题目
题目

MAT136H5 S 2025 - All Sections 6.1 preparation check

多重下拉选择题

In this question we work through the steps for solving the following problem: Find the interval of convergence and the radius of convergence of the power series ∞ ∑ n=0 8n n! xn  Solution outline: To find interval and radius of convergence, we usually start by using the Ratio Test: a) Evaluate the limit lim n→∞| an+1 an |=lim n→∞| 8n+1xn+1 (n+1)! 8nxn n! |      [ Select ] 1 / 4 0 1 infinity 8 1 / 8 b) What does the Ratio Test and your answer in (a) tell you about the series?    [ Select ] The series converges only when -1 < x < 1 The series converges only when -8 < x < 8 The series converges for all values of x The series only converges when x = 0 The series does not converge for any values of x c) What is the radius of convergence for the series?  R = infinity d) What is the interval of convergence for the series?    [ Select ] (-infinity, infinity) [-1, 1] (-8, 8) [-8, 8] x = 0 (-1, 1) Note: The interval of convergence is the set of values of x for which the series converges.  e) Consider your answer to (d). Does the series converge when x=3 ?    [ Select ] Yes There is not enough information No

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As we work through the given power series, a clear path is to recognize its form. The general term is a_n = (8^n / n!) x^n, so the series is sum_{n=0}^∞ (8^n x^n)/n! = sum_{n=0}^∞ ( (8x)^n / n!). Option by option reasoning: Option a) The choice '0' for lim_{n→∞} |a_{n+1}/a_n|' obviously targets the ratio test limit. Compute the ratio: |a_{n+1}/a_n| = | (8^{n+1} x^{n+1}/(n+1)!) / (8^n x^n / n!) | = | (8x)/(n+1) |. As n → ∞, this tends to 0 for every fixed x, since the denominator grows without bound while the numerator stays constant. Hence the ......Login to view full explanation

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Question text 5Marks a) Find the radii of convergence of the following series:[math: ∑n=1∞n2+2n3n2+2xn]\displaystyle \sum_{n=1}^{\infty} \dfrac{n^2+2n}{3n^2+2}x^n radius = Answer 1[input][math: ∑n=1∞n2+2n3n+2xn]\displaystyle \sum_{n=1}^{\infty} \dfrac{n^2+2n}{3^n+2}x^n radius = Answer 2[input]b) Suppose a power series [math: ∑n=1∞anxn]\displaystyle \sum_{n=1}^{\infty} a_n x^n is convergent for [math: x=−3]x= -3 and divergent for [math: x=5]x = 5.For [math: x=−1]x = -1, the power series Answer 3[select: , is convergent., is divergent., can be convergent or divergent depending on the coefficients.]For [math: x=−6]x = -6, the power series Answer 4[select: , is convergent., is divergent., can be convergent or divergent depending on the coefficients.]For [math: x=3]x = 3, the power series Answer 5[select: , is convergent., is divergent., can be convergent or divergent depending on the coefficients.]Notes Report question issue Question 8 Notes

Which of the following statements is not correct?

Which of the following statements is NOT true?

In this question we work through the steps for solving the following problem: Find the interval of convergence and the radius of convergence of the power series ∞ ∑ n=0 (x−4)n 3n2    Solution outline: To find interval and radius of convergence, we usually start by using the Ratio Test: a) Evaluate the limit lim n→∞| an+1 an |=lim n→∞| (x−4)n+1 3(n+1)2 (x−4)n 3n2 |      |x-4| b) What does the Ratio Test and your answer in (a) tell you about convergence when |x−4|<1  ?    [ Select ] Nothing (because the ratio test inconclusive) The power series converges The power series diverges c) What does the Ratio Test and your answer in (a) tell you about convergence when |x−4|>1  ?    [ Select ] The power series converges Nothing (because the ratio test inconclusive) The power series diverges d) What does the Ratio Test and your answer in (a) tell you about convergence when |x−4|=1 ?    [ Select ] The power series diverges The power series converges Nothing (because the ratio test inconclusive) e) For what values of x is |x−4|<1 ? (Solve the inequality)  [ Select ] x < 5 -4 < x < 5 -4 < x < 4 3 < x < 5 2 < x < 6 f) When x=5  the power series in the question becomes ∞ ∑ n=0 1 3n2  . Does this series converge or diverge?    [ Select ] Diverges Converges g) When x=3   the power series in the question becomes ∞ ∑ n=0 (−1)n 3n2   . Does this series converge or diverge?    [ Select ] Diverges Converges h) Summarizing the results we have found, for what values of x does the power series in the question converge? [ Select ] Converges for x in the interval (-4,4) Converges for x in the interval [3,5] Converges for x in the interval (-4,5] Converges for x in the interval [3,5) Converges for x in the interval (3,5] Converges for x in the interval [-4,4] Note: This is the interval of convergence.

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