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

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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.

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IMPORTANT -
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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

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

Final Answer

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.

Final Answer

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

close solution

\(\displaystyle{ a_n = \frac{n}{e^n} }\)

Problem Statement

\(\displaystyle{ a_n = \frac{n}{e^n} }\)

Final Answer

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.

Final Answer

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

close solution

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

Problem Statement

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

Final Answer

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.

\( e^{\ln(y)} = e^2 \)

\( y = e^2 \)

Final Answer

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

close solution

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

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

Final Answer

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.

Final Answer

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

close solution

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

Problem Statement

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

Final Answer

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.

Final Answer

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

close solution

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

Problem Statement

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

Final Answer

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.

Final Answer

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

close solution

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

Problem Statement

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

Final Answer

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

Final Answer

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

close solution

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

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

Final Answer

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.

topics and links

infinite series table

Divergence Test

Integral Test

Final Answer

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

close solution

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

Problem Statement

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

Final Answer

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.

topics and links

infinite series table

Alternating Series Test

L'Hôpital's Rule

Derivative Quotient Rule

Final Answer

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

close solution

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.
\[ A = 1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \frac{1}{7^2} - . . . \]

Final Answer

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.

topics and links

p-Series

Alternating Series

absolute convergence

Final Answer

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.

close solution

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 \( S \).)

Final Answer

\(\displaystyle{ A = \frac{\pi^2}{12} }\)

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 \( A = \pi^2 / 12 \).

Final Answer

\(\displaystyle{ A = \frac{\pi^2}{12} }\)

close solution

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 \[ | a_n - L | < \epsilon ~~~~~ \text{ for every } n > N \]

close solution

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 \(\displaystyle{ \lim_{n \to \infty}{\frac{1}{\sqrt{n}}} = 0 }\).

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