<s>[INST] How do I compute the taylor series for $(1+x+x^2)^\frac{1}{x}$ I tried rewriting $(1+x+x^2)^\frac{1}{x}$ as $e^{\frac{1}{x}\ln(1+x+x^2)}$ and then computing the taylor series of $\frac{1}{x}$ and $\ln(1+x+x^2)$ but I'm still not getting the correct answer..
 [/INST] This appears to be the approach you used, but without seeing your work, it is hard to tell where you were having trouble.
Use the series $\log(1+x)=x-\frac{x^2}2+\frac{x^3}3+\cdots$
$$
\begin{align}
\frac1x\log\left(1+x+x^2\right)
&=\frac1x\left(x+x^2-\frac{x^2+2x^3+x^4}2+\frac{x^3+3x^4+3x^5+x^6}3+\cdots\right)\\
&=1+\frac12x-\frac23x^2+\cdots
\end{align}
$$
Then use the series $e^x=1+x+\frac{x^2}2+\cdots$
$$
\begin{align}
\left(1+x+x^2\right)^{1/x}
&=e^{1+\frac12x-\frac23x^2+\cdots}\\
&=e\left(1+\left(\frac12x-\frac23x^2+\cdots\right)+\frac12\left(\frac12x-\frac23x^2+\cdots\right)^2+\cdots\right)\\
&=e\left(1+\frac12x-\frac{13}{24}x^2+\cdots\right)
\end{align}
$$
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