Integral Test - Calculus 2

1. Consider two series: $\sum_{n=2}^{\infty} \frac{1}{n \ln n}$ and $\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}$. Using the integral test, what can be concluded about the convergence of these series?
   Answer: The first series diverges and the second series converges When you encounter series with logarithmic terms, the integral test is often your best tool. The key insight is that small changes in the exponent of logarithmic terms can dramatically affect convergence behavior. For the first series $\sum_{n=2}^{\infty} \frac{1}{n \ln n}$, we test the corresponding integral $\int_2^{\infty} \frac{1}{x \ln x} dx$. Using the substitution $u = \ln x$ (so $du = \frac{1}{x}dx$), this becomes $\int_{\ln 2}^{\infty} \frac{1}{u} du = \ln u \Big|_{\ln 2}^{\infty} = \infty$. Since the integral diverges, the series diverges. For the second series $\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}$, we evaluate $\int_2^{\infty} \frac{1}{x (\ln x)^2} dx$. Using the same substitution, we get $\int_{\ln 2}^{\infty} \frac{1}{u^2} du = -\frac{1}{u} \Big|_{\ln 2}^{\infty} = \frac{1}{\ln 2}$. This integral converges, so the series converges. Answer choice A is wrong because only the second integral converges. Choice B incorrectly claims both diverge—while the rate of approach to zero matters, you must actually compute the integrals to determine convergence. Choice D reverses the correct conclusion. Study tip: Remember that $\frac{1}{n \ln n}$ behaves like the harmonic series (diverges), while adding any positive power to the logarithm, like $(\ln n)^2$, typically ensures convergence. The integral test makes these distinctions precise through substitution techniques.
2. Consider the series $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}$ where $p$ is a positive constant. For which values of $p$ does the integral test guarantee convergence of this series?
   Answer: $p > 1$ To apply the integral test, we evaluate $\int_2^{\infty} \frac{1}{x(\ln x)^p} dx$. Using substitution $u = \ln x$, $du = \frac{1}{x}dx$, this becomes $\int_{\ln 2}^{\infty} \frac{1}{u^p} du$. This integral converges when $p > 1$ and diverges when $p \leq 1$. Choice B is incorrect because when $p = 1$, the integral diverges. Choice C is too broad as it includes values $0 < p \leq 1$ where the series diverges. Choice D is too restrictive as convergence occurs for all $p > 1$, not just $p \geq 2$.
3. Consider the function $g(x) = \frac{x^2 + 1}{x^3 + x^2 + x}$ for $x \geq 1$. To determine the convergence of $\sum_{n=1}^{\infty} g(n)$ using the integral test, which statement correctly identifies why the integral test can or cannot be applied?
   Answer: The integral test can be applied because $g(x)$ is positive, continuous, and eventually decreasing For $x \geq 1$, $g(x) = \frac{x^2 + 1}{x(x^2 + x + 1)} > 0$ and is continuous. For large $x$, $g(x) \approx \frac{x^2}{x^3} = \frac{1}{x}$, which is decreasing. More rigorously, $g'(x) = \frac{-x^4 - 2x^3 - 3x^2 - 1}{(x^3 + x^2 + x)^2} < 0$ for all $x \geq 1$, confirming $g(x)$ is decreasing. Therefore, all conditions for the integral test are satisfied. Choice A incorrectly claims $g(x)$ is not decreasing. Choice C gives an irrelevant reason. Choice D is incorrect since the integral test only requires the function to be well-behaved for $x \geq 1$.
4. Using the reasoning of the Integral Test (or the p-series test), for which value of $k$ does the series $\sum_{n=1}^\infty \frac{n^k}{n^3+1}$ diverge?
   Answer: $k=2$ For large $n$, the term $a_n = \frac{n^k}{n^3+1}$ behaves like $\frac{n^k}{n^3} = \frac{1}{n^{3-k}}$. This is a p-series with $p = 3-k$. A p-series diverges if $p \le 1$. So, we need $3-k \le 1$, which means $k \ge 2$. Of the choices given, only $k=2$ satisfies this condition. For $k=2$, the integral $\int_1^\infty \frac{x^2}{x^3+1} dx$ diverges, confirming the series diverges.
