# binomial series - why is there a range of validity?

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in the form (1 + x)^n the expansion is only valid between -1<x<1 so that the terms get progressively smaller in the series

I dont get why do they need to get smaller?

I dont get why do they need to get smaller?

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in the form (1 + x)^n the expansion is only valid between -1<x<1 so that the terms get progressively smaller in the series

I dont get why do they need to get smaller?

**m.s124**)in the form (1 + x)^n the expansion is only valid between -1<x<1 so that the terms get progressively smaller in the series

I dont get why do they need to get smaller?

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**m.s124**)

in the form (1 + x)^n the expansion is only valid between -1<x<1 so that the terms get progressively smaller in the series

I dont get why do they need to get smaller?

The problem with infinite series is that they only converge and give us a meaningful finite value under very particular conditions.

The infinite series for (1+x)^n is a geometric series with common ratio x.

If |x| > 1 it means that every successive term is bigger in magnitude than the term before it. Summing these up blows the sum up to infinity.

If |x| < 1 it means that successive terms get smaller, and it turns out to be sufficient for convergence.

When n is a positive integer, this sum is finite therefore it will always give a finite value with no restriction on x.

Last edited by RDKGames; 1 month ago

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(Original post by

It is possible to express (1+x)^n as an infinite series of terms in increasing powers of x.

The problem with infinite series is that they only converge and give us a meaningful finite value under very particular conditions.

The infinite series for (1+x)^n is a geometric series with common ratio x.

If |x| > 1 it means that every successive term is bigger in magnitude than the term before it. Summing these up blows the sum up to infinity.

If |x| < 1 it means that successive terms get smaller, and it turns out to be sufficient for convergence.

When n is a positive integer, this sum is finite therefore it will always give a finite value with no restriction on x.

**RDKGames**)It is possible to express (1+x)^n as an infinite series of terms in increasing powers of x.

The problem with infinite series is that they only converge and give us a meaningful finite value under very particular conditions.

The infinite series for (1+x)^n is a geometric series with common ratio x.

If |x| > 1 it means that every successive term is bigger in magnitude than the term before it. Summing these up blows the sum up to infinity.

If |x| < 1 it means that successive terms get smaller, and it turns out to be sufficient for convergence.

When n is a positive integer, this sum is finite therefore it will always give a finite value with no restriction on x.

but the series still works outside that range... when |x|>1... so why does there have to be a range if it works for alll values of x

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so the range of validity are the values that it will make the series converge to a value?

but the series still works outside that range... when |x|>1... so why does there have to be a range if it works for alll values of x

**m.s124**)so the range of validity are the values that it will make the series converge to a value?

but the series still works outside that range... when |x|>1... so why does there have to be a range if it works for alll values of x

The range of validity is when the series converges, yes.

When |x| > 1 the infinite series does not converge.

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**m.s124**)

so the range of validity are the values that it will make the series converge to a value?

but the series still works outside that range... when |x|>1... so why does there have to be a range if it works for alll values of x

(1+x)^(-1) = 1 - x + x^2 - x^3 + x^4 - ...

The RHS converges if |x|<1 and converges to 1/(1+x).

If you tried setting x = -2 then the LHS is -1 and the RHS is 1+2+4+8+16+... which diverges to infinity.

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