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! :)


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

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

  • $\begingroup$ Your series is right for $ln(1-x)$ but I'm still not sure how to write it for $ln(1-7x^9)$. $\endgroup$ – Christina Apr 7 '14 at 21:41
  • $\begingroup$ Plug in $7x^9$ in place os $x$. $\endgroup$ – 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: $\frac{(-1)^{2n+1}7^nx^{9n}}{n}$


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