5. What is the result of applying the Integral Test to the series $\sum_{n=1}^\infty \frac{1}{\sqrt{n}(1+\sqrt{n})}$?
   Answer: The series diverges because the corresponding integral diverges. The integral is $\int_1^\infty \frac{1}{\sqrt{x}(1+\sqrt{x})} dx$. Use substitution with $u = 1+\sqrt{x}$, so $du = \frac{1}{2\sqrt{x}} dx$, or $2du = \frac{1}{\sqrt{x}} dx$. The integral becomes $\int_2^\infty \frac{2}{u} du = [2 \ln|u|]_2^\infty$. This evaluates to $\infty - 2\ln(2)$, which diverges. Therefore, the series diverges by the Integral Test.
6. If the Integral Test is successfully applied to a series $\sum_{n=1}^\infty a_n$ and shows convergence, which of the following statements is not necessarily true?
   Answer: The value of the series' sum is equal to the value of the corresponding integral. The Integral Test determines whether a series converges or diverges, but it does not find the sum. The value of the sum is generally not equal to the value of the integral. If a series converges, its tail (A) must also converge, its terms must approach zero (C), and its sequence of partial sums must be bounded and convergent (D).
7. A student claims that since $\int_1^{\infty} \frac{\sin^2 x}{x^2} dx$ converges, the series $\sum_{n=1}^{\infty} \frac{\sin^2 n}{n^2}$ must also converge by the integral test. Which statement best explains the validity of this reasoning?
   Answer: The reasoning is incorrect because $f(x) = \frac{\sin^2 x}{x^2}$ is not monotonic for $x \geq 1$ When you encounter a question about the integral test, remember that this test has specific conditions that must be satisfied - the function must be positive, continuous, and decreasing (monotonic) on the interval of interest. The integral test states that for a function $f(x)$ that is positive, continuous, and decreasing for $x \geq N$, the series $\sum_{n=N}^{\infty} f(n)$ and the integral $\int_N^{\infty} f(x) dx$ either both converge or both diverge. The critical issue here is that $f(x) = \frac{\sin^2 x}{x^2}$ is not monotonically decreasing for $x \geq 1$. While $\frac{1}{x^2}$ decreases, the $\sin^2 x$ factor oscillates between 0 and 1, causing the entire function to oscillate rather than decrease consistently. This violates the monotonicity requirement, making the integral test inapplicable. Choice A is wrong because $f(x)$ fails the monotonicity condition. Choice B misses the point entirely - having the same function values doesn't guarantee the integral test applies if the conditions aren't met. Choice C incorrectly suggests $\sin^2 x$ must be constant, which isn't a requirement of the integral test. The correct answer is D because the lack of monotonicity invalidates the student's reasoning. Study tip: Always verify all three conditions for the integral test: positive, continuous, AND decreasing. Oscillating functions, even if they involve decreasing factors, typically fail the monotonicity requirement.
8. Let $p(x)$ be a polynomial of degree $d > 1$ with a positive leading coefficient, and $p(x) > 0$ for $x \ge 1$. According to the Integral Test, for what values of $k$ does the series $\sum_{n=1}^\infty \frac{n^k}{p(n)}$ converge?
   Answer: $k - d < -1$ For large $x$, the function $f(x) = \frac{x^k}{p(x)}$ behaves like $\frac{x^k}{c x^d} = C x^{k-d}$ for some constants $c, C > 0$. We can apply the Integral Test by analyzing the convergence of $\int_1^\infty x^{k-d} dx$. This is a p-integral with power $p = -(k-d) = d-k$. The integral converges if this power is greater than 1, so $d-k > 1$. Rearranging this gives $k < d-1$, which is equivalent to $k-d < -1$.
9. For which of the following series are all the conditions of the Integral Test (positive, continuous, eventually decreasing) satisfied?
   Answer: $\sum_{n=2}^\infty \frac{1}{n \sqrt{\ln n}}$ A: The factorial function is not defined for non-integers, so a continuous function $f(x)$ cannot be formed. B: The terms $n \cos(1/n)$ do not approach 0, so the series diverges by the nth Term Test, and the corresponding function is not decreasing. C: The function $f(x) = (3-\sin x)/x$ is not guaranteed to be decreasing due to the oscillating numerator. D: The function $f(x) = \frac{1}{x \sqrt{\ln x}}$ is positive, continuous, and decreasing for $x \ge 2$ because the denominator is a product of positive, increasing functions.
10. Consider the series $\sum_{n=1}^\infty a_n$ where $a_n = \int_n^{n+1} \frac{1}{x^2+1} dx$. Which statement about the convergence of this series is correct?
