# Taylor series for $\frac{1}{(1+x)^t}$

I'm having some trouble finding the Taylor series for the following function at zero (Maclaurin series). \begin{equation} \frac{1}{(1+x)^t} \end{equation} Where $t$ is a constant that is greater than zero.

An approach is to obtain successive derivatives of $\displaystyle f(x)=\frac{1}{(1+x)^t}$ with respect to $x$ and then plug it in the Taylor series expansion near $0$. You easily find by induction that $$f^{(n)}(x)=(-1)^n\frac{t(t+1) \cdots (t+n-1)}{(1+x)^{t+n}}, \quad n=1,2,\ldots,$$ leading to
$$\frac{1}{(1+x)^t}=1+\sum_{n=1}^{\infty}(-1)^n\frac{t(t+1) \cdots (t+n-1)}{n!}x^n$$
for $x$ near $0$.
• You are welcome! You may write, for the third degree Taylor polynomial : $$\frac{1}{(1+x)^t}=1-t x+\frac{1}{2} t(t+1) x^2-\frac{1}{6} t(t+1)(t+2) x^3+\mathcal{O}(x^4)$$ – Olivier Oloa Feb 25 '15 at 13:03
• Please, if you mean those in my preceding comment, you just have to put $n=1,\,2,\,3$ in the formula for $f^{(n)}(x)$ given in my answer. – Olivier Oloa Feb 25 '15 at 13:13
• I'm not sure how you get the first term ($-tx$) because I do not get that when I plug in the value for $n$. – friedmanfanboy Feb 25 '15 at 13:17