Limits for sequences

Question 1: Definition of a Sequence (from file image_984037.png)

The Question: Which of the following represents a sequence?

Detailed Solution: By definition, a sequence is a function whose domain is the set of natural numbers ($\mathbb{N}$) or a similar set like the non-negative integers. We need to find the functions listed that have a domain of $\mathbb{N}$ or $\mathbb{R}^+$ (positive real numbers, which includes $\mathbb{N}$).

• $f_1: \mathbb{R} \to \mathbb{R}$: Domain is all real numbers. Not a sequence.
• $f_2: \mathbb{N} \to \mathbb{R}$: Domain is the natural numbers. This is a sequence.
• $f_3: \mathbb{R}^- \to \mathbb{R}$: Domain is negative real numbers. Not a sequence.
• $f_4: \mathbb{R}^+ \to \mathbb{R}$: Domain is positive real numbers. This is a broader category, but a function defined on $\mathbb{N}$ could be restricted from this. However, $f_2$ is more precise.
• $f_5: \mathbb{N} \to \mathbb{Z}$: Domain is the natural numbers. This is a sequence.

Final Answer: $f_2$ and $f_5$ represent sequences.

Question 2: Limit of a Rational Sequence (from file image_984037.png)

The Question: Let ${a_n}$ be a sequence defined as $a_n = \frac{3n^2+5n+2}{6n^2+2n+1}$. The limit of the sequence ${a_n}$ is __________.

Core Concept: To find the limit of a rational function of $n$ as $n \to \infty$, we divide the numerator and the denominator by the highest power of $n$ present in the denominator.

Detailed Solution:

1. The highest power of $n$ in the denominator is $n^2$.
2. Divide every term in the expression by $n^2$: $$a_n = \frac{\frac{3n^2}{n^2} + \frac{5n}{n^2} + \frac{2}{n^2}}{\frac{6n^2}{n^2} + \frac{2n}{n^2} + \frac{1}{n^2}} = \frac{3 + \frac{5}{n} + \frac{2}{n^2}}{6 + \frac{2}{n} + \frac{1}{n^2}}$$
3. Now, take the limit as $n \to \infty$. Any term with $n$ in the denominator will approach 0. $$\lim_{n\to\infty} a_n = \frac{3 + 0 + 0}{6 + 0 + 0} = \frac{3}{6} = \frac{1}{2}$$

Final Answer: The limit is 1/2.

Question 3: Identifying Subsequences (from file image_984037.png)

The Question: Consider a sequence $(1, 2, 3, 4, 5, 6, …)$, that is $a_n = n$. Which of the followings are subsequence of $a_n$?

Detailed Solution: The original sequence is ${1, 2, 3, 4, 5, 6, 7, 8, …}$. A subsequence must consist of terms from this sequence, in their correct order.

• $b_n=2$ for all $n$: This sequence is $(2, 2, 2, …)$. This is not a subsequence because you can’t pick the number 2 multiple times.
• $b_n=2n$ for all $n$: This sequence is $(2, 4, 6, 8, …)$. All these terms are in the original sequence, and they are in the correct order. This is a subsequence.
• $b_n=10$ and $b_n=2n$ for $n \ge 2$: This sequence is $(10, 4, 6, 8, …)$. This is not a subsequence because the terms are not in their original increasing order (10 comes after 4, 6, 8 in the original sequence).
• $b_n=2n+1$ for all $n$: This sequence is $(3, 5, 7, 9, …)$. All these terms are in the original sequence, and they are in the correct order. This is a subsequence.

Final Answer: The subsequences are $b_n=2n$ and $b_n=2n+1$.

Question 4: Subsequences of ${n^2}$ (from file image_983d54.png)

The Question: Consider a sequence ${n^2}$. Which of the following options are subsequences of the given sequence?

Detailed Solution: The original sequence is ${1^2, 2^2, 3^2, 4^2, …} = {1, 4, 9, 16, …}$. A subsequence must be a sequence of squares where the base is an increasing sequence of natural numbers.

• ${(n+5)^2}$: This generates the sequence ${6^2, 7^2, 8^2, …} = {36, 49, 64, …}$. This is a subsequence.
• ${(2n+1)^2}$: This generates the sequence of squares of odd numbers ${3^2, 5^2, 7^2, …} = {9, 25, 49, …}$. This is a subsequence.
• ${(2n+1)^3}$: This generates a sequence of cubes, not squares. Not a subsequence.
• ${n^3}$: A sequence of cubes, not squares. Not a subsequence.
• ${n^2}$: The sequence itself is trivially a subsequence of itself. Is a subsequence.
• ${(2n)^2}$: This generates the sequence of squares of even numbers ${2^2, 4^2, 6^2, …} = {4, 16, 36, …}$. This is a subsequence.
• ${(\frac{1}{2}n+1)^2}$: The base, $(\frac{1}{2}n+1)$, is not always an integer. For example, when $n=2$, the base is 2, but when $n=1$, the base is 1.5. A subsequence must pick terms from the original, so the base must be a natural number. Not a subsequence.

