Find Maclaurin series of $f(x)$ = $x\ln(2+x^2)$

Question

I have to find Maclaurin series for $x\ln(2+x^2)$, for what values of $x$ it converges and calculate $f^{(100)}(0)$. I know that $g(x)$ = $\ln(2+x)$ is equal to $$\ln 2 +\sum_{n=1}^{\infty} (-1)^{n+1}\cdot {{({x \over 2})^n}\over n}. $$ Therefore, $x\ln(2+x^2)$ will look like $$x\ln(x+2) + \sum_{n=1}^{\infty} (-1)^{n+1}\cdot {x^{2n+1}\over2^n\cdot n}. $$ Is it correct, and how to do the rest?

Answer 1

Hints: You have the right idea. And if you know for which $t$ the Maclaurin series for $\ln (1+t)$ converges, you can use this to get the values of $x$ for which the series for $\ln\left(1+x^2\right)$ converges. By the way, to find $f^{(100)}(0)$, recall that in general, if you know the coefficients $a_k$ for the Maclaurin series of some function $f(x)$ (as you do here for $f(x) = x\ln\left(2+x^2\right)$), then you know the derivatives of $f$ at $0$. I.e. if $$f(x) = \sum\limits_{k=0}^{\infty}a_k x^k,$$ then what is the formula for $a_k$ in terms of $f^{(k)}(0)$, and hence what is the formula for $f^{(k)}(0)$ in terms of $a_k$?