    Answer: The series converges, and its sum is $\int_1^\infty \frac{1}{x^2+1} dx$. This is a telescoping sum in disguise. The N-th partial sum is $S_N = \sum_{n=1}^N \int_n^{n+1} \frac{1}{x^2+1} dx = \int_1^2 + \int_2^3 + ... + \int_N^{N+1} = \int_1^{N+1} \frac{1}{x^2+1} dx$. The sum of the series is $S = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \int_1^{N+1} \frac{1}{x^2+1} dx = \int_1^\infty \frac{1}{x^2+1} dx$. Since this integral converges (to $\pi/4$), the series converges to the exact value of the integral.
11. Let $f(x)$ be a positive, continuous, and decreasing function for $x \ge 1$. If $\sum_{n=1}^\infty f(n)$ converges to $S$ by the Integral Test, which of the following provides the tightest guaranteed bound for $S$?
    Answer: $S < f(1) + \int_1^\infty f(x) dx$ The sum of the series can be visualized as the sum of areas of rectangles. For a decreasing function, the sum of right-hand rectangles is an underestimate of the integral, and the sum of left-hand rectangles is an overestimate. Specifically, $\sum_{n=2}^\infty f(n) < \int_1^\infty f(x) dx$. Adding the first term, $S = f(1) + \sum_{n=2}^\infty f(n) < f(1) + \int_1^\infty f(x) dx$. This is a standard result from the geometric argument for the Integral Test.
12. Consider $S_1 = \sum_{n=2}^\infty \frac{1}{n \sqrt{\ln n}}$ and $S_2 = \sum_{n=2}^\infty \frac{1}{n (\ln n)^{3/2}}$. Which statement is correct according to the Integral Test?
    Answer: $S_1$ diverges and $S_2$ converges. Both series are of the form $\sum \frac{1}{n(\ln n)^p}$. The Integral Test leads to the integral $\int \frac{1}{x(\ln x)^p} dx$, which after substitution $u = \ln x$ becomes $\int \frac{1}{u^p} du$. This p-integral converges if $p > 1$ and diverges if $p \le 1$. For $S_1$, $p=1/2$, so it diverges. For $S_2$, $p=3/2$, so it converges.
13. The series $\sum_{n=1}^\infty ne^{-n^2}$ is tested for convergence using the Integral Test. Which statement correctly describes the result?
    Answer: The series converges because the corresponding integral converges. The corresponding integral is $\int_1^\infty xe^{-x^2} dx$. Using u-substitution with $u = -x^2$, $du = -2x dx$, the integral becomes $-\frac{1}{2} \int_{-1}^{-\infty} e^u du = [-\frac{1}{2} e^{-x^2}]_1^\infty = 0 - (-\frac{1}{2}e^{-1}) = \frac{1}{2e}$. Since the integral converges to a finite value, the series also converges. The sum of the series is not equal to the value of the integral.
14. The series $\sum_{n=2}^{\infty} \frac{1}{n^2 \sqrt{\ln n}}$ is being tested for convergence using the integral test. After applying the substitution $u = \ln x$ to the corresponding integral, which of the following correctly represents the transformed integral?
    Answer: $\int_{\ln 2}^{\infty} \frac{1}{e^u \sqrt{u}} du$ Starting with $\int_2^{\infty} \frac{1}{x^2 \sqrt{\ln x}} dx$, we use substitution $u = \ln x$, so $x = e^u$ and $dx = e^u du$. The integral becomes $\int_{\ln 2}^{\infty} \frac{1}{(e^u)^2 \sqrt{u}} \cdot e^u du = \int_{\ln 2}^{\infty} \frac{e^u}{e^{2u} \sqrt{u}} du = \int_{\ln 2}^{\infty} \frac{1}{e^u \sqrt{u}} du$. Choice A stops before simplifying $\frac{e^u}{e^{2u}} = \frac{1}{e^u}$. Choice C incorrectly transforms $\sqrt{\ln x} = \sqrt{u}$ to $u^{3/2}$. Choice D omits the exponential factor entirely.
15. If the Integral Test is applied to $\sum_{n=N}^\infty a_n$ using the function $f(x)$, which of the following is a necessary condition for the test to be valid?
    Answer: $f(x)$ must be positive for all $x \ge N$. The three conditions for the Integral Test are that the function $f(x)$ corresponding to the series terms $a_n$ must be continuous, positive, and eventually decreasing. Being positive for all $x \ge N$ (where the test is being applied) is a necessary condition. The function only needs to be decreasing for $x \ge N$, not for all $x > 0$. The series must not be alternating, and there is no condition on the magnitude of the first term.
