Maclaurin series for a function

Question

Provided I have the function

\begin{equation*} f(x)=(1+x)^{1/x}, \end{equation*}

and I want to calculate a 3rd order Maclaurin series, how can that be done without taking direct derivatives (as this seems hard..). I know that

\begin{equation*} (1+x)^{1/x}=e^{ln(1+x)/x}, \end{equation*}

and the Maclaurin series for $e^x$ is easy to prove, so I think it's a good direction..

Now you should be able to calculate the derivatives of $e^{\frac{\ln(1+x)}{x}}$ — George1811, May 07 '15 at 09:11
still a bit complicated... It seems like there could be a nicer way here... — byteSlayer, May 07 '15 at 09:12

score 3 · Accepted Answer · answered May 07 '15 at 09:34

$$\frac{\ln(1+x)}x=1-\frac x2+\frac{x^2}3-\frac{x^3}4+o(x^3)$$ Substitute $-\frac x2+\frac{x^2}3-\frac{x^3}4$ to $u$ in the development of $\mathrm e^u$, first computing the succesive powers of $u$: \begin{align*} u^2&=\frac{x^2}4-\frac{x^3}3+o(x^3),\\ u^3&-\frac{x^3}8+o(x^3), \end{align*} so that $$(1+x)^{\tfrac 1x}=\mathrm e\Bigl(1+u+\frac{u^2}2+\frac{u^3}6+o(u)\Bigr)=\mathrm e\Bigl(1-\frac x2+\frac{11}{48}x^2-\frac 7{16}x^3+o(x^3)\Bigr).$$

Maclaurin series for a function

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