Representation of power series?

Question

Link to problem

I got $∑\frac{(3x^2)^{n+1}}{n+1}$ as my series. But the coefficients do not work for the power series. Why?

Answer 1

For

$$\;|3x^2|<1\iff x^2<\frac13\iff |x|<\frac1{\sqrt3}\;,$$

we get

$$\log(1-3x^2)=-\sum_{n=1}^\infty\frac{(3x^2)^n}n=-3x^2-\frac{9x^4}2-\frac{27x^6}3-\ldots$$

Answer 2

Using Taylor's theorem:

$$f(x)=f(0)+\frac{f'(0)}1x+\frac{f''(0)}{1\times2}x^2+\frac{f'''(0)}{1\times2\times3}x^3+\dots+\frac{f^{(n)}}{n!}x^n+\dots$$

So, we have

$$\frac{d}{dx}\ln(1-3x^2)=\frac{2x}{x^2-\frac13}=\frac4{x-\sqrt{\frac13}}-\frac4{x+\sqrt{\frac13}}$$

$$\frac{d^n}{dx^n}\ln(1-3x^2)=\frac{d^{n-1}}{dx^{n-1}}\frac{d}{dx}\ln(1-3x^2)$$

$$=\frac{d^{n-1}}{dx^{n-1}}\left(\frac4{x-\sqrt{\frac13}}-\frac4{x+\sqrt{\frac13}}\right)$$

$$=\frac{4\times(-1)^{n-1}(n-2)!}{\left(x-\sqrt{\frac13}\right)^n}-\frac{4\times(-1)^{n-1}(n-2)!}{\left(x+\sqrt{\frac13}\right)^n}$$

Now all that's left is to consider $x=0$ and $n=3,4,\dots$ and divide them by $n!$