)(n+5)}$

A + B = 0

5A + 3B = 1

The solution can easily be found to be $A = \frac{1}{2}, B = -\frac{1}{2}$ , and the series can then be rewritten:

$\sum_{x= 0}^{\infty } \frac{1}{n^2+8n+15} = \sum_{x= 0}^{\infty } \left ( \frac{ \frac{1}{2}}{n + 3} - \frac{ \frac{1}{2} }{n+ 5} \right )$

$= \frac{1}{2}\sum_{x= 0}^{\infty } \left ( \frac{1}{n + 3} - \frac{1}{n+ 5} \right )$

The series is telescoping and is equal to the following:

$= \frac{1}{2}\left [ \left ( \frac{1}{0 + 3} - \frac{1}{0+ 5} \right ) + \left ( \frac{1}{1 + 3} - \frac{1}{1+ 5} \right ) + \left ( \frac{1}{2 + 3} - \frac{1}{2+ 5} \right ) +...\right ]$

$= \frac{1}{2}\left [ \left ( \frac{1}{ 3} - \frac{1}{ 5} \right ) + \left ( \frac{1}{4} - \frac{1}{6} \right ) + \left ( \frac{1}{5} - \frac{1}{7} \right ) +...\right ]$

$= \frac{1}{2}\left [ \frac{1}{ 3} +\frac{1}{4} + \left ( \frac{1}{ 5} - \frac{1}{ 5} \right ) + \left ( \frac{1}{6} - \frac{1}{6} \right ) + ...\right ]$

$= \frac{1}{2}\left ( \frac{1}{ 3} +\frac{1}{4} +0 + 0 + 0+... \right ) = \frac{1}{2}\left ( \frac{1}{ 3} +\frac{1}{4} \right ) = \frac{1}{2} \cdot \frac{7}{12} = \frac{7}{24}$

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Example Question #32 : Types Of Series

Evaluate:

$\sum_{x= 0}^{\infty } \frac{1}{n^2+9n+20}$

Possible Answers:

$\frac{1}{9}$

$\frac{1}{4}$

$\frac{1}{20}$

$\frac{1}{5}$

$\frac{9}{20}$

Correct answer:

$\frac{1}{4}$

Explanation:

The method of partial fractions can be used to rewrite this expression:

$\frac{1}{n^2+9n+20} = \frac{1}{(n+4)(n+5)} = \frac{A}{n + 4} + \frac{B }{n+ 5}$

$\frac{1}{(n+4)(n+5)}$

$= \frac{A \left ( n+ 5 \right )}{(n+4)(n+5)} + \frac{B \left ( n + 4 \right )}{(n+4)(n+5)}$

$= \frac{A \left ( n+ 5 \right )+B \left ( n + 4 \right )}{(n+4)(n+5)}$

$= \frac{\left (A+B \right )n +5A+4B}{(n+4)(n+5)}$

A + B = 0

5A + 4B = 1

The solution can easily be found to be A = 1, B = -1 , and the series can be rewritten:

$\sum_{x= 0}^{\infty } \frac{1}{n^2+9n+20} = \sum_{x= 0}^{\infty } \left ( \frac{1}{n+4} - \frac{1}{n+5} \right )$

This is a telescoping series:

$\left ( \frac{1}{0+4} - \frac{1}{0+5} \right ) + \left ( \frac{1}{1+4} - \frac{1}{1+5} \right ) + \left ( \frac{1}{2+4} - \frac{1}{2+5} \right ) +...$

$= \left ( \frac{1}{ 4} - \frac{1}{ 5} \right ) + \left ( \frac{1}{5} - \frac{1}{6} \right ) + \left ( \frac{1}{6} - \frac{1}{7} \right ) +...$

$= \frac{1}{ 4} + \left ( - \frac{1}{ 5} + \frac{1}{5} \right ) + \left ( - \frac{1}{ 6} + \frac{1}{6} \right ) +...$

$\frac{1}{4} + 0 + 0 + ... = \frac{1}{4}$

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Example Question #1 : Harmonic Series

The Harmonic series is a special case of a -series, with equal to what?

Possible Answers:

$\infty$

Correct answer:

Explanation:

A -series is a series of the form $\sum_{k=1}^{\infty} \frac{1}{k^p}$ , and the Harmonic Series is $\sum_{k=1}^{\infty} \frac{1}{k}$ . Hence p=1 .

