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What is the radius of convergence and interval of convergence for the power series ∞∑n=0m!(2x-1)^m?(a) 3, 12(b) 1, 0.87(c) 2, 5.4(d) 0, 1/2I got this question during an interview.This question is from Discrete Probability in section Discrete Probability of Discrete Mathematics

Answer»

Right CHOICE is (d) 0, 1/2

Explanation: SUPPOSE, L=limn→∞ |(m+1)!(2x+1)^m+1/m!(2x+1)^m|

= limm→∞∣(m+1)m!(2x-1)/m!|

= |2x-1|limm→∞(m+1)

So, this power series will only CONVERGE if x=1/2. We know that every power series will converge for x=a and in this case a=1/2. Remember that we GET a from (x−a)^n. In this case, the radius of convergence is R=0 and the INTERVAL of convergence is x=1/2.



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