Question

What is the Binomial Expansion of `(1+r)^-1`?

Answer 1

If `abs(r) < 1` then
`(1+r)^(-1) = sum_(n=0)^oo(-1)^nr^n = 1 - r + r^2 - r^3 +...`

If `abs(r) > 1` then
`(1+r)^-1 = sum_(n=0)^oo(-1)^nr^(-n-1) = 1/r-1/(r^2)+1/(r^3)-...`

Explanation:

If `abs(r) < 1` then `sum_(n=0)^oo(-1)^nr^n` converges and

`(1+r)sum_(n=0)^oo(-1)^nr^n=1 + r - r + r^2 - r^2 + r^3 - r^3 +... = 1`

So `(1+r)^-1 = 1/(1+r) = sum_(n=0)^oo(-1)^nr^n`

If `abs(r) > 1` then `(1+r)^(-1) = 1/(1+r) = (1/r)/(1/r + 1) = (1/r)*(1+1/r)^(-1)`

and `abs(1/r) < 1`, so we can use our previous formula to get:

`(1+r)^(-1) = (1/r)sum_(n=0)^oo(-1)^nr^(-n)=sum_(n=0)^oo(-1)^nr^(-n-1)`

That just leaves the case `abs(r) = 1`, i.e. `r = +-1`

`(1+1)^-1 = 1/2`

`(1-1)^-1 = 1/0` is undefined