I'm trying to find the Maclaurin series for $f(x) =x\ln(1+x^3)$.
Unfortunately notes (and it seems online) are very inconsistent about whether the equality to memorize is $\ln(1+x)=\sum^{\infty}_{1} (-1)^{n-1}\frac{x^n}{n=1}$, or $=\sum^{\infty}_{n=1} (-1)^{n+1}\frac{x^n}{n}$, or $=\sum^{\infty}_{n=0} (-1)^{n}\frac{x^n}{n}$. Does the +/- matter in the first two? Are any of these correct/wrong?
Knowing that, I also have trouble subbing in the leading x. My answer key shows: $$ \ln(1+x^3)=\sum^{\infty}_{n=1} (-1)^{n+1} \frac{(x^3)^n}{n} $$ Which seems fine (assuming the first of my three assumed formulas are correct.But when it comes time to bring in the x, I don't know why the start point changes: $$ x \ln(1+x^3)=\sum^{\infty}_{n=0} (-1)^{n+1} \frac{x^{3n+1}}{n} $$ Why can't I just multiply the x in? Won't the denominator be zero on the first term?
The correct formula is $\ln(1+x) = \sum^{\infty}_{n=1} (-1)^{n+1}\frac{x^n}{n}$ for $|x| < 1$. Therefore $$x \ln(1+x^3) = \sum^{\infty}_{n=1} (-1)^{n+1}\frac{x^{3n+1}}{n}$$ for $|x|<1$. There is no reason to change the starting point in the sum.
If you want to remember the expression of $\ln(1+x)$ just remember that for $|x|<1$, $$\frac{1}{1-x} = \sum^{\infty}_{n=0} x^n$$ hence $$\frac{1}{1+x} = \sum^{\infty}_{n=0} (-1)^n x^n$$
Now, because $(\ln(1+x))'=\frac{1}{1+x}$, you can integrate the serie of $\frac{1}{1+x}$ to find the one of $\ln(1+x)$.