Solving for the Interval of Convergence
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 Preview
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 ...
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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