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Calculus 2 : Types of Series

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Example Questions

Example Question #1 : Types Of Series

Evaluate:

$\sum_{n=0}^{\infty } \frac{3^{n}}{\pi ^{n+1}}$

Possible Answers:

$\pi - \frac{1}{3}$

$\frac{1}{\pi - 3 \right )}$

The series diverges.

$\pi + \frac{1}{3}$

$3 - \frac{1}{\pi}$

Correct answer:

$\frac{1}{\pi - 3 \right )}$

Explanation:

This can be rewritten as

$\sum_{n=0}^{\infty } \frac{3^{n}}{\pi ^{n+1}}$

$= \sum_{n=0}^{\infty }\left ( \frac{1}{\pi} \cdot \frac{3^{n}}{\pi ^{n }} \right )$

$= \sum_{n=0}^{\infty }\frac{1}{\pi} \left ( \frac{3 }{\pi} \right ) ^{n }$

$\pi > 3$ , so $\frac{3}{\pi} < 1$ , making this a convergent geometric series with initial term $\frac{1}{\pi} \left ( \frac{3 }{\pi} \right ) ^{0 } = \frac{1}{\pi}$ and common ratio $\frac{3 }{\pi}$ . The sum is therefore

$= \frac{ \frac{1}{\pi}}{1 - \frac{3}{\pi }} = \frac{1}{\pi \left (1 - \frac{3}{\pi } \right )} = \frac{1}{\pi - 3 \right )}$ .

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Example Question #2 : Types Of Series

Evaluate:

$\sum_{n=0}^{\infty } \frac{e^{n}}{\pi ^{n }}$

Possible Answers:

$\frac{\pi}{ \pi - e}$

The series diverges.

$\frac{e}{ \pi - e}$

$\frac{\pi}{ \pi + e}$

$\frac{e}{ \pi + e}$

Correct answer:

$\frac{\pi}{ \pi - e}$

Explanation:

This can be rewritten as

$\sum_{n=0}^{\infty } \frac{e^{n}}{\pi ^{n }} = \sum_{n=0}^{\infty }\left ( \frac{e}{\pi } \right )^{n}$ .

This is a geometric series with initial term $\left (\frac{e}{\pi } \right )^{0} = 1$ and common ratio $\frac{e}{\pi } \right$ . Since $e < \pi$ , $\frac{e}{\pi } < 1$ , and the series converges to:

$\sum_{n=0}^{\infty } \frac{e^{n}}{\pi ^{n }} = \frac{1}{1 - \frac{e}{\pi }} = \frac{\pi}{ \pi - e}$

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Example Question #1 : Types Of Series

Evaluate:

$\sum_{n= 1}^{\infty } \left ( \frac{e-1}{e} \right )^n$

Possible Answers:

The series is not convergent.

e-1

$\frac{2e}{e+1}$

$\frac{2e}{e-1}$

Correct answer:

e-1

Explanation:

$\sum_{n= 1}^{\infty } \left ( \frac{e-1}{e} \right )^n$ is an infinite geometric series with initial term $\left ( \frac{e-1}{e} \right )^1 = \frac{e-1}{e}$ and common ratio $\frac{e-1}{e}$ . The sum is therefore

$\sum_{n= 1}^{\infty } \left ( \frac{e-1}{e} \right )^n$

$=\frac{ \frac{e-1}{e} }{1-\frac{e-1}{e}}$ $\frac{}{}$

$=\frac{ e \cdot \frac{e-1}{e} }{e \cdot \left ( 1-\frac{e-1}{e} \right )}$

$= \frac{e-1}{e-(e-1)} = \frac{e-1}{1} = e - 1$

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Example Question #2962 : Calculus Ii

Evaluate:

$\sum_{n=0}^{\infty } \left (-\frac{1}{e} \right )^n$

Possible Answers:

$\frac{e+1}{e\right )}$

$\frac{1}{e \right )}$

$\frac{e-1}{e\right )}$

$\frac{e}{e+1\right )}$

Correct answer:

$\frac{e}{e+1\right )}$

Explanation:

$\sum_{n=0}^{\infty } \left (-\frac{1}{e} \right )^n$ is a geometric series with initial term $\left (-\frac{1}{e} \right )^0 = 1$ and common ratio $-\frac{1}{e}$ . The sum of this series is

$\frac{1}{1-\left ( -\frac{1}{e} \right )}= \frac{1}{1+ \frac{1}{e} \right )} = \frac{e}{e+1\right )}$ .

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Example Question #3 : Types Of Series

How many terms of a geometric series must you know in order to uniquely define that series?

Possible Answers:

Four

Five

One

Two

Three

Correct answer:

Two

Explanation:

In order to uniquely define the geometric series, we need to know two things: the ratio between successive terms and at least one of the terms. Knowing one term doesn't give you the ratio of successive terms, but knowing two terms will give you the ratio. By the term generator for a geometric series $\small a_{n}=a_{1}r^{n-1}$ , you can see that you only need two terms to find the ratio $\small r$ .

