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    f'(x)=1/(x+1) (i)

    By the binomial theorem

    f'(x)=1-x+x^2-x^3 ... infinite expansion (ii)

    integrate (i) f(x)=ln(x+1)

    integrate (ii) f(x)=x-x^2/2+x^3/3-x^4/4 ... infinite expansion

    therefore ln(x+1)=x-x^2/2+x^3/3-x^4/4 ... infinite expansion

    Therefore ln(x)= (x-1)-(x-1)^2/2+(x-1)^3/3-(x-1)^4/4 ... infinite expansion

    therefore the summation of ((-1)^(x-1))/x from limits 1 to infinity is ln2

    Valid for limits 0<=x<=2
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    (Original post by MM04926412)
    f'(x)=1/(x+1) (i)

    By the binomial theorem

    f'(x)=1-x+x^2-x^3 ... infinite expansion (ii)

    integrate (i) f(x)=ln(x+1)

    integrate (ii) f(x)=x-x^2/2+x^3/3-x^4/4 ... infinite expansion

    therefore ln(x+1)=x-x^2/2+x^3/3-x^4/4 ... infinite expansion

    Therefore ln(x)= (x-1)-(x-1)^2/2+(x-1)^3/3-(x-1)^4/4 ... infinite expansion

    therefore the summation of ((-1)^(x-1))/x from limits 1 to infinity is ln2

    Valid for limits 0<=x<=2
    The bold line is the only one I would question, but I think it's okay. I'm fairly sure everything's valid.
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    If I ignore the limits of the expansion I can conclude that ln0=-1-1/2-1/3-1/4-1/5 ... infinitum does my answer actually mean anything here?

    Can I used this as a proof that the harmonic series is divergent?
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    The harmonic series is 1+1/2+1/3+1/4+1/5+1/6 etc.

    f'(x)=1/(x+1) (i)


    By the binomial theorem

    f'(x)=1-x+x^2-x^3 ... infinite expansion (ii)

    integrate (i) f(x)=ln(x+1)

    integrate (ii) f(x)=x-x^2/2+x^3/3-x^4/4 ... infinite expansion

    therefore ln(x+1)=x-x^2/2+x^3/3-x^4/4 ... infinite expansion

    Therefore ln(x)= (x-1)-(x-1)^2/2+(x-1)^3/3-(x-1)^4/4 ... infinite expansion

    Therefore ln0= -1-1/2-1/3... infinite

    therefore -ln0= 1+1/2+1/3+1/4

    Therefore if either ln0 or the harmonic series are divergent/undefined, so is the other

    ln0 is undefined

    the harmonic series is divergent
 
 
 
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