Ratio Test - Calculus 2

1. A student applies the ratio test to $\sum_{n=1}^{\infty} \frac{n^3 \cdot 5^n}{(n+2)!}$ and claims the limit is $5$. What is the error in the student's reasoning?
   Answer: The student ignored the factorial growth in the denominator, which dominates the polynomial and exponential growth For $a_n = \frac{n^3 \cdot 5^n}{(n+2)!}$, we have $\frac{a_{n+1}}{a_n} = \frac{(n+1)^3 \cdot 5^{n+1}}{(n+3)!} \cdot \frac{(n+2)!}{n^3 \cdot 5^n} = 5 \cdot \frac{(n+1)^3}{n^3} \cdot \frac{1}{n+3}$. As $n \to \infty$, this approaches $5 \cdot 1 \cdot 0 = 0$, not $5$. The factorial growth dominates both polynomial and exponential terms. Choice A is irrelevant since terms are positive. Choice B has the factorial expansion backwards. Choice D is unrelated to the calculation error.
2. Let $f(x) = \sum_{n=0}^\infty \frac{x^{2n}}{(n!)^2}$. Use the Ratio Test to find the interval of convergence.
   Answer: $(-\infty, \infty)$ When you encounter a power series and need to find its interval of convergence, the Ratio Test is your primary tool. This test determines where the series converges by examining the limit of consecutive term ratios. For $f(x) = \sum_{n=0}^\infty \frac{x^{2n}}{(n!)^2}$, we have $a_n = \frac{x^{2n}}{(n!)^2}$. Applying the Ratio Test: $\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x^{2(n+1)}}{((n+1)!)^2} \cdot \frac{(n!)^2}{x^{2n}}\right| = \left|\frac{x^2 \cdot (n!)^2}{((n+1)!)^2}\right| = \frac{|x|^2}{((n+1)^2)}$ Taking the limit as $n \to \infty$: $\lim_{n \to \infty} \frac{|x|^2}{(n+1)^2} = 0$ Since this limit is 0 for any finite value of $x$, the series converges for all real numbers. Answer A is wrong because the series clearly converges beyond just $x=0$ — the factorial squared in the denominator grows so rapidly that it dominates any polynomial growth in the numerator. Answers B and C are incorrect because they suggest finite intervals of convergence, which would occur if our limit equaled some positive constant, giving us a finite radius of convergence. The presence of $(n!)^2$ in the denominator creates such rapid growth that no finite $x$ value can make the series diverge. Strategy tip: When you see factorials in denominators of power series, expect infinite radius of convergence. Factorials grow much faster than exponentials or polynomials, overwhelming any finite $x$ values.
3. Consider $\sum_{n=0}^{\infty} \frac{x^n}{n^2 + 1}$ where $x > 0$. For what values of $x$ does the ratio test guarantee convergence?
   Answer: $0 < x < 1$ only, since the ratio test gives limit $x$ For $a_n = \frac{x^n}{n^2 + 1}$, we have $\frac{a_{n+1}}{a_n} = \frac{x^{n+1}}{(n+1)^2 + 1} \cdot \frac{n^2 + 1}{x^n} = x \cdot \frac{n^2 + 1}{(n+1)^2 + 1}$. As $n \to \infty$, this approaches $x \cdot 1 = x$. The ratio test guarantees convergence when $x < 1$ and divergence when $x > 1$. When $x = 1$, the test is inconclusive. Choice B incorrectly includes $x = 1$. Choice C ignores that $x$ can be arbitrarily large. Choice D incorrectly computes the limit.
4. The Ratio Test is applied to the series $\sum_{n=1}^{\infty} \frac{(x-3)^{2n}}{n \cdot 4^n}$. Which of the following conditions on $x$ guarantees absolute convergence based on the result of the test?
   Answer: $|x-3| < 2$ To apply the Ratio Test, we compute the limit $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$. Here, $a_n = \frac{(x-3)^{2n}}{n \cdot 4^n}$. The ratio is $\left| \frac{(x-3)^{2(n+1)}}{(n+1)4^{n+1}} \cdot \frac{n 4^n}{(x-3)^{2n}} \right| = \left| \frac{(x-3)^2}{4} \cdot \frac{n}{n+1} \right|$. As $n \to \infty$, the limit is $L = \frac{|x-3|^2}{4}$. For absolute convergence, we need $L < 1$, which means $\frac{|x-3|^2}{4} < 1$, or $|x-3|^2 < 4$. Taking the square root of both sides gives $|x-3| < 2$.
