Binomial expansion of (1+(1/x))^(-1)

[Appleton]

Expand the following functions as a series of ascending powers of x up to and including the term x^3. In each case give the range of values of x for which the expansion is valid.

(1+(1/x))^(-1)

The Attempt at a Solution

1 + (-1)(1/x) + (-1)(-2)(1/x^2)/2 + (-1)(-2)(-3)(1/x^3)/3!
= 1 - (1/x) + (1/x^2) - (1/x^3)
-1< 1/x <1
x<-1 and x>1

Which is invalid, but I can't see why. The way I deduced the ranges of values of x wasn't very rigorous so I suspect that might have something to do with it.

]

 

[Ray Vickson]

Expand the following functions as a series of ascending powers of x up to and including the term x^3. In each case give the range of values of x for which the expansion is valid.

(1+(1/x))^(-1)

The Attempt at a Solution

1 + (-1)(1/x) + (-1)(-2)(1/x^2)/2 + (-1)(-2)(-3)(1/x^3)/3!
= 1 - (1/x) + (1/x^2) - (1/x^3)
-1< 1/x <1
x<-1 and x>1

Which is invalid, but I can't see why. The way I deduced the ranges of values of x wasn't very rigorous so I suspect that might have something to do with it.

]

You were asked for ascending powers of $x$, that is, for the series in terms of $x, x^2, x^3, \ldots$. Instead, you gave descending powers $1/x =x^{-1}, 1/x^2 = x^{-2}, 1/x^3 = x^{-3}, \ldots$.

 

[Appleton]

Thank you, I get it now