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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
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 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

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 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
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 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
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 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

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17calculus > exam list > infinite series exam A

This page contains a complete infinite series exam with worked out solutions.

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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

1 hour

Questions

13

Total Points

100

Tools

Calculator

not allowed

Formula Sheet(s)

table of series tests

Other Tools

none

Instructions:
- Show all your work.
- For each problem, correct answers are worth 10%. 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).
- Give exact, simplified answers.

Section 1

3 questions - 20 points total

Do the following sequences $$\{ {a_n} \}$$ converge or diverge as $$n \to \infty$$? If the sequence converges, find it's limit. Justify your answers.

$$a_n=2+(-1)^n$$

Problem Statement

$$a_n=2+(-1)^n$$

The sequence $$\left\{ 2 + (-1)^n \right\}$$ diverges.

Problem Statement

$$a_n=2+(-1)^n$$

Solution

The sequence diverges since it oscillates between 1 and 3.
For example, if $$\epsilon=1$$, there is no number $$L$$ such that $$\abs{a_n-L} < \epsilon$$ for all sufficiently large $$n$$, since then we would have both $$\abs{1-L} < 1$$ (or $$0 < L < 2$$ ) and $$\abs{3-L} < 1$$ or $$(2 < L < 4)$$, which is impossible. So there is no $$L$$ that satisfies the definition of the limit.

The sequence $$\left\{ 2 + (-1)^n \right\}$$ diverges.

$$\displaystyle{ a_n = \frac{n}{e^n} }$$

Problem Statement

$$\displaystyle{ a_n = \frac{n}{e^n} }$$

The sequence $$\displaystyle{ \left\{ \frac{n}{e^n} \right\} }$$ converges to zero.

Problem Statement

$$\displaystyle{ a_n = \frac{n}{e^n} }$$

Solution

Looking at the limit $$\displaystyle{ \lim_{n \to \infty}{\frac{n}{e^n}} }$$, direct substitution yields $$\infty / \infty$$ which is indeterminate. So we can use L'Hôpital's Rule.

$$\displaystyle{ \lim_{n \to \infty}{\frac{n}{e^n}} = \lim_{n \to \infty}{\frac{1}{e^n}} = \frac{1}{\infty} = 0 }$$

Therefore, the sequence converges to zero.

The sequence $$\displaystyle{ \left\{ \frac{n}{e^n} \right\} }$$ converges to zero.

$$\displaystyle{ a_n = \left( 1 + \frac{2}{n} \right)^n }$$

Problem Statement

$$\displaystyle{ a_n = \left( 1 + \frac{2}{n} \right)^n }$$

The sequence $$\displaystyle{ \left\{ \left( 1 + \frac{2}{n} \right)^n \right\} }$$ converges to $$e^2$$.

Problem Statement

$$\displaystyle{ a_n = \left( 1 + \frac{2}{n} \right)^n }$$

Solution

Direct substitution yields $$\infty ^ {\infty}$$ which is indeterminate. We don't have a fraction, so we can't use L'Hôpital's Rule without doing some algebra.

 $$y = \displaystyle{ \lim_{n \to \infty}{ \left(1 + \frac{2}{n}\right)^n } }$$ Take the natural log of both sides. $$\ln(y) = \displaystyle{ \lim_{n \to \infty}{\ln \left(1 + \frac{2}{n}\right)^n }}$$ $$\ln(y) = \displaystyle{ \lim_{n \to \infty}{(n) \ln \left(1 + \frac{2}{n}\right) }}$$ $$\ln(y) = \displaystyle{\lim_{n \to \infty}{ \frac{\ln \left(1 + \frac{2}{n}\right)}{1/n} }}$$ $$\ln(y) = \displaystyle{\lim_{n \to \infty}{ \frac{\ln [(n+2)/n]}{n^{-1}} }}$$ Use L'Hôpital's Rule. $$\ln(y) = \displaystyle{\lim_{n \to \infty}{ \left[ \frac{-2n^{-2}(n)}{n+2} \cdot \frac{1}{-n^{-2}} \right] }}$$ $$\ln(y) = \displaystyle{\lim_{n \to \infty}{ \frac{2n}{n+2} }}$$ Use L'Hôpital's Rule again. $$\ln(y) = \displaystyle{\lim_{n \to \infty}{\frac{2}{1}}}$$ $$\ln(y) = 2$$ Undo the natural log by raising each side to the exponent of e. $$e^{\ln(y)} = e^2$$ $$y = e^2$$

