You CAN Ace Calculus

 Infinite Series

other exams from this semester

exam C1 - exam C2 - exam C3 - exam C4

### 17Calculus Subjects Listed Alphabetically

Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

Multi-Variable Calculus

 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

### Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

This is the third of four exams for semester C of calculus 2.

### Practice Exam Tips

Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.

- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.
- Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
- Use your calculator as little as possible except for graphing and checking your calculations.
- Work the entire exam before checking any solutions.
- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.

IMPORTANT -
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

Exam Details

Time

2.5 hours

Questions

9

Total Points

70

Tools

Calculator

no

Formula Sheet(s)

1 page, 8.5x11 or A4

Other Tools

none

General Instructions:
- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).

Specific Instructions:
- Unless otherwise instructed, determine the convergence or divergence of these series.
- If a series with positive and negative terms converges, state whether it converges conditionally or absolutely and show work supporting your answer.
- If possible, find the value that the series converges to.
- You may use any test on this site unless told to use a specific test.
- You may not use the idea of growth rates as your answer when evaluating limits.

[5 points] You must successfully use each of these tests at least once to solve the rest of the problems on this exam. By successfully we mean that the test proved either convergence or divergence. It is not successful if the test was inconclusive. You must use all these tests to get any points on this question, i.e. the only two scores possible on this question are 0 and 5.

 Integral Test Ratio Test Root Test Divergence Test Direct Comparison Test Limit Comparison Test

Problem Statement

[5 points] You must successfully use each of these tests at least once to solve the rest of the problems on this exam. By successfully we mean that the test proved either convergence or divergence. It is not successful if the test was inconclusive. You must use all these tests to get any points on this question, i.e. the only two scores possible on this question are 0 and 5.

 Integral Test Ratio Test Root Test Divergence Test Direct Comparison Test Limit Comparison Test

Solution

Here are the problems we used the tests on. Yours may be different.

 Integral Test [question 2] Ratio Test [question 8] Root Test [question 5] Divergence Test [question 3] Direct Comparison Test [question 4] Limit Comparison Test [question 4]

[15 points] $$\displaystyle{ \sum{ne^{-2n^2}} }$$ (use the integral test)

Problem Statement

[15 points] $$\displaystyle{ \sum{ne^{-2n^2}} }$$ (use the integral test)

The integral converges by the integral test.

Problem Statement

[15 points] $$\displaystyle{ \sum{ne^{-2n^2}} }$$ (use the integral test)

Solution

In order to use the integral test, we need a continuous function. The function $$f(x)=xe^{-2x^2}$$ is a continuous function with the same values as the same points as the sum. So we can use it to integrate.
The integral test requires that we integrate $$\int_k^{\infty}{f(x)~dx}$$. We need to determine a value for k where the function is decreasing and $$f(x)$$ is positive.
$$f(x)$$ is positive when x is positive and it is decreasing. Now we need to find the largest critical value so that the function is decreasing.

 $$\displaystyle{ f(x) = \frac{x}{e^{2x^2}} }$$ $$\displaystyle{ f'(x) = \frac{e^{2x^2}(1)-xe^{2x^2}(4x)}{e^{4x^2}} }$$ $$\displaystyle{ f'(x) = \frac{1-4x^2}{e^{2x^2}} }$$

The derivative is zero when the numerator is zero and the denominator is non-zero.
$$1-4x^2=0 \to 4x^2=1 \to x^2 = 1/4 \to x = \pm1/2$$
The largest critical value is $$1/2$$. We need to make sure that the function is decreasing above this critical value. Choose $$x=1$$ as a test value.
$$\displaystyle{ f'(1) = \frac{-3}{e^2} \lt 0 }$$ so the function is decreasing and $$f(x)$$ is positive.
We choose to integrate starting at 1. You can choose any real number greater than 1/2.

 $$\displaystyle{ \int_1^{\infty}{ x e^{-2x^2} ~dx } }$$ $$\displaystyle{ \lim_{b\to\infty}{ \int_1^{b}{ x e^{-2x^2} ~dx } } }$$

We will drop the limits of integration and integrate the indefinite integral.

 $$\displaystyle{ \int{ x e^{-2x^2} ~dx } }$$ Use integration by substitution. $$u = -2x^2 \to du = -4x~dx \to -du/4 = x~dx$$ $$\displaystyle{ \int{ e^u \frac{-du}{4} } }$$ $$\displaystyle{ \frac{-1}{4} e^u }$$ $$\displaystyle{ \frac{-1}{4} e^{-2x^2} }$$ Now apply the limits. $$\displaystyle{ \lim_{b\to\infty}{ \left[ \frac{-1}{4} e^{-2x^2} \right]_1^b } }$$ $$\displaystyle{ \lim_{b\to\infty}{ \left[ \frac{-1}{4} e^{-2b^2} - \frac{-1}{4}e^{-2} \right] } }$$ $$\displaystyle{ \frac{1}{4e^2} }$$

Since the limit is finite, the integral converges, therefore the series converges by the integral test.