5. For the series $\sum_{n=1}^{\infty} \frac{(n!)^2 \cdot 4^n}{(2n)! \cdot n}$, what happens when we apply the ratio test?
   Answer: The limit is $1$ and the ratio test is inconclusive, requiring another convergence test When you encounter a series with factorials and exponentials like this one, the ratio test is often your best tool since it handles factorials elegantly through cancellation. For the ratio test, you need to find $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$ where $a_n = \frac{(n!)^2 \cdot 4^n}{(2n)! \cdot n}$. Setting up the ratio: $\frac{a_{n+1}}{a_n} = \frac{((n+1)!)^2 \cdot 4^{n+1}}{(2(n+1))! \cdot (n+1)} \cdot \frac{(2n)! \cdot n}{(n!)^2 \cdot 4^n}$ Since $(n+1)! = (n+1) \cdot n!$ and $(2n+2)! = (2n+2)(2n+1)(2n)!$, this becomes: $\frac{(n+1)^2 \cdot 4 \cdot n}{(2n+2)(2n+1)(n+1)} = \frac{4(n+1)n}{(2n+2)(2n+1)} = \frac{4n(n+1)}{2(n+1)(2n+1)} = \frac{2n}{2n+1}$ Taking the limit: $\lim_{n \to \infty} \frac{2n}{2n+1} = \lim_{n \to \infty} \frac{2}{2 + \frac{1}{n}} = 1$ Since the limit equals 1, the ratio test is inconclusive, making D correct. Choice A incorrectly calculates the limit as $\frac{1}{4}$ (perhaps from mishandling the $4^n$ terms). Choice B gives 4 (likely from an algebra error with the factorial simplification). Choice C claims the limit is 0, which would require the numerator to shrink much faster than the denominator. Study tip: When the ratio test gives you a limit of 1, don't panic—try the root test next, or look for comparison tests. Many important series require multiple approaches to determine convergence.
6. The Ratio Test is applied to a power series $\sum_{n=0}^\infty c_n (x-a)^n$, and it is found that $\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = K$, where $K$ is a finite positive number. What is the radius of convergence, $R$?
   Answer: $R = 1/K$ When you encounter a power series with the Ratio Test, you're finding the radius of convergence by determining where the series converges absolutely. The key insight is understanding how the Ratio Test connects to the radius formula. For a power series $\sum_{n=0}^\infty c_n (x-a)^n$, when you apply the Ratio Test, you examine $\lim_{n \to \infty} \left| \frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n} \right| = |x-a| \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|$. This simplifies to $|x-a| \cdot K$, where $K$ is your given limit. The Ratio Test tells us the series converges when this expression is less than 1: $|x-a| \cdot K < 1$, which means $|x-a| < \frac{1}{K}$. Since the radius of convergence $R$ is the distance from the center $a$ where convergence occurs, we have $R = \frac{1}{K}$. This makes choice B correct. Choice A ($R = K$) confuses the coefficient limit with the radius directly. Choice C ($R = 1$) might seem tempting since the Ratio Test uses 1 as the convergence threshold, but this ignores the coefficient behavior entirely. Choice D ($R = \infty$) would only occur if $K = 0$, but the problem states $K$ is a finite positive number. Remember this inverse relationship: when the ratio of consecutive coefficients is large, the radius of convergence is small, and vice versa. The formula $R = \frac{1}{\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|}$ is fundamental for power series analysis.
