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Câu 1: Khảo sát sự hội tụ các tích phân suy rộng sau:
a. $\int_{0}^{+\infty} \frac{3 x \sqrt{x^{5}}+x^{3}+ 1}{x^{6}+ 4} d x$
b. $\int_{1}^{+\infty} \frac{2 +\sin (4 x+ 2)}{x^{3}+ 5} d x$
c. $\int_{1}^{+\infty} \frac{\ln \left(x^{2}+ 3 x\right)}{5 x^{5}+x} d x$
d. $\int_{1}^{+\infty} \frac{(x+ 2) \cos ^{2} x}{x^{3}+ 3 x} d x$
Câu 2: Tính các giới hạn sau:
a. $\lim _{x \rightarrow \infty} \frac{4 n- 1}{3 n+ 4}$
b. $\lim _{x \rightarrow \infty} \frac{\sqrt{n^{2}+ 2}+\sqrt{3 n^{2}}}{\sqrt[3]{n^{6}+ 2}-n}$
c. $\lim _{x \rightarrow \infty}\left(\sqrt{n^{2}+ 5}-n\right)$
d. $\lim _{k \rightarrow \infty} \sqrt[k]{\frac{2 k^{2}+k}{k^{5}+ 4 k}}$
Câu 3: Xét xem chuỗi nào là chuỗi CSN. Nếu là cấp số nhân thì tính tổng nếu nó hội tụ.
a. $\sum_{k= 1}^{+\infty} \frac{- 2}{4^{k+ 1}}$
b. $\sum_{k= 1}^{+\infty}(- 1)^{\frac{k}{2}} \frac{e^{k- 1}}{2^{3 k}}$
c. $\sum_{k= 1}^{+\infty}(- 1)^{\frac{k}{2}} \frac{k^{2}}{2^{k+ 1}}$
d. $-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\ldots \ldots$.
e. $\sum_{k= 1}^{+\infty}(- 1)^{\frac{5}{2}} \frac{5^{2 k- 1}}{T^{k+ 2}}$
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Step 1Problem 1: Investigate the convergence of the following improper integrals:
a. $$\int_{0}^{+\infty} \frac{3 x \sqrt{x^{5}}+x^{3}+1}{x^{6}+4} d x
Step 2: Determine if the degree of the numerator is less than or equal to the degree of the denominator.
b. $$\int_{1}^{+\infty} \frac{2+\sin (4 x+2)}{x^{3}+5} d x
- The degree of the numerator is 5 (highest power of x is x^(5)). - The degree of the denominator is 6. Since the degree of the numerator is less than the degree of the denominator, the limit as x approaches infinity of the ratio of the leading terms is 0. Therefore, the integral converges by the limit comparison test.
Final Answer
a. Converges. b. Converges. c. Converges. d. Converges. Problem 2: Evaluate the following limits: a. \lim _{x \rightarrow \infty} \frac{4 n- 1}{3 n+ 4} Step 1: Divide both the numerator and the denominator by n. \lim _{n \rightarrow \infty} \frac{4 -\frac{1}{n}}{3 +\frac{4}{n}} Step 2: Take the limit as n approaches infinity. \frac{4 - 0}{3 + 0} = \frac{4}{3} b. \lim _{x \rightarrow \infty} \frac{\sqrt{n^{2}+ 2}+\sqrt{3 n^{2}}}{\sqrt[3]{n^{6}+ 2}-n} Step 1: Divide both the numerator and the denominator by n. \lim _{n \rightarrow \infty} \frac{\sqrt{1 +\frac{2}{n^{2}}}+\sqrt{3}}{\sqrt[3]{1 +\frac{2}{n^{6}}}- 1} Step 2: Take the limit as n approaches infinity. \frac{\sqrt{1 + 0}+\sqrt{3}}{\sqrt[3]{1 + 0}- 1} = \frac{1 +\sqrt{3}}{1 - 1} = \text{DNE} c. \lim _{x \rightarrow \infty}\left(\sqrt{n^{2}+ 5}-n\right) Step 1: Multiply and divide by the conjugate. \lim _{n \rightarrow \infty} \frac{\left(\sqrt{n^{2}+ 5}-n\right)\left(\sqrt{n^{2}+ 5}+n\right)}{\sqrt{n^{2}+ 5}+n} Step 2: Simplify the expression. \lim _{n \rightarrow \infty} \frac{n^{2}+ 5 -n^{2}}{\sqrt{n^{2}+ 5}+n} = \lim _{n \rightarrow \infty} \frac{5}{\sqrt{n^{2}+ 5}+n} Step 3: Divide both the numerator and the denominator