what is the interval of convergence of summation k=0 to infinity of [(x-1)^(3k)]

Question

what is the interval of convergence of summation k=0 to infinity of [(x-1)^(3k)]/[(2k!)*(k^2)]. I tried to use the ratio test but got stuck.

Explanation / Answer

So using ratio test should give you: lim(k->infinity) [(x-1)^(3(k+1))]/[(2(k+1))!*((k+1)^2)] * [(2k!)*(k^2)]/[(x-1)^(3k)] = lim(k->infinity) [(x-1)^(3k+3))]/[(2k+2))!*((k+1)^2)] * [(2k!)*(k^2)]/[(x-1)^(3k)] = lim(k->infinity) [(x-1)^3]/[(2k+2)(2k+1)] * [(k^2)/(k+1)^2] = 0 [(x-1)^3]/[(2k+2)(2k+1)] goes to 0 as k goes to infinity [(k^2)/(k+1)^2] goes to 1 as k goes to infinity So the whole thing goes to 0. Notice how it doesn't matter what value x is so x can be any real number. Thus the interval of convergence is (-infinity, infinity)

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