The sequence $$\displaystyle{ \left\{ \left( 1 + \frac{2}{n} \right)^n \right\} }$$ converges to $$e^2$$.

Section 2

4 questions - 20 points total

Determine whether the following series converge or diverge. You may use any appropriate test provided you explain your answer.

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

Problem Statement

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

The series $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^3}{3^n}} }$$ converges by the Ratio Test.

Problem Statement

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

Solution

Let's go down through the infinite series table in order.
Let $$\displaystyle{ a_n = \frac{n^3}{3^n} }$$
Group 1: Divergence Test: Using L'Hôpital's Rule repeatedly, we can eventually get to the point where we can see that the limit $$\displaystyle{ \lim_{n \to \infty}{a_n} = 0 }$$, so the test is inconclusive.
Group 2: We don't have one of the special series, p-Series, Geometric Series, Alternating Series or Telescoping Series.
Group 3: So, let's try the Ratio Test.
We need to set up the limit $$\displaystyle{ \lim_{n \to \infty}{\frac{a_{n+1}}{a_n}} }$$.
Since $$\displaystyle{ a_n = \frac{n^3}{3^n} }$$, $$\displaystyle{ a_{n+1} = \frac{(n+1)^3}{3^{n+1}} }$$.

 $$\displaystyle{\lim_{n \to \infty}{\frac{a_{n+1}}{a_n}}}$$ $$\displaystyle{\lim_{n \to \infty}{ \frac{(n+1)^3}{3^{n+1}} \cdot \frac{3^n}{n^3} }}$$ $$\displaystyle{\lim_{n \to \infty}{ \frac{(n+1)^3}{n^3} \cdot \frac{3^n}{3^{n+1}} }}$$ $$\displaystyle{\lim_{n \to \infty}{ \left( \frac{n+1}{n} \right)^3 \cdot \frac{1}{3} }}$$ $$\displaystyle{\frac{1}{3} \lim_{n \to \infty}{ \left( 1 + \frac{1}{n} \right)^3 }}$$ $$\displaystyle{\frac{1}{3} (1 + 0)^3 = \frac{1}{3}}$$

Since $$\displaystyle{ \lim_{n \to \infty}{\frac{a_{n+1}}{a_n}} = 1/3 < 1 }$$ the series converges.

The series $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^3}{3^n}} }$$ converges by the Ratio Test.

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

Problem Statement

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

The series $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n}{n^2+1} } }$$ diverges by the Limit Comparison Test.

