Calculus 2 Quiz: Sequence Convergence And Limits

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Sequence Convergence And LimitsQuestion 1 of 20

Find the limit of the sequence $a_n = n \sin(\frac{\pi}{n})$ .

\pi

\infty

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Calculus 2 Quiz

Calculus 2 Quiz: Sequence Convergence And Limits

Practice Sequence Convergence And Limits in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Sequence Convergence And Limits, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Find the limit of the sequence $a_n = n \sin(\frac{\pi}{n})$ .

0
1
$\pi$ (correct answer)
$\infty$

Explanation: This limit is of the indeterminate form

\infty \cdot 0

. We can rewrite it as

a_n = \frac{\sin(\pi/n)}{1/n}

. This is now in the

0/0

form, suitable for L'Hôpital's rule. Let's consider the function

f(x) = \frac{\sin(\pi/x)}{1/x}

. Taking derivatives of the numerator and denominator with respect to

x

gives:

\lim_{x \to \infty} \frac{\cos(\pi/x) \cdot (-\pi/x^2)}{-1/x^2} = \lim_{x \to \infty} \pi \cos(\pi/x)

. As

x \to \infty

\pi/x \to 0

, and

\cos(\pi/x) \to \cos(0) = 1

. Thus, the limit is

\pi

. This is a variation of the fundamental trigonometric limit

\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1

Question 2

What is the limit of the sequence $a_n = \left(\frac{n-1}{n+1}\right)^n$ ?

1
$e^{-2}$ (correct answer)
$e^{-1}$
$e$

Explanation: This limit is of the indeterminate form

1^\infty

. We can rewrite the base as

\frac{n-1}{n+1} = \frac{n+1-2}{n+1} = 1 - \frac{2}{n+1} = 1 + \frac{-2}{n+1}

. The expression becomes

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

. This resembles the standard limit form

\lim_{m \to \infty} (1 + x/m)^m = e^x

. We can rewrite our expression as

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

. As

n \to \infty

, the first term approaches

e^{-2}

and the second term approaches

1^{-1}=1

. Thus, the limit is

e^{-2}

Question 3

Which of the following statements about a sequence $\{a_n\}$ of real numbers is always true?

If $\lim_{n \to \infty} a_n = 0$ , then the series $\sum a_n$ converges.
If the sequence $\{a_n\}$ is bounded, then it converges.
If the sequence $\{a_n\}$ is not bounded, then it diverges. (correct answer)
If the sequence $\{a_n\}$ is monotonic, then it converges.

Explanation: A convergent sequence must be bounded. The contrapositive of this statement is that if a sequence is not bounded, it must diverge. This is always true. Choice A is false; the harmonic series

a_n = 1/n

provides a counterexample. Choice B is false; the sequence

a_n = (-1)^n

is bounded but diverges. Choice D is false; the sequence

a_n = n

is monotonic but diverges.

Question 4

Let a sequence be defined by $a_1 = 0$ and $a_{n+1} = \frac{a_n^2 + 2}{3}$ . Which statement correctly describes the sequence?

The sequence is increasing and converges to 1. (correct answer)
The sequence is decreasing and converges to 1.
The sequence is increasing and converges to 2.
The sequence diverges.

Explanation: First, let's find potential limits by setting

L = \frac{L^2+2}{3}

, which gives

L^2 - 3L + 2 = 0

, or

(L-1)(L-2)=0

. Potential limits are 1 and 2. Let's check the first few terms:

a_1=0, a_2=2/3, a_3=22/27, ...

. The sequence appears to be increasing. We can prove by induction that

a_n < 1

for all

n

. The base case

a_1=0<1

is true. Assume

a_k < 1

, then

a_{k+1} = (a_k^2+2)/3 < (1^2+2)/3 = 1

. So the sequence is bounded above by 1. To check if it is increasing, consider

a_{n+1}-a_n = \frac{a_n^2-3a_n+2}{3} = \frac{(a_n-1)(a_n-2)}{3}

. Since we know

a_n < 1

, both

(a_n-1)

and

(a_n-2)

are negative, making their product positive. Thus

a_{n+1} > a_n

. The sequence is increasing and bounded above, so it converges. Since it's bounded by 1, its limit must be 1.