The integral converges by the integral test.

[5 points] $$\displaystyle{ 3 + \frac{5}{2} + \frac{7}{3} + \frac{9}{4} + \ldots }$$

Problem Statement

[5 points] $$\displaystyle{ 3 + \frac{5}{2} + \frac{7}{3} + \frac{9}{4} + \ldots }$$

The series diverges by the divergence test.

Problem Statement

[5 points] $$\displaystyle{ 3 + \frac{5}{2} + \frac{7}{3} + \frac{9}{4} + \ldots }$$

Solution

 $$\displaystyle{ 3 + \frac{5}{2} + \frac{7}{3} + \frac{9}{4} + \ldots }$$ $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2n+1}{n} } }$$ Let's try the divergence test. $$\displaystyle{ \lim_{n\to\infty}{ \frac{2n+1}{n} } }$$ $$\displaystyle{ \lim_{n\to\infty}{ \left[ 2 + \frac{1}{n} \right] } = 2 }$$

Since the limit is not zero, the series diverges.

The series diverges by the divergence test.

[5 points] $$\displaystyle{ \sum{ \frac{2n+1}{n^2+n} } }$$

Problem Statement

[5 points] $$\displaystyle{ \sum{ \frac{2n+1}{n^2+n} } }$$

The series diverges by the direct comparison test.

Problem Statement

[5 points] $$\displaystyle{ \sum{ \frac{2n+1}{n^2+n} } }$$

Solution

Compare with the divergent p-series $$\sum{ 1/n }$$. If we assume the original series also diverges, we can use the direct comparison test and try to show that $$t_n \lt a_n$$.

 $$\displaystyle{ t_n = \frac{1}{n} }$$ and $$\displaystyle{ a_n = \frac{2n+1}{n^2+n} }$$ $$t_n \lt a_n$$ $$\displaystyle{ \frac{1}{n} \lt \frac{2n+1}{n^2+n} }$$ $$\displaystyle{ \frac{n^2+n}{n} \lt 2n+1 }$$ $$\displaystyle{ n+1 \lt 2n+1 }$$ $$\displaystyle{ n \lt 2n }$$ $$\displaystyle{ 1 \lt 2 }$$

The last inequality $$1 \lt 2$$ holds for all n. So the series diverges by the direct comparison test with the divergent p-series $$\sum{1/n}$$.
After we finish all the problems, we realize that we have not used the limit comparison test. The instructions do not state that we can use only one test per problem. So for this problem, we will prove it again using the limit comparison test.
We will use the same test series $$t_n=1/n$$.

 $$\displaystyle{ \lim_{n\to\infty}{ \frac{a^n}{t^n} } }$$ $$\displaystyle{ \lim_{n\to\infty}{ \frac{2n+1}{n^2+n} \frac{n}{1} } }$$ $$\displaystyle{ \lim_{n\to\infty}{ \frac{2n^2+n}{n^2+n} } = 2 }$$

Since the limit is greater than zero, the series diverges by the limit comparison test.

The series diverges by the direct comparison test.

[5 points] $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{2^{n+1}}{n^n} } }$$

Problem Statement

[5 points] $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{2^{n+1}}{n^n} } }$$

The series converges by the root test.

Problem Statement

[5 points] $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{2^{n+1}}{n^n} } }$$

Solution

Rewrite the sum $$\displaystyle{ 2 \sum_{n=2}^{\infty}{ \frac{2^{n}}{n^n} } }$$.
Try the root test.

 $$\displaystyle{ \lim_{n\to\infty}{ \sqrt[n]{ |a_n| } } }$$ $$\displaystyle{ \lim_{n\to\infty}{ \sqrt[n]{ \frac{2^n}{n^n} } } }$$ $$\displaystyle{ \lim_{n\to\infty}{ \frac{2}{n} } = 0 }$$

Since the limit is less than 1, the series converges by the root test.

The series converges by the root test.

[10 points] $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2}{e^n}} }$$

Problem Statement

[10 points] $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2}{e^n}} }$$

The series converges by the Ratio Test.

Problem Statement

[10 points] $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2}{e^n}} }$$

Solution

If we evaluate the limit $$\displaystyle{ \lim_{n\to\infty}{a_n} }$$, we get zero. So the divergence test won't help. Let's try the ratio test.

 $$\displaystyle{ a_n = \frac{n^2}{e^n} }$$ and $$\displaystyle{ a_{n+1} = \frac{(n+1)^2}{e^{n+1}} }$$ For the ratio test, we need to evaluate $$\displaystyle{ \lim_{n\to\infty}\left| \frac{a_{n+1}}{a_n} \right| }$$ $$\displaystyle{ \lim_{n\to\infty}\left| \frac{(n+1)^2}{e^{n+1}} \frac{e^n}{n^2} \right| }$$ $$\displaystyle{ \lim_{n\to\infty}\left| \frac{(n+1)^2}{n^2} \frac{e^n}{e^{n+1}} \right| }$$ $$\displaystyle{ \lim_{n\to\infty}\left| \left[ 1+\frac{1}{n} \right]^2 \cdot \frac{1}{e} \right| = \frac{1}{e} < 1 }$$

Since the limit is less than one, the series converges.