Problem Statement

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

Solution

Let's go down through the infinite series tests in the order they appear in the infinite series table.
Group 1: Divergence Test: Using L'Hôpital's Rule, the limit goes to zero, so the test is inconclusive.
Group 2: This series is not one of the special series.
Group 3: Start with the Ratio Test. Looking at the series, our coefficients and constants are all one, so we might think that the limit would equal one, which is correct.
Next, we can try a comparison test. But what do we compare this series to? Well, as n gets very large, the $$1$$ in the denominator becomes almost negligible compared to the $$n^2$$. So let's compare this series to $$\displaystyle{ \sum{\frac{n}{n^2}} = \sum{\frac{1}{n}} }$$, which we know is a p-Series with $$p=1$$, so it diverges.
We can use either the Limit Comparison Test or the Direct Comparison Test. Let's try the Limit Comparison Test.
We need to set up the limit $$\displaystyle{ \lim_{n \to \infty}{\frac{a_n}{t_n}} }$$ where $$\displaystyle{ a_n = \frac{n}{n^2+1} }$$ and $$\displaystyle{ t_n = \frac{1}{n} }$$ which is our divergent test series. If this limit is finite and positive then the original series will also diverge. Let's see what happens.
$$\begin{array}{rcl} \displaystyle{\lim_{n \to \infty}{\frac{a_n}{t_n}}} & = & \displaystyle{\lim_{n \to \infty}{ \frac{n}{n^2+1} \cdot \frac{n}{1} }} \\ & = & \displaystyle{\lim_{n \to \infty}{ \frac{n^2}{n^2 + 1} }} \\ & = & \displaystyle{\lim_{n \to \infty}{ \frac{2n}{2n} } = 1} \end{array}$$
In the second step, we used L'Hôpital's Rule.
Since the limit is finite and positive and the test series diverges, the original series also diverges.

The series $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n}{n^2+1} } }$$ diverges by the Limit Comparison Test.

$$\displaystyle{ \sum_{n=1}^{\infty}{n \sin\left( \frac{1}{n} \right) } }$$

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{n \sin\left( \frac{1}{n} \right) } }$$

The series $$\displaystyle{ \sum_{n=1}^{\infty}{n \cdot \sin\left( \frac{1}{n} \right) } }$$ diverges by the Divergence Test.

Problem Statement

$$\displaystyle{ \sum_{n=1}^{\infty}{n \sin\left( \frac{1}{n} \right) } }$$

Solution

There are several ways to work this one. The most straight-forward is to go through the infinite series table in order.
Group 1 - Divergence Test: Evaluate the limit $$\displaystyle{ \lim_{n \to \infty}{ n \cdot \sin\left( \frac{1}{n} \right) } }$$. Direct substitution gives us $$\infty \cdot 0$$ which is indeterminate. So we can use L'Hôpital's Rule if we get the limit in a fraction.

 $$\displaystyle{\lim_{n \to \infty}{ n \cdot \sin\left( \frac{1}{n} \right) } }$$ $$\displaystyle{\lim_{n \to \infty}{ \frac{\sin(1/n)}{1/n} }}$$ $$\displaystyle{\lim_{n \to \infty}{ \frac{\sin(n^{-1})}{n^{-1}} }}$$ $$\displaystyle{\lim_{n \to \infty}{ \frac{\cos(n^{-1}) (-1)(n^{-2})}{(-1)n^{-2}} }}$$ $$\displaystyle{\lim_{n \to \infty}{ \cos(n^{-1}) } = \cos(0) = 1}$$

So, since the limit does not equal zero, the series diverges.
Note: Another way to evaluate this limit is to substitute $$x = 1/n$$ and as $$n \to \infty$$, $$x \to 0$$. So the limit is $$\displaystyle{ \lim_{n \to \infty}{ n \cdot \sin\left( \frac{1}{n} \right) } = \lim_{n \to \infty}{ \frac{\sin\left( \frac{1}{n} \right)}{1/n} } = \lim_{x \to 0}{\frac{\sin(x)}{x}} }$$
The last limit is a special trig limit which is equal to one. If you didn't remember this, you could have used L'Hôpital's Rule again here and arrived at the same answer.

The series $$\displaystyle{ \sum_{n=1}^{\infty}{n \cdot \sin\left( \frac{1}{n} \right) } }$$ diverges by the Divergence Test.

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

Problem Statement

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

The Telescoping Series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}\right) } }$$ converges.