7. The Ratio Test is applied to $\sum_{n=1}^\infty \frac{(n+2)!}{n! 10^n}$. What is the conclusion?
   Answer: The series converges because the limit of the ratio is $1/10$. When you encounter a series with factorials and exponential terms like this one, the Ratio Test is often your best tool because it handles these expressions elegantly through cancellation. To apply the Ratio Test, you need to find $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$ where $a_n = \frac{(n+2)!}{n! \cdot 10^n}$. First, identify $a_{n+1} = \frac{(n+3)!}{(n+1)! \cdot 10^{n+1}}$. Now compute the ratio: $\frac{a_{n+1}}{a_n} = \frac{(n+3)!}{(n+1)! \cdot 10^{n+1}} \cdot \frac{n! \cdot 10^n}{(n+2)!}$ Simplifying: $\frac{(n+3)!}{(n+2)!} = n+3$ and $\frac{n!}{(n+1)!} = \frac{1}{n+1}$, so: $\frac{a_{n+1}}{a_n} = \frac{(n+3)}{(n+1)} \cdot \frac{1}{10} = \frac{n+3}{10(n+1)}$ Taking the limit: $\lim_{n \to \infty} \frac{n+3}{10(n+1)} = \frac{1}{10}$ Since this limit is less than 1, the series converges, confirming answer D. Answer A is wrong because while $a_n$ not approaching 0 would cause divergence, we haven't shown that here. Answer B incorrectly states the limit as 10 instead of 1/10 – a common error from inverting the ratio. Answer C mistakes the limit value; the limit is 1/10, not 1, so the test is conclusive. Study tip: With factorial and exponential series, always look for telescoping cancellations when computing ratios. The algebra often simplifies dramatically, leaving you with a clean limit to evaluate.
8. A common error in testing a series $\sum a_n$ is to calculate $\lim_{n \to \infty} a_n$ and use it as the result of the Ratio Test. Consider the series $\sum_{n=1}^\infty \frac{n}{n^2+1}$. Which statement below is true?
   Answer: The Ratio Test is inconclusive, but the common error would incorrectly suggest convergence. When you encounter series convergence problems, it's crucial to distinguish between different tests and apply them correctly. The Ratio Test examines $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$, while the common error mentioned simply looks at $\lim_{n \to \infty} a_n$. For the series $\sum_{n=1}^\infty \frac{n}{n^2+1}$, let's apply the Ratio Test properly. We have $a_n = \frac{n}{n^2+1}$, so: $\frac{a_{n+1}}{a_n} = \frac{\frac{n+1}{(n+1)^2+1}}{\frac{n}{n^2+1}} = \frac{(n+1)(n^2+1)}{n((n+1)^2+1)}$ After algebraic simplification, this ratio approaches 1 as $n \to \infty$, making the Ratio Test inconclusive. The "common error" refers to using $\lim_{n \to \infty} a_n$ instead. Here, $\lim_{n \to \infty} \frac{n}{n^2+1} = \lim_{n \to \infty} \frac{1/n}{1+1/n^2} = 0$. Someone making this error might think the series converges because the limit is zero. Looking at the answer choices: Choice C correctly identifies that the Ratio Test is inconclusive and the common error suggests convergence. Choice A is wrong because the Ratio Test doesn't show divergence. Choice B is wrong because the Ratio Test doesn't show convergence. Choice D is wrong because the common error (getting limit = 0) would suggest convergence, not divergence. Study tip: Always remember that the Ratio Test uses the ratio of consecutive terms, not the limit of the terms themselves. When the ratio limit equals 1, try other tests like the Limit Comparison Test.
9. What is the result of applying the Ratio Test to the series $\sum_{n=1}^{\infty} \frac{n^n}{n! e^n}$?
   Answer: The limit of the ratio is 1, so the test is inconclusive. Let $a_n = \frac{n^n}{n! e^n}$. The ratio $\frac{a_{n+1}}{a_n}$ is $\frac{(n+1)^{n+1}}{(n+1)! e^{n+1}} \cdot \frac{n! e^n}{n^n} = \frac{(n+1)^{n+1}}{(n+1)n! e \cdot e^n} \cdot \frac{n! e^n}{n^n} = \frac{(n+1)^n}{e \cdot n^n} = \frac{1}{e} \left(\frac{n+1}{n}\right)^n = \frac{1}{e} \left(1 + \frac{1}{n}\right)^n$. The limit as $n \to \infty$ is $L = \lim_{n \to \infty} \frac{1}{e} \left(1 + \frac{1}{n}\right)^n$. Using the well-known limit $\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$, we find $L = \frac{1}{e} \cdot e = 1$. Since the limit is 1, the Ratio Test is inconclusive.
10. When applying the Ratio Test to a series $\sum a_n$, the limit is calculated as $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{k}$, where $k$ is a positive constant from the series definition. Which condition on $k$ guarantees divergence of the series based on this test?
    Answer: $0 < k < 1$ The Ratio Test states that a series diverges if the limit $L$ is greater than 1. Given $L = 1/k$, the condition for divergence is $1/k > 1$. Since $k$ is a positive constant, we can multiply both sides by $k$ without changing the inequality direction, which gives $1 > k$. Combining this with the given condition that $k$ is positive, the range for $k$ that guarantees divergence is $0 < k < 1$.
