Sequence Convergence & Limits - Calculus 2

1. Consider the sequence $a_n = \frac{(-1)^n (n^2 - 1)}{n^2 + n + 1}$. Which of the following is true?
   Answer: The sequence diverges. 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.
2. What is the limit of the sequence defined by $a_n = n^{1/(n+1)}$?
   Answer: 1 This limit is of the indeterminate form $\infty^0$. Let $L = \lim_{n \to \infty} n^{1/(n+1)}$. We can find the limit of its natural logarithm: $\ln(L) = \lim_{n \to \infty} \ln(n^{1/(n+1)}) = \lim_{n \to \infty} \frac{\ln(n)}{n+1}$. This is an $\infty/\infty$ indeterminate form, so we can use L'Hôpital's rule. Taking the derivative of the numerator and denominator with respect to $n$ gives $\lim_{n \to \infty} \frac{1/n}{1} = 0$. Since $\ln(L) = 0$, we have $L = e^0 = 1$.
3. Find the limit of the sequence $a_n = n \sin(\frac{\pi}{n})$.
   Answer: $\pi$ 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 $\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1$.
4. The sequence $\{a_n\}$ is defined such that $a_{2k} = \frac{k+1}{k}$ for even indices and $a_{2k-1} = \frac{4k-5}{4k-3}$ for odd indices. What can be concluded about the sequence $\{a_n\}$?
   Answer: The sequence converges to 1. A sequence converges if and only if all of its subsequences converge to the same limit. Here, the sequence is partitioned into even and odd terms. We find the limit of each subsequence. For the even terms: $\lim_{k \to \infty} a_{2k} = \lim_{k \to \infty} \frac{k+1}{k} = 1$. For the odd terms: $\lim_{k \to \infty} a_{2k-1} = \lim_{k \to \infty} \frac{4k-5}{4k-3} = \frac{4}{4} = 1$. Since both the subsequence of even terms and the subsequence of odd terms converge to the same limit, 1, the entire sequence $\{a_n\}$ converges to 1.
5. Evaluate the limit $\lim_{n \to \infty} \frac{3^{n+1} + n^4}{n! + 3^n}$.
   Answer: 0 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.
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?
   Answer: $|a_n - L| < \epsilon$ 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 $\epsilon$.
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 $\lim_{n \to \infty} h_n$?
   Answer: $\lim_{n \to \infty} h_n = 0$ by Stolz-Cesàro theorem applied to the ratio of partial sums to $n$ 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.
8. Let $a_n = \int_n^{n+1} \frac{\sin(x)}{x} dx$. Find $\lim_{n \to \infty} a_n$.
   Answer: 0 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, $\lim_{n \to \infty} a_n = 0$.
9. 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?
   Answer: The original expression can be simplified algebraically, and L'Hôpital's rule is unnecessary. 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.
10. Find the limit of the sequence defined by $a_n = \frac{\ln(n^2 + e^n)}{n}$.
    Answer: 1 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 $\frac{0+1}{0+1} = 1$.
11. Find the limit of the sequence $a_n = (\sqrt{n^2+n} - n)(-1)^n$.
    Answer: The sequence diverges. This is an alternating sequence. We first examine the limit of its magnitude, $|a_n| = \sqrt{n^2+n} - n$. This is an $\infty - \infty$ indeterminate form. We multiply by the conjugate: $|a_n| = \frac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n} = \frac{n}{\sqrt{n^2+n}+n}$. Taking the limit, we get $\lim_{n \to \infty} \frac{n}{\sqrt{n^2+n}+n} = \lim_{n \to \infty} \frac{1}{\sqrt{1+1/n}+1} = \frac{1}{1+1} = \frac{1}{2}$. Since $\lim_{n \to \infty} |a_n| = 1/2 \neq 0$, the sequence $\{a_n\}$ does not converge to 0. The terms oscillate between values approaching 1/2 and -1/2, so the sequence diverges.
12. Which of the following statements about a sequence $\{a_n\}$ of real numbers is always true?
    Answer: If the sequence $\{a_n\}$ is not bounded, then it diverges. 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.
13. Find the limit of the sequence $a_n = n(\sqrt{n^2+4} - n)$.
    Answer: 2 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 $n$: $\lim_{n \to \infty} \frac{4}{\frac{\sqrt{n^2+4}}{n} + 1} = \lim_{n \to \infty} \frac{4}{\sqrt{1+4/n^2} + 1} = \frac{4}{\sqrt{1+0} + 1} = \frac{4}{2} = 2$.
14. Find the limit of the sequence $a_n = \sum_{k=1}^n \frac{1}{n+k}$.
    Answer: $\ln(2)$ 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: $\int_0^1 \frac{1}{1+x} dx = [\ln|1+x|]_0^1 = \ln(2) - \ln(1) = \ln(2)$.
15. Find the limit of the sequence $a_n = \frac{5^n + (-2)^n}{5^n - (-2)^n}$.
    Answer: 1 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 $\lim_{n \to \infty} a_n = \frac{1+0}{1-0} = 1$.
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?
    Answer: 3 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 $L=3$.
17. Find the limit of the sequence $a_n = \frac{1}{n} \ln\left(\frac{n!}{n^n}\right)$.
    Answer: -1 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.
18. A sequence is defined by $a_1 = 1$ and $a_{n+1} = \frac{3a_n + 2}{a_n + 2}$. If the sequence converges, what is its limit?
    Answer: 2 Let the limit of the sequence be $L$. Taking the limit of both sides of the recurrence relation gives $L = \frac{3L + 2}{L + 2}$. We solve this equation for $L$: $L(L+2) = 3L+2 \implies L^2 + 2L = 3L+2 \implies L^2 - L - 2 = 0$. This factors into $(L-2)(L+1) = 0$, giving possible limits of $L=2$ and $L=-1$. Since $a_1=1$ is positive, and the recurrence relation will produce a positive $a_{n+1}$ if $a_n$ is positive, all terms of the sequence are positive. Therefore, the limit must be $L=2$.
19. What is the limit of the sequence $a_n = \left(\frac{n-1}{n+1}\right)^n$?
    Answer: $e^{-2}$ 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 $e^{-2}$.
20. Consider the sequence $b_n = \frac{\ln(n^3 + n^2)}{\ln(n^4 + 2n^3 + 1)}$. To find $\lim_{n \to \infty} b_n$, which approach correctly identifies the limit?
    Answer: Rewrite as $\frac{3\ln(n) + \ln(1 + 1/n)}{4\ln(n) + \ln(1 + 2/n + 1/n^4)}$ which approaches $\frac{3}{4}$ Using logarithm properties: $b_n = \frac{\ln(n^3) + \ln(1 + 1/n)}{\ln(n^4) + \ln(1 + 2/n + 1/n^4)} = \frac{3\ln(n) + \ln(1 + 1/n)}{4\ln(n) + \ln(1 + 2/n + 1/n^4)}$. As $n \to \infty$, $\ln(1 + 1/n) \to 0$ and $\ln(1 + 2/n + 1/n^4) \to 0$, so the limit becomes $\frac{3\ln(n)}{4\ln(n)} = \frac{3}{4}$. Choice A is incorrect because you can't factor constants out of logarithms that way. Choice B applies L'Hôpital's rule incorrectly. Choice D ignores the different powers in the logarithmic arguments.

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