Problem Statement

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

Solution

Without too much work, this series is crying out Telescoping Series. Let's set up a table to see if we can find a pattern.
$$\begin{array}{ccc} n & & a_n \\ 1 & & \displaystyle{\frac{1}{1} - \frac{1}{\sqrt{2}}} \\ 2 & & \displaystyle{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}} \\ 3 & & \displaystyle{\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}}} \\ 4 & & \displaystyle{\frac{1}{\sqrt{4}} - \frac{1}{\sqrt{5}}} \end{array}$$
With only 4 terms, we can see that the last term of each line cancels with the first term in the next line.
This gives us $$\displaystyle{ 1 - \lim_{n \to \infty}{\frac{1}{\sqrt{n+1}}} = 1 }$$
So the series converges (specifically to 1).

topics

Telescoping Series

The Telescoping Series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}\right) } }$$ converges.

Section 3

2 questions - 20 points total

Determine whether the following series converge or diverge and justify your answer.

$$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n \ln(n)} } }$$

Problem Statement

$$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n \ln(n)} } }$$

The series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n~\ln(n)} } }$$ diverges by the Integral Test.

Problem Statement

$$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n \ln(n)} } }$$

Solution

Going through the infinite series table in order will help us determine what rule to use here.
Group 1: Divergence Test: The limit here is obviously zero, so this test is inconclusive.
Group 2: This series is not one of the special series.
Group 3: From experience you probably know that, using the Ratio Test or a comparison test, would yield a very difficult limit or inequality to evaluate due to the $$\ln(n)$$ term. So, let's try the Integral Test.
We have $$\displaystyle{ f(n) = \frac{1}{n~\ln(n)} }$$ which is a discrete function. To use the integral test, we need a continuous function. So let $$\displaystyle{ f(x) = \frac{1}{x~\ln(x)} }$$.
The function $$f(x)$$ is continuous and positive on $$2 \leq x < \infty$$. It is also decreasing on this interval since the denominator is increasing and the numerator is constant. Therefore, by the integral test, the series converges or diverges with the integral $$\displaystyle{ \int_{2}^{\infty}{\frac{1}{x~\ln(x)} dx } }$$. $$\displaystyle{ \int_{2}^{\infty}{\frac{1}{x~\ln(x)} dx } = \lim_{b \to \infty}{ \int_{2}^{b}{\frac{1}{x~\ln(x)} dx} }}$$
Using the technique of substitution, we let $$u = \ln(x) \to du = dx/x$$.
$$\displaystyle{ \int{\frac{1}{x~\ln(x)} dx} = }$$ $$\displaystyle{ \int{\frac{1}{u} du} = \ln(u) = \ln( \ln(x) ) }$$

$$\displaystyle{\lim_{b \to \infty}{ \int_{2}^{b}{\frac{1}{x~\ln(x)} dx} } = }$$ $$\displaystyle{\lim_{b \to \infty}{ \left[ \ln( \ln(x) ) \right]_2^b } = }$$ $$\displaystyle{\lim_{b \to \infty}{ [ \ln(\ln(b)) - \ln(\ln(2)) ] } = }$$ $$\displaystyle{ \infty - \ln(\ln(2)) = \infty }$$

Since the integral diverges, the series also diverges.

infinite series table

Divergence Test

Integral Test

The series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n~\ln(n)} } }$$ diverges by the Integral Test.

$$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n \ln(n)}{n} } }$$

Problem Statement

$$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n \ln(n)}{n} } }$$

The series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n \ln(n)}{n} } }$$ converges by the Alternating Series Test.

Problem Statement

$$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n \ln(n)}{n} } }$$

Solution

We don't really need to go through the tests in the infinite series table since we can see that this is an alternating series. So, let's try the Alternating Series Test.
For convergence, we need to verify that both of the following conditions hold.
Condition 1: $$\displaystyle{ \lim_{n \to \infty}{a_n} = 0 }$$
Condition 2: $$0 < a_{n+1} < a_n$$
Let's test condition 1.
$$\displaystyle{\lim_{n \to \infty}{a_n} = \lim_{n \to \infty}{\frac{\ln(x)}{x}} = \lim_{n \to \infty}{\frac{1/x}{1}} = 0 }$$
We used L'Hôpital's Rule to show that condition 1 holds. Now, let's test condition 2.
$$\displaystyle{ a_n = \frac{\ln(n)}{n} \text{ and } a_{n+1} = \frac{\ln(n+1)}{n+1} }$$