Final Answer: The subsequences are ${(n+5)^2}$, ${(2n+1)^2}$, ${n^2}$, ${(2n)^2}$.

Question 5: Alternating Sequence (from file image_983d54.png)

The Question: Consider a sequence $(1, -1, 1, -1, …)$, that is $a_n=1$ if $n$ is odd and $a_n=-1$ if $n$ is even. Which of the following options are correct?

Detailed Solution:

• Limit of ${a_n}$ is 1 / -1 / both: The terms of the sequence alternate between 1 and -1 and never settle on a single value. Therefore, the sequence diverges. These statements are FALSE.
• Limit of the sequence, $a_n$, does not exist.: TRUE. Since the sequence does not approach a single finite number, it does not have a limit.
• $(1, 1, 1, …)$ is a subsequence of $a_n$: TRUE. We can form this subsequence by picking only the terms at odd positions ($a_1, a_3, a_5, …$), which are all equal to 1.

Final Answer: The correct statements are:

• $(1, 1, 1, …)$, that is $b_n=1$ for all $n$, is a subsequence of $a_n$.
• Limit of the sequence, $a_n$, does not exist.

Question 6: Limit of a Subsequence (from file image_983cd4.png)

The Question: Let ${a_n}$ be a sequence converging to the limit $l$. Then $\lim_{n \to \infty} a_{3n}$ is __________.

Core Concept: A fundamental theorem of convergent sequences states that if a sequence converges to a limit $L$, then every possible subsequence of that sequence must also converge to the same limit $L$.

Detailed Solution:

1. We are given that the sequence ${a_n}$ converges to $l$.
2. The sequence ${a_{3n}}$ represents the subsequence formed by taking every third term of the original sequence ($a_3, a_6, a_9, …$).
3. According to the theorem, since the original sequence converges to $l$, this subsequence must also converge to $l$.

Final Answer: The limit is l.

Question 7: Limit with Radicals (from file image_983cd4.png)

The Question: Consider the sequence ${a_n}$, defined as $a_n = \sqrt{2n+1} - \sqrt{2n}$. Which of the following option(s) is(are) true?

Core Concept: To find the limit of an expression in the form $\infty - \infty$ involving square roots, we multiply by the conjugate. The conjugate of $(A-B)$ is $(A+B)$.

Detailed Solution:

1. The expression is $a_n = \sqrt{2n+1} - \sqrt{2n}$.
2. Multiply the expression by 1 in the form of its conjugate divided by itself: $$a_n = (\sqrt{2n+1} - \sqrt{2n}) \times \frac{\sqrt{2n+1} + \sqrt{2n}}{\sqrt{2n+1} + \sqrt{2n}}$$
3. The numerator becomes a difference of squares, $(A-B)(A+B) = A^2 - B^2$: $$a_n = \frac{(\sqrt{2n+1})^2 - (\sqrt{2n})^2}{\sqrt{2n+1} + \sqrt{2n}} = \frac{(2n+1) - (2n)}{\sqrt{2n+1} + \sqrt{2n}} = \frac{1}{\sqrt{2n+1} + \sqrt{2n}}$$
4. Now, take the limit as $n \to \infty$: $$\lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{1}{\sqrt{2n+1} + \sqrt{2n}}$$
5. As $n$ approaches infinity, the denominator approaches infinity. A constant divided by infinity approaches 0.

Final Answer:

• ${a_n}$ is convergent. (TRUE)
• Limit of ${a_n}$ is 0. (TRUE)
• The other options are false.

Question 8: Limits involving $n^{1/n}$ (from file image_983cd4.png)

The Question: Let ${a_n}$ be a sequence defined by $a_n=n^{1/n}$, has limit 1. Then which of the following option(s) is(are) true?

Core Concept: We can use limit laws and the given fact that $\lim_{n \to \infty} n^{1/n} = 1$.

Detailed Solution:

• $\lim_{n \to \infty} \ln(n^{1/n})$: Since $\ln(x)$ is a continuous function, we can bring the limit inside: $\ln(\lim_{n \to \infty} n^{1/n}) = \ln(1) = 0$. This statement is TRUE.
• $\lim_{n \to \infty} \ln(1+n^{1/n})$: Again, bring the limit inside: $\ln(1 + \lim_{n \to \infty} n^{1/n}) = \ln(1+1) = \ln(2)$. The statement says the limit is 0. This is FALSE.
• $\lim_{n \to \infty} \frac{\ln 2 - \frac{1}{n}}{\ln(1+n^{1/n})}$: We evaluate the limit of the numerator and denominator separately.
  • Limit of numerator: $\lim_{n \to \infty} (\ln 2 - \frac{1}{n}) = \ln 2 - 0 = \ln 2$.
  • Limit of denominator: $\lim_{n \to \infty} \ln(1+n^{1/n}) = \ln(2)$, as shown above.
  • The limit of the fraction is $\frac{\ln 2}{\ln 2} = 1$. This statement is TRUE.
• $\lim_{n \to \infty} (4n^{3/n} - 1)$: We can rewrite $n^{3/n}$ as $(n^{1/n})^3$.
  • The limit becomes $\lim_{n \to \infty} (4(n^{1/n})^3 - 1) = 4(\lim_{n \to \infty} n^{1/n})^3 - 1 = 4(1)^3 - 1 = 4 - 1 = 3$. This statement is TRUE.