Question 5

A student wants to find the limit of $a_n = \frac{n + \cos(n)}{n}$ . They argue that since this is an $\infty/\infty$ form, they can use L'Hôpital's rule on $f(x) = \frac{x+\cos(x)}{x}$ . The derivative of the numerator is $1-\sin(x)$ and the derivative of the denominator is 1. Since $\lim_{x \to \infty} (1-\sin(x))$ does not exist, they conclude the original sequence diverges. What is wrong with this reasoning?

L'Hôpital's rule cannot be applied if the limit of the derivatives does not exist.
The original expression can be simplified algebraically, and L'Hôpital's rule is unnecessary. (correct answer)
The form is not indeterminate because $\cos(n)$ is bounded.
The student should have applied L'Hôpital's rule a second time.

Explanation: The student's conclusion that the sequence diverges is incorrect. The best approach is to simplify the expression algebraically:

a_n = \frac{n}{n} + \frac{\cos(n)}{n} = 1 + \frac{\cos(n)}{n}

. We can then use the Squeeze Theorem on the

\frac{\cos(n)}{n}

term. Since

-1 \le \cos(n) \le 1

, we have

\frac{-1}{n} \le \frac{\cos(n)}{n} \le \frac{1}{n}

. As

n \to \infty

, this term is squeezed to 0. Thus,

\lim_{n \to \infty} a_n = 1+0=1

. The student's reasoning is flawed because L'Hôpital's rule being inconclusive (when the limit of derivatives DNE) does not imply the original limit diverges. The fundamental error was choosing a complicated and ultimately inconclusive method instead of a simple algebraic one.

Question 6

The sequence $\{a_n\}$ converges to $L$ . According to the formal definition of a limit, this means that for any $\epsilon > 0$ , there exists an integer $N$ such that for all $n > N$ , which of the following inequalities holds?

$|a_n - N| < \epsilon$
$|a_n - L| < \epsilon$ (correct answer)
$|a_n - \epsilon| < L$
$|L - N| < \epsilon$

Explanation: The formal definition of the limit of a sequence states that a sequence

\{a_n\}

converges to a limit

L

if for every positive number

\epsilon

, there is an integer

N

such that for all

n > N

, the distance between

a_n

and

L

is less than

\epsilon

. This is expressed mathematically as

|a_n - L| < \epsilon

. The other options confuse the roles of the limit

L

, the index

N

, and the error tolerance

\epsilon

Question 7

A sequence $\{g_n\}$ satisfies $0 < g_n \leq \frac{1}{n}$ for all $n \geq 1$ , and $\sum_{n=1}^{\infty} g_n$ converges. If $h_n = \frac{g_1 + g_2 + \cdots + g_n}{n}$ , what can be concluded about $\lim_{n \to \infty} h_n$ ?

$\lim_{n \to \infty} h_n = 0$ by the Cesàro mean theorem since $\lim_{n \to \infty} g_n = 0$
$\lim_{n \to \infty} h_n$ may not exist because the convergence of $\sum g_n$ doesn't guarantee convergence of averages
$\lim_{n \to \infty} h_n = 0$ by Stolz-Cesàro theorem applied to the ratio of partial sums to $n$ (correct answer)
$\lim_{n \to \infty} h_n$ exists and equals $\sum_{n=1}^{\infty} g_n$ by the definition of series convergence

Explanation: Since

\sum_{n=1}^{\infty} g_n

converges, we have

\lim_{n \to \infty} g_n = 0

. Let

S_n = g_1 + g_2 + \cdots + g_n

. Then

h_n = \frac{S_n}{n}

. Since

\{S_n\}

converges to

\sum_{n=1}^{\infty} g_n

, by Stolz-Cesàro theorem:

\lim_{n \to \infty} \frac{S_n}{n} = \lim_{n \to \infty} \frac{S_{n+1} - S_n}{(n+1) - n} = \lim_{n \to \infty} g_{n+1} = 0

. Choice A mentions Cesàro mean theorem incorrectly. Choice B is wrong because convergent series do guarantee this limit exists and equals 0. Choice D confuses the sum of the series with the limit of averages.