11. For what values of the positive constant $c$ does the series $\sum_{n=1}^\infty \frac{(cn)^n}{n!}$ converge, according to the Ratio Test?
    Answer: $0 < c < 1/e$ When you encounter a series with factorials in the denominator and exponential-like terms in the numerator, the Ratio Test is typically your best tool for determining convergence. For the series $\sum_{n=1}^\infty \frac{(cn)^n}{n!}$, apply the Ratio Test by computing $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$. Here, $a_n = \frac{(cn)^n}{n!}$, so: $\frac{a_{n+1}}{a_n} = \frac{(c(n+1))^{n+1}}{(n+1)!} \cdot \frac{n!}{(cn)^n} = \frac{c^{n+1}(n+1)^{n+1}}{(n+1) \cdot n!} \cdot \frac{n!}{c^n n^n} = c \cdot \frac{(n+1)^n}{n^n} = c\left(1 + \frac{1}{n}\right)^n$ As $n \to \infty$, we know that $\left(1 + \frac{1}{n}\right)^n \to e$. Therefore, $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = ce$. The Ratio Test tells us the series converges when this limit is less than 1, so we need $ce < 1$, which gives us $c < \frac{1}{e}$. Since we're told $c$ is positive, the answer is $0 < c < \frac{1}{e}$. Choice A is wrong because large values of $c$ make the ratio exceed 1. Choice B uses the wrong threshold—it should be $\frac{1}{e} \approx 0.368$, not 1. Choice C completely reverses the inequality direction. Study tip: When applying the Ratio Test to series with factorials, watch for the limit $\left(1 + \frac{1}{n}\right)^n \to e$—it appears frequently and is key to finding the correct convergence threshold.
12. Let $\sum a_n$ and $\sum b_n$ be two series with positive terms. If $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 0.5$ and $\lim_{n \to \infty} \frac{b_{n+1}}{b_n} = 1.5$, what can be concluded about the series $\sum c_n = \sum (a_n + b_n)$ by applying the Ratio Test?
    Answer: The limit of the ratio for $\sum c_n$ is 1.5, so the series diverges. When you encounter a question about the Ratio Test applied to a sum of series, you need to understand how limits behave when series are combined. The Ratio Test examines $\lim_{n \to \infty} \frac{c_{n+1}}{c_n}$ where $c_n = a_n + b_n$. To find this limit, we need to analyze $\frac{c_{n+1}}{c_n} = \frac{a_{n+1} + b_{n+1}}{a_n + b_n}$. As $n \to \infty$, we can determine which term dominates. Since $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 0.5 < 1$, the terms $a_n$ decrease exponentially fast. Meanwhile, $\lim_{n \to \infty} \frac{b_{n+1}}{b_n} = 1.5 > 1$ means $b_n$ grows exponentially. For large $n$, the $b_n$ terms completely dominate both the numerator and denominator, so $\frac{a_{n+1} + b_{n+1}}{a_n + b_n} \approx \frac{b_{n+1}}{b_n}$. Therefore, $\lim_{n \to \infty} \frac{c_{n+1}}{c_n} = 1.5$. Since this limit exceeds 1, the Ratio Test tells us $\sum c_n$ diverges, making C correct. A is wrong because limit properties do apply here through dominance analysis. B incorrectly adds the two ratios (0.5 + 1.5 = 2.0), but ratios don't add this way. D incorrectly averages the ratios to get 1.0, which is mathematically invalid. Study tip: When applying the Ratio Test to sums of series, identify which series dominates by comparing their ratio limits to 1. The faster-growing series determines the behavior of the combined series.
13. For the series $\sum_{n=1}^{\infty} \frac{n! p^n}{n^n}$, the Ratio Test is used to determine the range of values of $p$ for which the series converges. What is this range?
    Answer: $|p| < e$ Let $a_n = \frac{n! p^n}{n^n}$. The ratio is $\left| \frac{(n+1)! p^{n+1}}{(n+1)^{n+1}} \cdot \frac{n^n}{n! p^n} \right| = \left| \frac{(n+1) p \cdot n^n}{(n+1)^{n+1}} \right| = \left| p \cdot \frac{(n+1) n^n}{(n+1)(n+1)^n} \right| = |p| \left( \frac{n}{n+1} \right)^n = |p| \frac{1}{\left(1 + \frac{1}{n}\right)^n}$. As $n \to \infty$, the limit is $L = |p| \cdot \frac{1}{e} = \frac{|p|}{e}$. For convergence, we require $L < 1$, which implies $\frac{|p|}{e} < 1$, or $|p| < e$.