$$\displaystyle{ 0 < a_{n+1} < a_n \to 0 < \frac{\ln(n+1)}{n+1} < \frac{\ln(n)}{n} }$$
Since $$\ln(n)$$ and $$n$$ are both positive, we can drop the $$0 <$$ part of the inequality.
However, it is difficult to establish the inequality $$\displaystyle{ \frac{\ln(n+1)}{n+1} < \frac{\ln(n)}{n} }$$
So, we need to use another technique. The idea of the inequality is that the terms are decreasing. We can set up a continuous function $$\displaystyle{ f(x) = \frac{\ln(x)}{x} }$$ and show that it is decreasing, i.e. the slope is negative. Let's calculate the derivative and see if it works.
$$\displaystyle{ f'(x) = \frac{x \cdot 1/x - 1 (\ln(x))}{x^2} = \frac{1 - \ln(x)}{x^2} }$$
We used the Quotient Rule here.
Looking at the derivative, the denominator $$x^2$$ is always positive, so the numerator will determine the sign. We want the slope to be negative, so $$1 - \ln(x) < 0$$ or $$1 < \ln(x)$$. This inequality holds for $$x > e$$. So as long as $$n \geq 3$$, the terms are decreasing. Therefore, condition 2 holds and so the alternating series converges.

infinite series table

Alternating Series Test

L'Hôpital's Rule

Derivative Quotient Rule

The series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(-1)^n \ln(n)}{n} } }$$ converges by the Alternating Series Test.

Section 4

4 questions - 40 points total

Does this series diverge, converge conditionally or converge absolutely? Justify your answer.
$A = 1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \frac{1}{7^2} - . . .$

Problem Statement

Does this series diverge, converge conditionally or converge absolutely? Justify your answer.
$A = 1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \frac{1}{7^2} - . . .$

The series $$\displaystyle{ A = 1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \frac{1}{7^2} - . . . = \sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{n^2}} }$$ converges absolutely.

Problem Statement

Does this series diverge, converge conditionally or converge absolutely? Justify your answer.
$A = 1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \frac{1}{7^2} - . . .$

Solution

First let's get a closed form for the series. Since this is an alternating series, we need a term that looks like $$(-1)^n$$. And looking at the denominator, it looks like $$n^2$$. We also need to determine where to start $$n$$, $$0$$ or $$1$$. If we start with zero, we will get a zero in the denominator, which is not good. So we want to start at $$n = 1$$. Since the first term is positive, the power on the $$-1$$ term needs to be $$n + 1$$. So our series is $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{n^2}} }$$.
We know that absolute value of the alternating series, $$\displaystyle{ \sum{\frac{1}{n^2}} }$$, is a convergent p-Series. So the Alternating Series converges absolutely.

p-Series

Alternating Series

absolute convergence

The series $$\displaystyle{ A = 1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \frac{1}{7^2} - . . . = \sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{n^2}} }$$ converges absolutely.

Let $$\displaystyle{S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + }$$ $$\displaystyle{ \frac{1}{4^2} + \frac{1}{5^2} + }$$ $$\displaystyle{\frac{1}{6^2} + \frac{1}{7^2} + . . . }$$
Given that $$S = \pi^2/6$$, find the sum A given in the previous question.
(Hint: Consider $$S - A$$ and express it in terms of $$S$$.)

