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I've run into something confusing.

The problem is that I have to find the power series representation of $g(x)$ using the given $f(x)$, specifically $g(x) = \ln(1 - 3x)$ using $f(x) = \frac{1}{1-x}$.

Now, I did it, but my result is incorrect: $$\ln(1-3x) = \int \frac{dx}{1-3x} = \int \sum 3^{k}x^{k} = \sum \frac{3^{k}x^{k+1}}{k+1}$$

The answer at the back of my textbook is: $$\sum \frac{3^k x^k}{k}$$

My result for the interval of convergence was however correct, which leads me to believe that there is maybe a mistake in the book's answer?

Any help is appreciated, thanks!

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You're simply neglecting the chain rule: $$ \int\frac{dx}{1-3x} = \frac {-1} 3 \ln(1-3x) + C. $$

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  • $\begingroup$ Sorry to keep going with this, but the $-\frac{1}{3}$, though I understand where it came from, is yielding a completely different answer. I've been studying for 5 straight hours now so I could be doing it completely wrong, but this is what I've got: $$-\frac{1}{3} \int \sum 3^{k}x^{k}$$ Which leads to... $$-\frac{1}{3} \sum \frac{3^{k}x^{k+1}}{k+1}$$ $\endgroup$ – KingDan Apr 9 '14 at 22:48
  • $\begingroup$ you have to move the constant to the other side so it'll look like this: $$ \ln(1-3x) = -3 \int \frac{dx}{1-3x} = -3 \int \left( \sum_{k=0}^{\infty} 3^k x^k \right)\, dx = - \sum_{k=0}^{\infty} \frac{3^{k+1} x^{k+1}}{k+1} $$ To arrive at the answer in the book, just shift the index $$ - \sum_{k=0}^{\infty} \frac{3^{k+1} x^{k+1}}{k+1} = -\sum_{k=1}^{\infty} \frac{3^k x^k}{k} $$ $\endgroup$ – Dylan Apr 12 '14 at 23:01
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Note that $$\ln(1-3x) = -3\int \dfrac{dx}{1-3x}$$ and then shift the index of summation.

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Use the chain rule: $$\ln(1-3x)=\int\dfrac{dx}{1-3x}$$ Let $u=1-3x,du=-3dx$, thus giving $$\ln(1-3x)=\dfrac{-1}{3}\int\dfrac{du}{u}=\dfrac{-1}{3}\ln|1-3x|+C$$ Thus, $$\ln(1-3x)=\int\dfrac{dx}{1-3x}=\sum\dfrac{3^kx^k}{k}$$

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