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Example Question #1 : Arithmetic And Geometric Series

Assume the term generator for an arithmetic sequence is $\small a_{n}=3n-2$ . What is the sum of the first $\small 20$ terms of this sequence $(\small \small a_{1}+a_{2}...+a_{20})$ ?

Possible Answers:

$\small 1180$

$\small 59$

$\small 58$

$\small 590$

$\small 118$

Correct answer:

$\small 590$

Explanation:

The sum formula for $\small n$ terms of an arithmetic series is $\small (a_{1}+a_{n})(n/2)$ .

For $\small 20$ terms, this formula becomes $\small \small \small (a_{1}+a_{20})(20/2)$ .

Using our term generator for $\small a_{1}$ and $\small a_{20}$ , this formula becomes

$\small \small \small \small \small (3(1)-2+3(20)-2)(20/2)=(59)(10)=590$ .

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Example Question #2964 : Calculus Ii

What value does the series $\small \sum_{j=2}^\infty (-1)^j\left(\frac{1}{2}\right)^{j+1}$ approach?

Possible Answers:

$\small \frac{1}{12}$

$\small \infty$

$\small -\infty$

$\small \frac{17}{12}$

Correct answer:

$\small \frac{1}{12}$

Explanation:

We can evaluate the infinite series by recognizing it as a geometric series times some constant.

Let's manipulate this series:

$\small \small \small \sum_{j=2}^\infty (-1)^j\left(\frac{1}{2}\right)^{j+1}=\sum_{j=2}^\infty (-1)^2(-1)^{j-2}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^{j-2}$

$\small \small \small =(-1)^2\left(\frac{1}{2}\right)^3\sum_{j=2}^\infty (-1)^{j-2}\left(\frac{1}{2}\right)^{j-2}=\frac{1}{8}\sum_{j=0}^\infty \left(\frac{-1}{2}\right)^{j}$ .

Now it suffices to evaluate $\small \small \small \sum_{j=0}^\infty \left(\frac{-1}{2}\right)^{j}$ , which we can recognize as the power series of $\small \frac{1}{1-x}$ with $\small x=-\frac{1}{2}$ , which is

$\small \small \small \sum_{j=0}^\infty \left(\frac{-1}{2}\right)^{j}=\left( \frac{1}{1-(-1/2)}\right)=\left(\frac{1}{1.5} \right )=\frac{2}{3}$ .

So we have

$\small \small \sum_{j=2}^\infty (-1)^j\left(\frac{1}{2}\right)^{j+1}=\frac{1}{8}\cdot \frac{2}{3}=\frac{1}{12}$ .

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Example Question #4 : Types Of Series

Determine whether the series is arithmetic. If so, find the common difference.

1+8+27+64+125+216+...

Possible Answers:

Series is not arithmetic

Correct answer:

Series is not arithmetic

Explanation:

If a series is arithmetic, then there exists a common difference between each pair of consecutive terms in the series.

For this series

$a_1=1,\, a_2=8,\,a_3=27,\,a_4=64$

Because

a_2-a_1=8-1=7

and

a_3-a_2=27-8=19

we find that there does NOT exist a common difference and as such,

the series is not arithmetic.

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Example Question #5 : Types Of Series

Determine whether or not the geometric series converges. If it converges, find the sum of the sequence.

$1+\pi+\pi^2 +\pi^3 +\pi^4 +\pi^5+...$

Possible Answers:

Series does not converge.

$\pi^{100}$

$100000\pi$

Correct answer:

Series does not converge.

Explanation:

To determine the convergency of a geometric series, we must find the absolute value of the common ratio.

A geometric series will converge if the absolute value of the common ratio is less than one, or

|r|<1

In this problem, we see that

$r=\pi \approx 3.14$

And because $|\pi|>1$

we conclude that the series does not converge to a finite sum.

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Example Question #1 : Arithmetic And Geometric Series

Calculate the sum of the following series: $\sum^{500}_{k=1}2k+3$

Possible Answers:

234000

240000

252000

250000

Correct answer:

252000

Explanation:

This is an arithmetic series.

Its general form is $\sum^{n}_{k=1}a_k$ .

To calculate the sum of very large series such as these, use the formula $\frac{n}{2}(a_1 + a_n)$

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

Solution:

$a_{1} = 2*1 + 3 = 5$

$a_{500} = 2 * 500 +3 = 1000 + 3 = 1003$

n =500

$\sum^{500}_{k=1}2k+3= \frac{n}{2}(a_{1}+a_{500}) = \frac{500}{2}(5+1003)=250*1008=252000$

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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 : Types Of Series

Example Question #2 : Types Of Series

Example Question #1 : Types Of Series

Example Question #2962 : Calculus Ii

Example Question #3 : Types Of Series

Example Question #1 : Arithmetic And Geometric Series

Example Question #2964 : Calculus Ii

Example Question #4 : Types Of Series

Example Question #5 : Types Of Series

Example Question #1 : Arithmetic And Geometric Series

All Calculus 2 Resources

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