14. For the series $\sum_{n=1}^{\infty} \frac{n! \cdot a^n}{n^n}$ where $a > 0$, the ratio test gives $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{a}{e}$. For which values of $a$ does this series converge?
    Answer: $0 < a < e$ only, since the ratio test requires the limit to be less than 1 Using Stirling's approximation or careful ratio analysis: $\frac{a_{n+1}}{a_n} = \frac{(n+1)! \cdot a^{n+1}}{(n+1)^{n+1}} \cdot \frac{n^n}{n! \cdot a^n} = a \cdot \frac{n+1}{n+1} \cdot \left(\frac{n}{n+1}\right)^n = a \cdot \left(1 + \frac{1}{n}\right)^{-n}$. As $n \to \infty$, $\left(1 + \frac{1}{n}\right)^{-n} \to \frac{1}{e}$, so the limit is $\frac{a}{e}$. For convergence by ratio test, we need $\frac{a}{e} < 1$, so $a < e$. Choice B incorrectly includes the boundary case. Choice C reverses the inequality. Choice D ignores the ratio test conclusion.
15. The series $\sum_{n=1}^{\infty} a_n$ satisfies $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{3}{4}$. If we define $b_n = (a_n)^2$, what can we conclude about $\sum_{n=1}^{\infty} b_n$ using the ratio test?
    Answer: The series $\sum b_n$ converges since $\lim_{n \to \infty} \left| \frac{b_{n+1}}{b_n} \right| = \frac{9}{16} < 1$ Since $b_n = (a_n)^2$, we have $\frac{b_{n+1}}{b_n} = \frac{(a_{n+1})^2}{(a_n)^2} = \left(\frac{a_{n+1}}{a_n}\right)^2$. Therefore, $\lim_{n \to \infty} \left| \frac{b_{n+1}}{b_n} \right| = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16} < 1$, so $\sum b_n$ converges. Choice B uses the original ratio instead of squaring it. Choice C incorrectly suggests the limit is 1. Choice D inverts and then squares incorrectly.
16. For which of the following series is the Ratio Test guaranteed to be inconclusive?
    Answer: $\sum_{n=1}^\infty \frac{n^3 - 4n}{n^5 + n^2 + 1}$ The Ratio Test helps determine series convergence by examining $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$. When this limit equals 1, the test is inconclusive—it can't determine whether the series converges or diverges. For option B, $\sum_{n=1}^\infty \frac{n^3 - 4n}{n^5 + n^2 + 1}$, let's find the ratio limit. The general term behaves like $\frac{n^3}{n^5} = \frac{1}{n^2}$ for large n. When you compute $\left|\frac{a_{n+1}}{a_n}\right|$, you get approximately $\frac{n^2}{(n+1)^2}$, which approaches 1 as $n \to \infty$. This makes the Ratio Test inconclusive, though the series actually converges by comparison to $\sum \frac{1}{n^2}$. Option A involves exponential decay ($3^n$ in denominator), so the ratio approaches $\frac{1}{3} < 1$, indicating convergence. Option C has factorials in the numerator and exponentials in the denominator—using Stirling's approximation shows the ratio approaches $\frac{1}{e} < 1$, so it converges. Option D has factorials in the denominator growing much faster than the exponential numerator, making the ratio approach 0, indicating convergence. The key insight is that polynomial ratios (like in B) typically yield ratio limits of 1, making the test inconclusive, while exponential or factorial terms usually produce definitive results. When studying series, remember that the Ratio Test works best with factorials and exponentials but often fails with polynomial or rational functions.
17. The Ratio Test is applied to the series $\sum_{n=1}^\infty \frac{n^k}{2^n}$ for a fixed real constant $k$. What is the value of $L = \lim_{n \to \infty} |a_{n+1}/a_n|$?
