# Maclaurin Series for a natural logarithm

Can anyone please help me with this question?

Find the Maclaurin series and the interval of convergence for $f(x) = \ln(1-7x^9)$

I thought the answer was

$$\sum_{n=1}^{\infty} (-1)^n \frac{7x^{9n}}{n}$$

but it seems that my homework assignment website will not accept that answer. I also am not sure how to find the interval of convergence. I know that $\ln|1-x|$ converges for $|x| < 1$, but I cannot figure out the interval of convergence for my current problem.

Any help or insight is greatly appreciated! :)

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## 2 Answers

$$\ln(1-x)=-(x+x^2/2+x^3/3+\cdots)$$ doesn't it?

As for the interval of convergence, we would need $$|7x^9|<1$$ which is $$|x|<\left({1\over 7}\right)^{{1\over 9}}$$

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Your series is right for $ln(1-x)$ but I'm still not sure how to write it for $ln(1-7x^9)$. –  Christina Apr 7 '14 at 21:41
Plug in $7x^9$ in place os $x$. –  user140943 Apr 7 '14 at 21:46

this may not matter anymore but it doesnt work because you have (-1)^n. It should be (-1)^(2n+1) in order to start off positive instead of negative. I had the same problem and that was the solution. Here was my full answer: (-1)^(2n+1)(7)^n(x^(9n))/n

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