Final Answer:

• $\lim_{n \to \infty} \ln(n^{1/n}) = 0$
• $\lim_{n \to \infty} \frac{\ln 2 - \frac{1}{n}}{\ln(1+n^{1/n})} = 1$
• $\lim_{n \to \infty} (4n^{3/n} - 1) = 3$

Question 9: The Squeeze Theorem (from file image_983c7b.png)

The Question: If ${a_n}$ and ${b_n}$ be two sequences such that $|a_n| \le |b_n|$ for all $n \ge m$, where $m \in \mathbb{N}$ and $\lim_{n \to \infty} b_n = 0$, then find the limit of $a_n$.

Detailed Solution:

1. We are given $\lim_{n \to \infty} b_n = 0$. This implies that $\lim_{n \to \infty} |b_n| = 0$ as well.
2. We are given the inequality $|a_n| \le |b_n|$. We also know that the absolute value must be non-negative, so $0 \le |a_n|$.
3. Combining these, we can “squeeze” the sequence $|a_n|$: $$0 \le |a_n| \le |b_n|$$
4. Now we take the limit of all parts as $n \to \infty$.
   • $\lim_{n \to \infty} 0 = 0$.
   • $\lim_{n \to \infty} |b_n| = 0$.
5. By the Squeeze Theorem, since $|a_n|$ is squeezed between two sequences that both go to 0, the limit of $|a_n|$ must also be 0.
6. If $\lim_{n \to \infty} |a_n| = 0$, then it must be that $\lim_{n \to \infty} a_n = 0$.

Final Answer: The limit of $a_n$ is 0.

Question 10: Limit of an Alternating Sequence (from file image_983c7b.png)

The Question: Consider a sequence $(-1, \frac{1}{2}, -\frac{1}{3}, \frac{1}{4}, -\frac{1}{5}, …)$, that is $a_n = \frac{(-1)^n}{n}$. The limit of sequence ${a_n}$ is __________.

Core Concept: The Squeeze Theorem is perfect for alternating sequences whose magnitude goes to zero.

Detailed Solution:

1. The sequence is $a_n = \frac{(-1)^n}{n}$. The terms bounce between positive and negative while getting smaller.
2. We can bound the sequence. The smallest possible value is when the numerator is -1, and the largest is when it’s 1. $$-\frac{1}{n} \le \frac{(-1)^n}{n} \le \frac{1}{n}$$
3. Now, we find the limits of the outer “squeezing” sequences:
   • $\lim_{n \to \infty} -\frac{1}{n} = 0$.
   • $\lim_{n \to \infty} \frac{1}{n} = 0$.
4. Since our sequence $a_n$ is squeezed between two sequences that both approach 0, by the Squeeze Theorem, its limit must also be 0.

Final Answer: The limit is 0.

Question 11: Monotonic Sequences (from file image_983c7b.png)

The Question: Which of the following option(s) is(are) true?

Detailed Solution: A sequence is increasing if $a_{n+1} \ge a_n$ and decreasing if $a_{n+1} \le a_n$.

• ${a_n} = {\frac{3n-5}{4n+2}}$ is a decreasing sequence.

  • Let’s check the first few terms: $a_1 = \frac{-2}{6} = -0.33$. $a_2 = \frac{1}{10}=0.1$. $a_3 = \frac{4}{14} \approx 0.28$. The terms are increasing. FALSE.

• ${-\frac{1}{6n-5}}$ is an increasing sequence.

  • As $n$ increases, the denominator $6n-5$ increases.
  • The fraction $\frac{1}{6n-5}$ therefore decreases.
  • Multiplying by -1 reverses the direction, so $-\frac{1}{6n-5}$ is increasing. TRUE.

• ${\sqrt{n+1} - \sqrt{n}}$ is an increasing sequence.

  • Let’s simplify by multiplying by the conjugate: $\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$.
  • As $n$ increases, the denominator increases, which means the overall fraction decreases. FALSE.

• ${\sqrt[n]{2n}}$ is an increasing sequence.

  • Let’s test the first few terms:
    • $a_1 = \sqrt[1]{2(1)} = 2$.
    • $a_2 = \sqrt[2]{2(2)} = \sqrt{4} = 2$.
    • $a_3 = \sqrt[3]{2(3)} = \sqrt[3]{6} \approx 1.817$.
  • The sequence goes from 2 to 2 to 1.817. It is not increasing. FALSE.

Final Answer: The only true statement is ${-\frac{1}{6n-5}}$ is an increasing sequence.