My question is as follows:

> 
> (g) Show that Taylor's series for the function $f$ in terms of $t$, where
> 
> 
> $$f(t) = \frac{1}{(t+1)}$$
> 
> 
> in the vicinity of $t=1$, is given by
> 
> 
> $$\begin{align}f(t) = \frac{1}{(t+1)} &= \sum\_{n=0}^\infty (-1)^n t^n \\ &= 1 - t + t^2 - t^3 + t^4 + \cdots \end{align}$$
> 
> 
> 

I have shown by induction that for $f(x) = \frac{1}{x+1}$,the $n^{th}$ derivative is

$$f^{(n)}(x) = \frac{(-1)^n n!}{(x+1)^{n+1}}$$

Plugging this result into the Taylor series expansion for $f(x) = \frac{1}{x+1}$ at $x=1$,

$$\frac{1}{x+1} = \sum\_{n=0}^{\infty} \frac{1}{n!} f^{(n)}(x)\bigg|\_{x=1} (x-1)^n $$

I have simplified the expression into

$$ \sum\_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}} (x-1)^n$$

I was expecting the expression to simplify into $\sum\_{n=0}^{\infty} (-1)^n x^{n}$, but I am not getting anywhere.

How do I continue from here to get the expected expression, without using Maclaurin Series?