Question 8

Find the limit of the sequence $a_n = n(\sqrt{n^2+4} - n)$ .

0
2 (correct answer)
4
$\infty$

Explanation: The expression is of the indeterminate form

\infty \cdot (\infty - \infty)

. To resolve this, we multiply the term in the parenthesis by its conjugate:

\sqrt{n^2+4} - n = \frac{(\sqrt{n^2+4} - n)(\sqrt{n^2+4} + n)}{\sqrt{n^2+4} + n} = \frac{(n^2+4)-n^2}{\sqrt{n^2+4} + n} = \frac{4}{\sqrt{n^2+4} + n}

. Substituting back,

a_n = n \cdot \frac{4}{\sqrt{n^2+4} + n} = \frac{4n}{\sqrt{n^2+4} + n}

. Now, we can find the limit by dividing the numerator and denominator by

n

\lim_{n \to \infty} \frac{4}{\frac{\sqrt{n^2+4}}{n} + 1} = \lim_{n \to \infty} \frac{4}{\sqrt{1+4/n^2} + 1} = \frac{4}{\sqrt{1+0} + 1} = \frac{4}{2} = 2

Question 9

Find the limit of the sequence $a_n = \sum_{k=1}^n \frac{1}{n+k}$ .

0
1
$\ln(2)$ (correct answer)
The sequence diverges.

Explanation: This sequence can be interpreted as a Riemann sum. We can factor out

1/n

from the sum:

a_n = \sum_{k=1}^n \frac{1}{n(1+k/n)} = \frac{1}{n} \sum_{k=1}^n \frac{1}{1+k/n}

. This is the right Riemann sum for the function

f(x) = \frac{1}{1+x}

on the interval

[0, 1]

with

\Delta x = 1/n

. As

n \to \infty

, this sum converges to the definite integral:

\int_0^1 \frac{1}{1+x} dx = [\ln|1+x|]_0^1 = \ln(2) - \ln(1) = \ln(2)

Question 10

Find the limit of the sequence defined by $a_n = \frac{\ln(n^2 + e^n)}{n}$ .

0
1 (correct answer)
2
$\infty$

Explanation: To find the limit, we can consider the corresponding function

f(x) = \frac{\ln(x^2 + e^x)}{x}

and apply L'Hôpital's rule, as the limit is of the indeterminate form

\infty/\infty

. The derivative of the numerator is

\frac{2x + e^x}{x^2 + e^x}

and the derivative of the denominator is 1. The new limit is

\lim_{x \to \infty} \frac{2x + e^x}{x^2 + e^x}

. This is still

\infty/\infty

. We can divide the numerator and denominator by the fastest-growing term,

e^x

, to get

\lim_{x \to \infty} \frac{2x/e^x + 1}{x^2/e^x + 1}

. Since

\lim_{x \to \infty} x^p/e^x = 0

for any power p, the limit simplifies to

\frac{0+1}{0+1} = 1

Question 11

Let the sequence $a_n = \arctan(\ln(n))$ . What is the limit of this sequence as $n \to \infty$ ?

0
$\pi/4$
$\pi/2$ (correct answer)
The sequence diverges.

Explanation: To find the limit of a composition of functions, we evaluate the limit from the inside out. First, we find the limit of the inner part:

\lim_{n \to \infty} \ln(n) = \infty

. Next, we evaluate the limit of the outer function as its argument approaches infinity. Let

u = \ln(n)

. The problem becomes finding

\lim_{u \to \infty} \arctan(u)

. The range of the arctangent function is

(-\pi/2, \pi/2)

, and it has a horizontal asymptote at

y = \pi/2

as its input goes to infinity. Therefore, the limit of the sequence is

\pi/2

Question 12

Let $a_n = \frac{n! \cos^2(n\pi/3)}{n^n}$ . What is the limit of this sequence as $n \to \infty$ ?

0 (correct answer)
1/2
1
The sequence diverges.

