My book mentioned that

"If -1<x<1, then (1+x)^n = 1 + nx + n(n-1)(x^2)/2! + n(n-1)(n-2)(x^3)/3! + ... + n(n-1)(n-2)...(n-r+1)(x^r)/r! + ..."

Why -1<x<1? Why is this restriction on x necessary?

Thanks in advance.

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- August 31st 2012, 02:12 AMstartanewwwbinomial theorem for all powers
My book mentioned that

"If -1<x<1, then (1+x)^n = 1 + nx + n(n-1)(x^2)/2! + n(n-1)(n-2)(x^3)/3! + ... + n(n-1)(n-2)...(n-r+1)(x^r)/r! + ..."

Why -1<x<1? Why is this restriction on x necessary?

Thanks in advance. - August 31st 2012, 02:49 AMProve ItRe: binomial theorem for all powers
- August 31st 2012, 02:58 AMstartanewwwRe: binomial theorem for all powers
In that case, shouldn't the restriction be "x =/= +infinity or -infinity (when n is just a small number)"?

If -1<x<1, will the value of (1+x)^n be very small?

I really don't understand.. - August 31st 2012, 03:11 AMProve ItRe: binomial theorem for all powers
The series depends on x. But the series itself is only convergent if it follows specific conditions, which means that it will only be convergent for particular values of x (the ones which will satisfy the conditions). I suggest you research convergence tests for infinite series.

- August 31st 2012, 04:10 AMPlatoRe: binomial theorem for all powers
- August 31st 2012, 04:21 AMProve ItRe: binomial theorem for all powers
The OP was actually referring to the GENERALISED Binomial Theorem, which is an infinite series.