    Answer: $1/2$ When you encounter a Ratio Test problem, you're examining the behavior of consecutive terms in a series to determine convergence. The Ratio Test finds $L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$, and if $L < 1$, the series converges. For the series $\sum_{n=1}^\infty \frac{n^k}{2^n}$, we have $a_n = \frac{n^k}{2^n}$. Let's calculate the ratio: $\frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)^k}{2^{n+1}}}{\frac{n^k}{2^n}} = \frac{(n+1)^k}{2^{n+1}} \cdot \frac{2^n}{n^k} = \frac{(n+1)^k}{n^k} \cdot \frac{2^n}{2^{n+1}} = \left(\frac{n+1}{n}\right)^k \cdot \frac{1}{2}$ As $n \to \infty$, we know that $\frac{n+1}{n} = 1 + \frac{1}{n} \to 1$. Therefore, $\left(\frac{n+1}{n}\right)^k \to 1^k = 1$ regardless of the value of $k$. This gives us $L = 1 \cdot \frac{1}{2} = \frac{1}{2}$, confirming answer (C). Looking at the wrong answers: (A) $0$ would suggest the exponential dominates completely, ignoring the polynomial factor's contribution to the ratio. (B) $k/2$ incorrectly assumes the exponent $k$ affects the limiting ratio, but polynomial growth rates don't impact the exponential's dominance. (D) $1$ forgets about the factor of $\frac{1}{2}$ that comes from the exponential terms. Key insight: In ratio tests involving exponentials and polynomials, the exponential base determines the limit—polynomial powers wash out in the ratio calculation.
18. For the series $\sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!} k^n$, what is the range of values for the constant $k$ for which the Ratio Test guarantees absolute convergence?
    Answer: $|k| < 4$ Let $a_n = \frac{(n!)^2 k^n}{(2n)!}$. We compute the limit of the ratio $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$. The ratio simplifies to $\left| k \cdot \frac{((n+1)!)^2}{(2n+2)!} \cdot \frac{(2n)!}{(n!)^2} \right| = \left| k \cdot \frac{(n+1)^2}{(2n+2)(2n+1)} \right|$. As $n \to \infty$, the fraction $\frac{(n+1)^2}{(2n+2)(2n+1)} = \frac{n^2+2n+1}{4n^2+6n+2}$ approaches $1/4$. Thus, $L = \frac{|k|}{4}$. The series converges when $L < 1$, so $\frac{|k|}{4} < 1$, which implies $|k| < 4$.
19. Consider the power series $\sum_{n=1}^\infty \frac{(-1)^n (x+2)^n}{n \cdot 3^n}$. The Ratio Test is used to find the radius of convergence. What is the radius?
    Answer: 3 When you encounter a power series with the form $\sum_{n=1}^\infty a_n(x-c)^n$, the Ratio Test helps you find the radius of convergence by examining $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$. For this series $\sum_{n=1}^\infty \frac{(-1)^n (x+2)^n}{n \cdot 3^n}$, identify $a_n = \frac{(-1)^n}{n \cdot 3^n}$. To apply the Ratio Test, calculate: $\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(-1)^{n+1}}{(n+1) \cdot 3^{n+1}} \cdot \frac{n \cdot 3^n}{(-1)^n}\right| = \frac{n}{(n+1) \cdot 3}$ Taking the limit: $\lim_{n \to \infty} \frac{n}{(n+1) \cdot 3} = \lim_{n \to \infty} \frac{1}{3(1 + 1/n)} = \frac{1}{3}$ The radius of convergence is the reciprocal of this limit: $R = \frac{1}{1/3} = 3$. This makes (D) 3 correct. Now for the wrong answers: (A) ∞ would occur if the limit were 0, which happens when coefficients decrease faster than exponentially. (B) 1/3 is the actual limit value—a common mistake is confusing this with the radius itself. (C) 1 might tempt you if you forget the factor of 3 in the denominator of $a_n$. Study tip: Remember the Ratio Test gives you $\frac{1}{R}$, so always take the reciprocal of your limit to find the actual radius. Also, watch for series where the center isn't at $x = 0$—here it's at $x = -2$, but this doesn't affect the radius calculation.
20. The series $\sum_{n=1}^\infty \frac{(-10)^n}{n!}$ is analyzed using the Ratio Test. What is the strongest conclusion that can be drawn directly from the application of this test?
    Answer: The series converges absolutely. Let $a_n = \frac{(-10)^n}{n!}$. The limit of the ratio is $L = \lim_{n \to \infty} \left| \frac{(-10)^{n+1}}{(n+1)!} \cdot \frac{n!}{(-10)^n} \right| = \lim_{n \to \infty} \left| \frac{-10}{n+1} \right| = \lim_{n \to \infty} \frac{10}{n+1} = 0$. Since $L = 0 < 1$, the Ratio Test proves that the series converges absolutely. Conditional convergence is a weaker conclusion, and the test is conclusive, not inconclusive.

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