by n. \lim _{n \rightarrow \infty} \frac{\frac{5}{n}}{\sqrt{1 +\frac{5}{n^{2}}}+ 1} Step 4: Take the limit as n approaches infinity. \frac{0}{1 + 1} = 0 d. \lim _{k \rightarrow \infty} \sqrt[k]{\frac{2 k^{2}+k}{k^{5}+ 4 k}} Step 1: Divide both the numerator and the denominator by k^5. \lim _{k \rightarrow \infty} \sqrt[k]{\frac{2 k^{- 3}+k^{- 4}}{1 + 4 k^{- 4}}} Step 2: Take the limit as k approaches infinity. \sqrt[k]{0 + 0} = 0 a. \frac{4}{3} b. DNE c. 0 d. 0 Problem 3: Determine if the following series converge or diverge. If they converge, find the sum. a. \sum_{k= 1}^{+\infty} \frac{- 2}{4^{k+ 1}} Step 1: Rewrite the series as a geometric series. \sum_{k= 1}^{+\infty} \frac{- 2}{4^{k+ 1}} = -\frac{1}{2} \sum_{k= 1}^{+\infty} \frac{1}{4^{k}} Step 2: Identify the first term and the common ratio. First term (a) = - 1 / 2. Common ratio (r) = 1 / 4. Step 3: Check if the series converges or diverges. The series converges because |r| < 1. Step 4: Find the sum of the series. S = \frac{a}{1 -r} = \frac{-\frac{1}{2}}{1 -\frac{1}{4}} = -\frac{1}{2} \cdot \frac{4}{3} = -\frac{2}{3} b. \sum_{k= 1}^{+\infty}(- 1)^{\frac{k}{2}} \frac{e^{k- 1}}{2^{3 k}} Step 1: Determine if the series converges or diverges. The series converges by the alternating series test because |(- 1)^(k/ 2) e^(k- 1)/ 2^(3k)| approaches 0 as k approaches infinity. Step 2: Find the sum of the series. The sum of the series is not easily found because the series does not have a simple form. c. \sum_{k= 1}^{+\infty}(- 1)^{\frac{k}{2}} \frac{k^{2}}{2^{k+ 1}} Step 1: Determine if the series converges or diverges. The series converges by the alternating series test because |(- 1)^(k/ 2) k^2 / 2^(k+ 1)| approaches 0 as k approaches infinity. Step 2: Find the sum of the series. The sum of the series is not easily found because the series does not have a simple form. d. -\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\ldots \ldots Step 1: Rewrite the series as a geometric series. -\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\ldots \ldots = \sum_{k= 1}^{+\infty} \frac{(- 1)^{k+ 1}}{2^{k}} Step 2: Identify the first term and the common ratio. First term (a) = - 1 / 2. Common ratio (r) = - 1 / 2. Step 3: Check if the series converges or diverges. The series converges because |r| < 1. Step 4: Find the sum of the series. S = \frac{a}{1 -r} = \frac{-\frac{1}{2}}{1 -\frac{1}{2}} = -\frac{1}{2} \cdot \frac{2}{1} = - 1 e. \sum_{k= 1}^{+\infty}(- 1)^{\frac{5}{2}} \frac{5^{2 k- 1}}{T^{k+ 2}} Step 1: Determine if the series converges or diverges. The series converges by the alternating series test because |(- 1)^(5 / 2) 5^(2k- 1)/T^(k+ 2)| approaches 0 as k approaches infinity. Step 2: Find the sum of the series. The sum of the series is not easily found because the series does not have a simple form. a. Sum: - 2 / 3 b. Converges, but sum is not easily found. c. Converges, but sum is not easily found. d. Sum: - 1 e. Converges, but sum is not easily found.
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