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Example Question #2 : Harmonic Series

Which of the following tests will help determine whether $\sum_{x=1}^{\infty}\frac{1}{x}$ is convergent or divergent, and why?

Possible Answers:

$\textup{}$ Nth Term Test: The series diverge because the limit as goes to infinity is zero.

Root Test: Since the limit as approaches to infinity is zero, the series is convergent.

$\textup{}$ Divergence Test: Since limit of the series approaches zero, the series must converge.

Integral Test: The improper integral determines that the harmonic series diverge.

P-Series Test: The summation converges since p=1 .

Correct answer:

Integral Test: The improper integral determines that the harmonic series diverge.

Explanation:

The series $\sum_{x=1}^{\infty}\frac{1}{x}$ is a harmonic series.

The Nth term test and the Divergent test may not be used to determine whether this series converges, since this is a special case. The root test also does not apply in this scenario.

According the the P-series Test, $\sum_{x=1}^{\infty}\frac{1}{x^p}$ must converge only if p>1 . Therefore this could be a valid test, but a wrong definition as the answer choice since the series diverge for $p \leq1$ .

This leaves us with the Integral Test.

$\int_{1}^{\infty}\frac{1}{x}= \lim_{b \to \infty}\int_{1}^{b}\frac{1}{x}dx=\lim_{b \to \infty}( ln(b)-ln(1))=\infty$

Since the improper integral diverges, so does the series.

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Example Question #2 : Harmonic Series

Determine whether the following series converges or diverges:

$\sum_{n=0}^{\infty }(-1)^n(\frac{5}{n})$

Possible Answers:

The series (absolutely) converges

The series conditionally converges

The series may (absolutely) converge, diverge, or conditionally converge

The series diverges

Correct answer:

The series (absolutely) converges

Explanation:

Given just the harmonic series, we would state that the series diverges. However, we are given the alternating harmonic series. To determine whether this series will converge or diverge, we must use the Alternating Series test.

The test states that for a given series $\sum_{n=0}^{\infty }a_{n}$ where $a_{n}=(-1)^n(b_{n})$ or $a_{n}=(-1)^{n+1}(b_{n})$ where $b_{n}\geq0$ for all n, if $\lim_{n\rightarrow \infty }b_{n}=0$ and $b_{n}$ is a decreasing sequence, then $\sum_{n=0}^{\infty }a_{n}$ is convergent.

First, we must evaluate the limit of $b_{n}$ as n approaches infinity:

$\lim_{n\rightarrow \infty }\frac{5}{n}=5\lim_{n\rightarrow \infty }\frac{n}{n}(\frac{n^-1}{1})=0$

The limit equals zero because the numerator of the fraction equals zero as n approaches infinity.

Next, we must determine if $b_{n}$ is a decreasing sequence. $\frac{1}{n}< \frac{1}{n+1}$ , thus the sequence is decreasing.

Because both parts of the test passed, the series is (absolutely) convergent.

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Example Question #2 : Harmonic Series

Consider the alternating series

$\sum_{i=1}^\infty\frac{(-1)^{i-1}}{2^i}$ .

Which of the following tests for convergence is NOT conclusive?

Possible Answers:

The root test

The alternating series test

The limit test for divergence

The ratio test

Correct answer:

The limit test for divergence

Explanation:

Let

$a_n=\frac{(-1)^{n-1}}{2^n}$

be the nth summand in the series. The limit test for divergence states that

$\lim_{n\rightarrow\infty}a_n\neq 0$

implies that the series diverges.

However,

$\lim_{n\rightarrow\infty}\frac{(-1)^{n-1}}{2^n}=0$ ,

so the test is inconclusive.

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Example Question #5 : Harmonic Series

Does the following series converge?

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

Possible Answers:

Cannot be determined

Yes

Correct answer:

Explanation:

No the series does not converge. The given problem is the harmonic series, which diverges to infinity.

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Calculus 2 : Types of Series

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Sequences

Example Question #31 : Types Of Series

Example Question #32 : Types Of Series

Example Question #31 : Types Of Series

Example Question #32 : Types Of Series

Example Question #1 : Harmonic Series

Example Question #2 : Harmonic Series

Example Question #2 : Harmonic Series

Example Question #2 : Harmonic Series

Example Question #5 : Harmonic Series

All Calculus 2 Resources

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