fp2 ln(1+x) mclaurin series question

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thebrahmabull
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#1
As I have marked in the attachment, the book shows that the rth term of the series is (xto the power r)/r but this means if we substitute r as 0 the term is undefined?

Is the book still right?

My second question: I can see the series is valid for x>-1 because ln (1-1) =ln 0 but how can I tell the series is valid for x<1?
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joostan
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(Original post by thebrahmabull)
As I have marked in the attachment, the book shows that the rth term of the series is (xto the power r)/r but this means if we substitute r as 0 the term is undefined?

Is the book still right?

My second question: I can see the series is valid for x>-1 because ln (1-1) =ln 0 but how can I tell the series is valid for x<1?
The series will diverge for |x|&gt;1, but converges for x=1.
You could do so by noting that \dfrac{d}{dx}(\ln(1+x))=\dfrac{1}{1+x}, which you can express as the sum of a GP, and infer from there that the given series is convergent for |x|&lt;1 and check x=-1,1 as special cases.
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tiny hobbit
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#3
(Original post by thebrahmabull)
As I have marked in the attachment, the book shows that the rth term of the series is (xto the power r)/r but this means if we substitute r as 0 the term is undefined?

Is the book still right?

My second question: I can see the series is valid for x>-1 because ln (1-1) =ln 0 but how can I tell the series is valid for x<1?
The first term is x and the next involves x^2 so the general term only applies from r=1 onwards, so r will never equal 0
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