16. The series $\sum_{n=2}^\infty \frac{1}{n(\ln n)^p}$ is analyzed using the Integral Test. For which values of the constant $p$ does the corresponding improper integral converge?
    Answer: $p > 1$ To apply the Integral Test, we evaluate $\int_2^\infty \frac{1}{x(\ln x)^p} dx$. We use the substitution $u = \ln x$, so $du = \frac{1}{x} dx$. The limits of integration become $\ln 2$ and $\infty$. The integral transforms to $\int_{\ln 2}^\infty \frac{1}{u^p} du$. This is a p-integral, which is known to converge if and only if the power $p > 1$. Therefore, the original series converges for $p > 1$.
17. A student attempts to prove the convergence of $\sum_{n=1}^\infty \frac{2+\cos(n)}{n^2}$ using the Integral Test with $f(x) = \frac{2+\cos(x)}{x^2}$. Why is this application of the Integral Test invalid?
    Answer: The function $f(x)$ is not eventually decreasing. To apply the Integral Test, the function $f(x)$ must be positive, continuous, and eventually decreasing. The function $f(x) = \frac{2+\cos(x)}{x^2}$ is positive and continuous for $x \ge 1$. However, due to the oscillating $\cos(x)$ term in the numerator, the function is not monotonically decreasing for all $x$ greater than some $N$. The derivative contains an oscillating term that prevents it from staying negative. Therefore, a key condition of the Integral Test is not met.
18. For the series $\sum_{n=1}^{\infty} \frac{1}{n^{1+1/n}}$, a student attempts to apply the integral test by evaluating $\int_1^{\infty} \frac{1}{x^{1+1/x}} dx$. What is the primary obstacle in using this approach?
    Answer: The integral $\int_1^{\infty} \frac{1}{x^{1+1/x}} dx$ is too difficult to evaluate analytically The integral test is a powerful tool for determining series convergence, but it requires that you can actually evaluate or analyze the corresponding integral. Sometimes the biggest obstacle isn't whether the function meets the test's conditions, but whether the integral itself is manageable. Let's examine why option D is correct. The integral $\int_1^{\infty} \frac{1}{x^{1+1/x}} dx$ presents a significant analytical challenge because the exponent $1+\frac{1}{x}$ is not constant—it's a function of $x$. This creates an integral that doesn't fit standard forms and lacks obvious substitutions or integration techniques. While the integral might converge or diverge, determining which requires advanced methods beyond typical calculus courses, making the integral test impractical here. Now for the incorrect options: Option A is wrong because $f(x) = \frac{1}{x^{1+1/x}}$ is clearly positive for all $x \geq 1$ since we're taking the reciprocal of a positive quantity. Option B is incorrect—while checking if the function is eventually decreasing requires some work, this isn't the primary obstacle; the function's behavior can be analyzed using derivatives. Option C is false because the function is continuous at $x = 1$; as $x \to 1$, we get $\frac{1}{1^{1+1/1}} = \frac{1}{1^2} = 1$, which matches the function's value at $x = 1$. When applying the integral test, always consider whether the resulting integral is actually solvable with available techniques. If not, explore alternative convergence tests like comparison or ratio tests.
19. The Integral Test is applied to $\sum_{n=1}^\infty \frac{1}{n^p}$. The corresponding integral $\int_1^\infty \frac{1}{x^p} dx$ is found to converge. What must be true about $p$?
    Answer: The conclusion is valid for any $p > 1$, and the series converges. The integral test establishes the p-series test. The function $f(x) = 1/x^p$ is positive, continuous, and decreasing for $x \ge 1$ as long as $p > 0$. The integral $\int_1^\infty x^{-p} dx$ converges if $-p < -1$, which simplifies to $p > 1$. This result holds for any real number $p > 1$, not just integers or rationals. While verifying the function is decreasing is required, this is straightforward for $p > 0$.
20. Consider the series $\sum_{n=1}^\infty \frac{\ln n}{n}$. Applying the Integral Test leads to the improper integral $\int_1^\infty \frac{\ln x}{x} dx$. What is the correct evaluation of this integral and the conclusion about the series?
    Answer: The integral diverges, so the series diverges. To evaluate $\int_1^\infty \frac{\ln x}{x} dx$, use the substitution $u = \ln x$, so $du = \frac{1}{x} dx$. The integral becomes $\int_0^\infty u du = [\frac{u^2}{2}]_0^\infty$, which diverges to $\infty$. Therefore, by the Integral Test, the series diverges. The fact that the first term is zero does not invalidate the test; convergence is determined by the tail of the series, so we can start the test at $n=2$.

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