Explanation: This limit can be determined using the Squeeze Theorem. We know that

0 \le \cos^2(n\pi/3) \le 1

for all

n

. Therefore, the sequence

a_n

is bounded by

0 \le a_n \le \frac{n!}{n^n}

. We need to find the limit of the upper bound.

\lim_{n \to \infty} \frac{n!}{n^n} = \lim_{n \to \infty} \frac{1}{n} \cdot \frac{2}{n} \cdots \frac{n}{n}

. Since

\frac{k}{n} \le 1

for

k \le n

, we have

0 \le \frac{n!}{n^n} \le \frac{1}{n}

. As

n \to \infty

\frac{1}{n} \to 0

. Since

a_n

is squeezed between 0 and a sequence that converges to 0, the limit of

a_n

must also be 0.

Question 13

Find the limit of the sequence $a_n = \frac{5^n + (-2)^n}{5^n - (-2)^n}$ .

1 (correct answer)
-1
0
The sequence diverges.

Explanation: The dominant term in both the numerator and the denominator is

5^n

, as it grows much faster than

(-2)^n

. To find the limit, we can divide both the numerator and the denominator by this dominant term:

a_n = \frac{1 + (-2/5)^n}{1 - (-2/5)^n}

. As

n \to \infty

, the term

(-2/5)^n

approaches 0 because its base has an absolute value less than 1. Therefore, the limit is

\lim_{n \to \infty} a_n = \frac{1+0}{1-0} = 1

Question 14

For what values of $r$ does the sequence $a_n = n \cdot r^n$ converge?

$-1 < r < 1$ (correct answer)
$-1 \le r < 1$
$-1 < r \le 1$
$-1 \le r \le 1$

Explanation: We analyze the convergence based on

r

. If

|r| > 1

|r^n|

grows exponentially, and multiplying by

n

makes it diverge faster. If

r=1

a_n=n

, which diverges. If

r=-1

a_n = n(-1)^n

, which oscillates with increasing magnitude and diverges. If

|r|<1

, we have an

\infty \cdot 0

form. The exponential term

r^n

goes to zero much faster than

n

goes to infinity. Using L'Hôpital's rule on

f(x) = x/r^{-x}

, we find the limit is 0. Therefore, the sequence converges only for

|r|<1

, which is the interval

-1 < r < 1

. The endpoints

r=1

and

r=-1

are common distractors as they are included in the convergence interval for the simpler sequence

b_n=r^n

Question 15

Evaluate the limit $\lim_{n \to \infty} \frac{3^{n+1} + n^4}{n! + 3^n}$ .

0 (correct answer)
1
3
$\infty$

Explanation: To evaluate this limit, we compare the rates of growth of the terms. In the numerator, the exponential term

3^{n+1}

grows much faster than the polynomial term

n^4

. In the denominator, the factorial term

n!

grows much faster than the exponential term

3^n

. Therefore, the limit is determined by the ratio of the fastest-growing terms:

\lim_{n \to \infty} \frac{3^{n+1}}{n!}

. It is a standard result that factorials grow faster than any exponential function, so

\lim_{n \to \infty} \frac{c^n}{n!} = 0

for any constant

c

. Thus, the limit is 0.

Question 16

The sequence $\{a_n\}$ is defined by $a_1=1$ and $a_{n+1} = \sqrt{6+a_n}$ . It can be shown that the sequence is increasing and bounded above by 3. What is the limit of the sequence?

-2
$\sqrt{6}$
3 (correct answer)
6

Explanation: The problem states that the sequence is increasing and bounded above, so by the Monotonic Sequence Theorem, it converges to a limit

L

. We can find

L

by taking the limit of the recurrence relation:

\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \sqrt{6+a_n}

, which gives

L = \sqrt{6+L}

. Squaring both sides yields

L^2 = 6+L

, or

L^2 - L - 6 = 0

. This factors as

(L-3)(L+2) = 0

, so the possible limits are

L=3

and

L=-2

. Since

a_1=1

is positive and the square root function always produces a non-negative result, all terms of the sequence are positive. Thus, the limit must be positive, so

L=3

Question 17

Consider the sequence $a_n = \frac{(-1)^n (n^2 - 1)}{n^2 + n + 1}$ . Which of the following is true?

