# 1)*1k4хх-1 in the series In (1 x)4 = 4to obtain a power series for Incentered at x2. What is the interval of convergence for the new power series?Replace x by2k2k 1Choose the correct power series below.(-1)** 1(x-2)*-1x-1*4EA.В.kkk 1k 1(-1) 1x-2)*2k(-1)** (x- 1)*42kKk 1k 1The interval of convergence is (0,4] .(Simplify your answer. Type your answer in interval notation.)

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In the screenshot it shows a problem with a value to substitute into a given series. In the answer explanation, it mentioned that the interval of the original series is -1<x<=1. Why is this/ how can this interval be found from this alternating series? help_outlineImage Transcriptionclose1)*1k 4 х х -1 in the series In (1 x)4 = 4 to obtain a power series for In centered at x2. What is the interval of convergence for the new power series? Replace x by 2 k 2 k 1 Choose the correct power series below. (-1)** 1(x-2)* -1x-1* 4E A. В. k k k 1 k 1 (-1) 1x-2)* 2k (-1)** (x- 1)* 4 2kK k 1 k 1 The interval of convergence is (0,4] . (Simplify your answer. Type your answer in interval notation.) fullscreen
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Step 1

Given that

Step 2

Substitute x by x/2-1in the above series and simplify it help_outlineImage Transcriptionclose(-1)8*4 -4M 1 2 14 In 1+ k 2 (-1) - 4Σ 2 In 2 k k-1 (-1) (x-2) 2 k k+1 n 2 fullscreen
Step 3

Now to find the interval of convergenc... help_outlineImage Transcriptionclose(-1)f (x-2) 2 k k+1 a 1) x-2) 2 (k+1) +1+ (-1)9 x-2) 23-1 (k+1) k+1-1 (-1)d (x-2) 2 k (-1)0 (x-2)f 2 3(k+1) 2.k )(-1) (x-2) a (-1)° (x-2) k 2(k+1) _ 1 k (-1)) (x-2) Dividing the numerator and denominator by k k 1 22 k k + 2) 1 2 1+ k fullscreen

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