The series converges by the Ratio Test.

[10 points] $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(n+1)} } }$$

Problem Statement

[10 points] $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(n+1)} } }$$

This is a telescoping series that converges to 1/2.

Problem Statement

[10 points] $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n(n+1)} } }$$

Solution

Comparing this to the series $$\sum{1/n^2}$$ using the direct comparision test or limit comparison test, we can show that this series converges. However, this looks a lot like a telescoping series, in which case we can not only determine that it converges but also what it converges to. Let's try it.

 Use partial fractions to rewrite $$a_n$$. $$\displaystyle{ \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1} }$$ Solving for A and B, gives us $$\displaystyle{ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} }$$

This looks promising. Let's build a table of the first few terms to see if we can see a pattern.

$$n$$

$$a_n$$

2

$$\displaystyle{ \frac{1}{2} - \frac{1}{3} }$$

3

$$\displaystyle{ \frac{1}{3} - \frac{1}{4} }$$

4

$$\displaystyle{ \frac{1}{4} - \frac{1}{5} }$$

n

$$\displaystyle{ \frac{1}{n} - \frac{1}{n+1} }$$

Okay, so notice that the nth sum is $$\displaystyle{ S_n = \frac{1}{2} - \frac{1}{n+1} }$$. If we take the limit as n goes to infinity, this will tell us what the series converges to.
$$\displaystyle{ \lim_{n\to\infty}{S_n} = \lim_{n\to\infty}{ \left[ \frac{1}{2} - \frac{1}{n+1} \right] } = \frac{1}{2} }$$

This is a telescoping series that converges to 1/2.

[10 points] $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}11^n}{n!} } }$$

Problem Statement

[10 points] $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}11^n}{n!} } }$$

The series converges absolutely by the ratio test.

Problem Statement

[10 points] $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}11^n}{n!} } }$$

Solution

Because of the factorial, the ratio test will work well.

 $$\displaystyle{ a_n = \frac{11^n}{n!} }$$ and $$\displaystyle{ a_n = \frac{11^{n+1}}{(n+1)!} }$$ $$\displaystyle{ \lim_{n\to\infty}{ \left| \frac{a_{n+1}}{a_n} \right| } }$$ $$\displaystyle{ \lim_{n\to\infty}{ \left| \frac{11^n}{n!} \frac{(n+1)!} {11^{n+1}}\right| } }$$ $$\displaystyle{ \lim_{n\to\infty}{ \left| \frac{11}{n+1} \right| } = 0 < 1 }$$

Since the limit is less than one, the series converges.
In this case, we showed that $$\sum|a_n|$$ converges. So series $$\sum(-1)^{n+1}a_n$$ also converges and by the absolute convergence theorem, it converges absolutely.

The series converges absolutely by the ratio test.

[5 points] $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n+2}} } }$$

Problem Statement

[5 points] $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n+2}} } }$$

This is a geometric series that converges to $$\displaystyle{ \frac{2e^{-2}}{e-2} }$$.

Problem Statement

[5 points] $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n+2}} } }$$

Solution

Let's rewrite this.

 $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n+2}} } }$$ $$\displaystyle{ \frac{1}{e^2} \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n}} } }$$ $$\displaystyle{ \frac{1}{e^2} \sum_{n=1}^{\infty}{ \left[ \frac{2}{e} \right]^n } }$$

This looks a lot like a geometric series. We know that $$\displaystyle{ \sum_{n=0}^{\infty}{ r^n } = \frac{1}{1-r} }$$. Our sum starts at 1, so let's rewrite this as $$\displaystyle{ r^0 + \sum_{n=1}^{\infty}{ r^n } = \frac{1}{1-r} \to }$$ $$\displaystyle{ \sum_{n=1}^{\infty}{ r^n } = \frac{1}{1-r} - 1 }$$
In our problem $$r = 2/e < 1$$, so $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^n}{e^{n+2}} } = \frac{1}{e^2}\left[ \frac{1}{1-2/e} - 1 \right] }$$
Simplifying, we get $$\displaystyle{ \frac{1}{e^2}\left[ \frac{2}{e-2} \right] = \frac{2e^{-2}}{e-2} }$$

This is a geometric series that converges to $$\displaystyle{ \frac{2e^{-2}}{e-2} }$$.