# Solving for the Interval of Convergence

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Question: Find the open interval of convergence and test the endpoints for absolute and conditional convergence:

(x-2)^n / (2^n)(n^2)

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##### Solution Summary

In a step-wise response of about 145 words, this solution illustrates how an open interval of convergence is found. All calculations required are included.

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Solution. Let Fn(x)=(x-2)^n/{(2^n)(n^2)}. Let y=(x-2)/2, then

Fn(x)=Gn(y)=y^n/n^2. Consider series

G1(y)+G2(y)+...+Gn(y)+... (1)

The coefficient of the nth term is a(n)=1/n^2. Obviously, the limit L of a(n+1)/a(n) is 1 as n tends to ...

###### Education

- BSc , Wuhan Univ. China
- MA, Shandong Univ.

###### Recent Feedback

- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
- "Thank you"
- "Thank you very much for your valuable time and assistance!"

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