Problem Statement

Let $$\displaystyle{S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + }$$ $$\displaystyle{ \frac{1}{4^2} + \frac{1}{5^2} + }$$ $$\displaystyle{\frac{1}{6^2} + \frac{1}{7^2} + . . . }$$
Given that $$S = \pi^2/6$$, find the sum A given in the previous question.
(Hint: Consider $$S - A$$ and express it in terms of $$S$$.)

$$\displaystyle{ A = \frac{\pi^2}{12} }$$

Problem Statement

Let $$\displaystyle{S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + }$$ $$\displaystyle{ \frac{1}{4^2} + \frac{1}{5^2} + }$$ $$\displaystyle{\frac{1}{6^2} + \frac{1}{7^2} + . . . }$$
Given that $$S = \pi^2/6$$, find the sum A given in the previous question.
(Hint: Consider $$S - A$$ and express it in terms of $$S$$.)

Solution

By the linearity of limits and sums, if the series $$\sum{a_n}$$ and $$\sum{b_n}$$ converge, then $$\sum{(a_n - b_n) }$$ also converges and $$\sum{(a_n - b_n) } = \sum{a_n} - \sum{b_n}$$. Also, for any constant $$c$$, we have $$\sum{ca_n} = c \sum{a_n}$$.
Since the series for $$S$$ and $$A$$ both converge, so we can form $$S-A$$.

 Follow the hint and form $$S-A$$. $$\displaystyle{\left( 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + . . . \right) - \left( 1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - . . . \right)}$$ Pair up terms that match to see what cancels. $$\displaystyle{(1 - 1) + \left( \frac{1}{2^2} + \frac{1}{2^2} \right) + \left( \frac{1}{3^2} - \frac{1}{3^2} \right) + \left( \frac{1}{4^2} + \frac{1}{4^2} \right) + \left( \frac{1}{5^2} - \frac{1}{5^2} \right) + . . . }$$ $$\displaystyle{2 \left( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \frac{1}{8^2} + . . . \right)}$$ Factor $$\displaystyle{\frac{2}{2^2} \left( 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + . . . \right)}$$ Notice that the term on the right is $$S$$. $$\displaystyle{\frac{1}{2}S}$$

So now we have $$S - A = S/2$$ and solving for $$A$$, we get $$A = S/2$$. Since $$S = \pi^2/6$$, it follows that $$A = \pi^2 / 12$$.

$$\displaystyle{ A = \frac{\pi^2}{12} }$$

State the definition for a sequence $$\{ a_n \}$$ to converge to a limit $$L$$ as $$n \to \infty$$.

Problem Statement

State the definition for a sequence $$\{ a_n \}$$ to converge to a limit $$L$$ as $$n \to \infty$$.

Solution

A sequence $$\{ a_n \}$$ converges to a limit $$L$$ if, for every $$\epsilon > 0$$ there exists a number $$N$$ such that $| a_n - L | < \epsilon ~~~~~ \text{ for every } n > N$

If $$\displaystyle{ a_n = \frac{1}{\sqrt{n}} }$$ for $$n = 1, 2, 3, . . .$$ prove from the definition that $$\displaystyle{ \lim_{n \to \infty}{a_n} = 0 }$$.

Problem Statement

If $$\displaystyle{ a_n = \frac{1}{\sqrt{n}} }$$ for $$n = 1, 2, 3, . . .$$ prove from the definition that $$\displaystyle{ \lim_{n \to \infty}{a_n} = 0 }$$.

Solution

Given $$\epsilon > 0$$, let $$N = 1/\epsilon^2$$. Then if $$n > N$$, we have
$$\begin{array}{rcl} \displaystyle{\left| \frac{1}{\sqrt{n}} - 0 \right|} & = & \displaystyle{\frac{1}{\sqrt{n}}} \\ & < & \displaystyle{\frac{1}{\sqrt{N}}} \\ & < & \epsilon \end{array}$$
Since $$1/\sqrt{N} = \epsilon$$, it follows that $$\displaystyle{ \lim_{n \to \infty}{\frac{1}{\sqrt{n}}} = 0 }$$.