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Geometric Series

Why

It is believable that 1/2,1/4,1/8, has a convergent series. And likewise with 1/3,1/9,1/27, What of ak=xk for xR.

Definition

Let xR. The geometric series of x is the series of the sequence (ak) defined by ak=xk.

Characterization of convergence

Does the geometric series of x converge? In other words, does (sn) defined by sn=k=1nxk have a limit.

For x=1 and x=1, we have seen (see Real Series) that the series diverges. However for the cases x=1/2 and x=1/3 the geometric series converges.

If |x|<1, then the geometric series of x converges and

limnk=1nxk=x1x

If |x|1 then the geometric series of x diverges.

Define sn=k=1nxk. Then

xsn=x(x1+x2++xn)=x2+x3++xn+1=snx+xn+1.

From which we deduce, sn(1x)=x(1xn). If x1, then

sn=11x(1xn)

If |x|<1, then using the algebra of limits (see  Real Limit Algebra) we deduce

limn11x(1xn)=11x(10)=11x,

since limkxk=0 for |x|<1.

If x=1 or x=1, then we have seen that (sn) diverges.1


  1. Future editions will include the trivial account about the case |x|>1. ↩︎
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