# Maclaurin Series for 1/|1+x| as the latter is not differentiable at x=−1. However, would it be appropriate for me to refer 1/|1+x| as 'not a smooth' curve (discontinuity at x=1) hence the non-existence of a Maclaurin or in any case a Taylor Series?

Maclaurin series for $\frac{1}{|1+x|}$
However, would it be appropriate for me to refer $\frac{1}{|1+x|}$ as 'not a smooth' curve (discontinuity at $x=1$) hence the non-existence of a Maclaurin or in any case a Taylor Series?
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Karli Moreno
In order to write down a Taylor series for $f$ centered at $a$, we only need $f$ to have derivatives of all orders at $a$. (This, in turn, requires that for every $n$ the derivative ${f}^{\left(n\right)}$ be defined in some neighborhood of $a$, so that ${f}^{\left(n+1\right)}\left(a\right)$ makes sense.) The function $f\left(x\right)=1/|1+x|$ meets this requirement at every point $a\ne -1$.
For example, with $a=0$ the series is
$\begin{array}{}\text{(1)}& \sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}{x}^{n}\end{array}$
which, not coincidentally, is the same series as for 1/(1+x).
I have not yet considered the question of whether the series actually converges to $f$. This can be addressed using the geometric series formula: when $|x|<1$, the series (1) has the sum $1/\left(1+x\right)$, which in this range agrees with $f\left(x\right)$.