Question
Asked Nov 17, 2019
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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?

1)*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.)
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1)*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.)

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Expert Answer

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Step 1

Given that

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k+1 -1)° In(1+x)4-)x k k1

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Step 2

Substitute x by x/2-1in the above series and simplify it

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(-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

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Step 3

Now to find the interval of convergenc...

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(-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

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