Topics You Need To Understand For This Page
Calculus Main Topics
Tools
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 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 its limit. Justify your answers.
Question 1 
\(a_n=2+(1)^n\) 


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


The sequence \(\displaystyle{ \left\{ \frac{n}{e^n} \right\} }\) converges to zero. 
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 \( 0 \).
Question 2 Final Answer 
The sequence \(\displaystyle{ \left\{ \frac{n}{e^n} \right\} }\) converges to zero. 
Question 3 
\(\displaystyle{ a_n = \left( 1 + \frac{2}{n} \right)^n }\) 


The sequence \(\displaystyle{ \left\{ \left( 1 + \frac{2}{n} \right)^n \right\} }\) converges to \( e^2 \). 
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.
\(\displaystyle{
\begin{array}{rclcl}
y & = & \lim_{n \to \infty}{ \left(1 + \frac{2}{n}\right)^n } \\
\ln(y) & = & \lim_{n \to \infty}{\ln \left(1 + \frac{2}{n}\right)^n } & & \text{Take the natural log of both sides} \\
& = & \lim_{n \to \infty}{(n) \ln \left(1 + \frac{2}{n}\right) } \\
& = & \lim_{n \to \infty}{ \frac{\ln \left(1 + \frac{2}{n}\right)}{1/n} } \\
& = & \lim_{n \to \infty}{ \frac{\ln [(n+2)/n]}{n^{1}} } \\
& = & \lim_{n \to \infty}{ \left[ \frac{2n^{2}(n)}{n+2} \cdot \frac{1}{n^{2}} \right] } & & \text{ L'Hôpital's Rule} \\
\\
& = & \lim_{n \to \infty}{ \frac{2n}{n+2} } \\
& = & \lim_{n \to \infty}{\frac{2}{1}} & & \text{L'Hôpital's Rule again} \\
\ln(y) & = & 2 \\
e^{\ln(y)} & = & e^2 & & \text{Undo the natural log by raising each side to the exponent of e} \\
y & = & e^2
\end{array}}\)
Question 3 Final Answer 
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.
Question 4 
\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^3}{3^n}} }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^3}{3^n}} }\) converges by the Ratio Test. 
Let's go down through the infinite series table in order.
Let \(\displaystyle{ a_n = \frac{n^3}{3^n} }\)
Group 1: nthTerm 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, pSeries, 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 \(\displaystyle{ a_n = \frac{n^3}{3^n} }\), \(\displaystyle{ a_{n+1} = \frac{(n+1)^3}{3^{n+1}} }\).
\(\displaystyle{
\begin{array}{rcl}
\lim_{n \to \infty}{\frac{a_{n+1}}{a_n}} & = & \lim_{n \to \infty}{ \frac{(n+1)^3}{3^{n+1}} \cdot \frac{3^n}{n^3} } \\
& = & \lim_{n \to \infty}{ \frac{(n+1)^3}{n^3} \cdot \frac{3^n}{3^{n+1}} } \\
& = & \lim_{n \to \infty}{ \left( \frac{n+1}{n} \right)^3 \cdot \frac{1}{3} } \\
& = & \frac{1}{3} \lim_{n \to \infty}{ \left( 1 + \frac{1}{n} \right)^3 } \\
& = & \frac{1}{3} (1 + 0)^3 = \frac{1}{3}
\end{array}}\)
Since \(\displaystyle{ \lim_{n \to \infty}{\frac{a_{n+1}}{a_n}} = 1/3 < 1 }\) the series converges.
Question 4 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^3}{3^n}} }\) converges by the Ratio Test. 
Question 5 
\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n}{n^2+1} } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n}{n^2+1} } }\) diverges by the Limit Comparison Test. 
Let's go down through the infinite series tests in the order they appear in the infinite series table.
Group 1: nthTerm 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 pSeries 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.
\(\displaystyle{
\begin{array}{rcl}
\lim_{n \to \infty}{\frac{a_n}{t_n}} & = & \lim_{n \to \infty}{ \frac{n}{n^2+1} \cdot \frac{n}{1} } \\
& = & \lim_{n \to \infty}{ \frac{n^2}{n^2 + 1} } \\
& = & \lim_{n \to \infty}{ \frac{2n}{2n} } = 1
\end{array}
}\)
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.
Question 5 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n}{n^2+1} } }\) diverges by the Limit Comparison Test. 
Question 6 
\(\displaystyle{ \sum_{n=1}^{\infty}{n \sin\left( \frac{1}{n} \right) } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{n \cdot \sin\left( \frac{1}{n} \right) } }\) diverges by the nthTerm Test. 
There are several ways to work this one. The most straightforward is to go through the infinite series table in order.
Group 1  nthTerm 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.
\(\displaystyle{
\begin{array}{rcl}
\lim_{n \to \infty}{ n \cdot \sin\left( \frac{1}{n} \right) } & = & \lim_{n \to \infty}{ \frac{\sin(1/n)}{1/n} } \
& = & \lim_{n \to \infty}{ \frac{\sin(n^{1})}{n^{1}} } \
& = & \lim_{n \to \infty}{ \frac{\cos(n^{1}) (1)(n^{2})}{(1)n^{2}} } \
& = & \lim_{n \to \infty}{ \cos(n^{1}) } = \cos(0) = 1
\end{array}
}\)
So, since the limit does not equal zero, the series diverges.
Note: Another way to solve this problem 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 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.
Question 6 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{n \cdot \sin\left( \frac{1}{n} \right) } }\) diverges by the nthTerm Test. 
Question 7 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left( \frac{1}{\sqrt{n}}  \frac{1}{\sqrt{n+1}}\right) } }\) 