The sequence converges to 1.
The sequence converges to 0.
The sequence diverges. (correct answer)
The sequence converges to -1.

Explanation: For an alternating sequence to converge, the limit of the absolute value of its terms must be 0. Let's examine

\lim_{n \to \infty} |a_n| = \lim_{n \to \infty} \frac{n^2 - 1}{n^2 + n + 1}

. Since the degrees of the numerator and denominator are equal, the limit is the ratio of the leading coefficients, which is

1/1 = 1

. Because

\lim_{n \to \infty} |a_n| = 1 \neq 0

, the sequence does not converge. The terms for even

n

approach 1, while the terms for odd

n

approach -1. The sequence oscillates between values close to 1 and -1 and therefore diverges.

Question 18

A sequence $\{a_n\}$ is known to be non-increasing. Which of the following additional conditions is not sufficient to prove that $\{a_n\}$ converges?

$a_n \ge 0$ for all $n$ .
The corresponding series $\sum a_n$ converges.
The subsequence $\{a_{2n}\}$ converges.
$a_n$ is a sequence of integers. (correct answer)

Explanation: By the Monotonic Sequence Theorem, a non-increasing (monotonic) sequence converges if and only if it is bounded below. Option A provides a lower bound of 0, so it's sufficient. Option B implies that

\lim_{n \to \infty} a_n = 0

, so the sequence converges, which is sufficient. Option C is also sufficient; if a subsequence of a monotonic sequence converges, the entire sequence must converge to the same limit. Option D is not sufficient. Consider the sequence

a_n = -n

. It is a non-increasing sequence of integers, but it is not bounded below and diverges to

-\infty

Question 19

Let $a_n = \int_n^{n+1} \frac{\sin(x)}{x} dx$ . Find $\lim_{n \to \infty} a_n$ .

0 (correct answer)
1
$\pi$
The sequence diverges.

Explanation: We can use the Squeeze Theorem. For

x

in the interval

[n, n+1]

where

n \ge 1

, we know

-1 \le \sin(x) \le 1

and

\frac{1}{n+1} \le \frac{1}{x} \le \frac{1}{n}

. This gives the bounds

\frac{-1}{n} \le \frac{\sin(x)}{x} \le \frac{1}{n}

. Integrating this inequality over the interval

[n, n+1]

of length 1, we get

\int_n^{n+1} \frac{-1}{n} dx \le a_n \le \int_n^{n+1} \frac{1}{n} dx

, which simplifies to

\frac{-1}{n} \le a_n \le \frac{1}{n}

. As

n \to \infty

, both

\frac{-1}{n}

and

\frac{1}{n}

approach 0. By the Squeeze Theorem,

\lim_{n \to \infty} a_n = 0

Question 20

Find the limit of the sequence $a_n = \frac{1}{n} \ln\left(\frac{n!}{n^n}\right)$ .

-1 (correct answer)
0
1
$-\infty$

Explanation: We can use the properties of logarithms to rewrite the expression:

a_n = \frac{1}{n}(\ln(n!) - \ln(n^n)) = \frac{1}{n}(\sum_{k=1}^{n} \ln(k) - n\ln(n)) = \frac{1}{n}\sum_{k=1}^{n} \ln(k) - \ln(n)

. The term

\frac{1}{n}\sum_{k=1}^{n} \ln(k)

can be seen as an approximation of the integral

\frac{1}{n}\int_1^n \ln(x) dx

. Evaluating the integral gives

\frac{1}{n}[x\ln(x) - x]_1^n = \frac{1}{n}(n\ln(n) - n + 1) = \ln(n) - 1 + \frac{1}{n}

. Substituting this back gives

a_n \approx (\ln(n) - 1 + \frac{1}{n}) - \ln(n) = -1 + \frac{1}{n}

. As

n \to \infty

, this approaches -1. A more formal approach uses Stirling's approximation for

\ln(n!)

, which also yields the limit -1.