Maclaurin Series Expansion $\ln(8+x^3)$.

Question

Find Maclaurin Series Expansion of function $\ln(8+x^3)$. For which $x$ series converges? For which $ x $ ,$ f(x)$ equals sum of this series?

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$$ (\ln(8+x^3))'=\frac{3x^2}{8+x^3} =\frac{\frac{3}{8}x^2}{1-(-\frac{x^3}{8})}=$$ $$ =\sum_{n=1}^{\infty}\frac{3x^2}{8}(-1)^n\frac{x^{3n}}{8^n}$$ and $$ \sum_{n=1}^{\infty}\int\frac{3x^2}{8}(-1)^n\frac{x^{3n}}{8^n}dx=$$ $$=\sum_{n=1}^{\infty}\frac{(-1)^n3}{8^{n+1}}\int x^{3n+2}dx=$$ $$=\sum_{n=1}^{\infty}\frac{(-1)^n3}{8^{n+1}}\frac{x^{3n+3}}{3n+3}+C$$

for $x=0$ $$\ln(8+0^3)= \sum_{n=1}^{\infty}\frac{(-1)^n3}{8^{n+1}}\frac{0^{3n+3}}{3n+3}+C $$ $$ C=0 $$

$8+x^3>0 $

$x>-2 $

and

$|\frac{-x^3}{8}|<1$

$x<2$

I have to find if it converges only at $x=2$, $x=-2$ is out of domain? For which $x$ series converges? For which $ x $ ,$ f(x)$ equals sum of this series? From Abel theorem $\forall$ x which converges $f(x)$=sum of series?

Answer 1

HINT:

$$\ln(8+x^3)=\ln8+\ln\left(1+\dfrac{x^3}8\right)$$

See Why does the taylor series of $\ln (1 + x)$ only approximate it for $-1<x \le 1$?

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