The Telescoping Series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left( \frac{1}{\sqrt{n}}  \frac{1}{\sqrt{n+1}}\right) } }\) converges. 
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.
\(\displaystyle{
\begin{array}{ccc}
n & & a_n \\
1 & & \frac{1}{1}  \frac{1}{\sqrt{2}} \\
2 & & \frac{1}{\sqrt{2}}  \frac{1}{\sqrt{3}} \\
3 & & \frac{1}{\sqrt{3}}  \frac{1}{\sqrt{4}} \\
4 & & \frac{1}{\sqrt{4}}  \frac{1}{\sqrt{5}}
\end{array}}\)
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 \( 1 \)).
Question 7 Final Answer 
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.
Question 8 
\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n \ln(n)} } }\) 


The series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n~\ln(n)} } }\) diverges by the Integral Test. 
Going through the infinite series table in order while help us determine what rule to use here.
Group 1: nthTerm 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. 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.
Question 8 Final Answer 
The series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n~\ln(n)} } }\) diverges by the Integral Test. 
Question 9 
\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(1)^n \ln(n)}{n} } }\) 


The series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{(1)^n \ln(n)}{n} } }\) converges by the Alternating Series Test. 
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.
Question 9 Final Answer 
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 
Question 10 
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.

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 pSeries. So the alternating series converges absolutely.
Question 10 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.

Question 11 
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 question 10. (Hint: Consider \( S  A \) and express it in terms of \( S \).)



\( \displaystyle{ A = 1  \frac{1}{2^2} + }\) \(\displaystyle{\frac{1}{3^2}  \frac{1}{4^2} + \frac{1}{5^2}  }\) \(\displaystyle{\frac{1}{6^2} + \frac{1}{7^2}  . . . = }\) \(\displaystyle{\frac{\pi^2}{12}}\)

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, we get
\(\displaystyle{
\begin{array}{rcl}
S  A & = & \left( 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + . . . \right) 
\left( 1  \frac{1}{2^2} + \frac{1}{3^2}  \frac{1}{4^2} + \frac{1}{5^2}  . . . \right) \\
& = & (1  1) + \left( \frac{1}{2^2} + \frac{1}{2^2} \right) + \left( \frac{1}{3^2}  \frac{1}{3^2} \right) + \left( \frac{1}{4^2} + \frac{1}{4^2} \right) + \left( \frac{1}{5^2}  \frac{1}{5^2} \right) + . . . \\
& = & 2 \left( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \frac{1}{8^2} + . . . \right) \\
& = & \frac{2}{2^2} \left( 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + . . . \right) \\
& = & \frac{1}{2}S
\end{array}
}\)
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 \( A = \pi^2 / 12 \).
Question 11 Final Answer 
\( \displaystyle{ A = 1  \frac{1}{2^2} + }\) \(\displaystyle{\frac{1}{3^2}  \frac{1}{4^2} + \frac{1}{5^2}  }\) \(\displaystyle{\frac{1}{6^2} + \frac{1}{7^2}  . . . = }\) \(\displaystyle{\frac{\pi^2}{12}}\)

Question 12 
State the definition for a sequence \( \{ a_n \} \) to converge to a limit \( L \) as \( n \to \infty \).


A sequence \( \{ a_n \} \) converges to a limit \( L \) if, for every \( \epsilon > 0 \) there exists a number \(N\) such that
\[  a_n  L  < \epsilon ~~~~~ \text{ for every } n > N \]
Question 13 
If \(\displaystyle{ a_n = \frac{1}{\sqrt{n}} }\) for \( n = 1, 2, 3, . . .\) prove from the definition that \(\displaystyle{ \lim_{n \to \infty}{a_n} = 0 }\).


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