Calculation with infinitely many operands

Recently I have come across the equation

xxxx...=2

which I found out to be solvable using substitution:

u=xxxx...=2
xu=2
x2=2
x=2

Thinking about that, I tried to apply the same concept to other operations

xxxx  ... =2
u=xxxx  ... =2
xu=2
2x=2
x=1

This solution is rather strange, as 111... should of course always equal 1, but the equation says something else. Same thing goes for the addition:

x+x+x+x+ ... =2
u=x+x+x+ ... =2
x+u=2
x+2=2
x=0

Which basically says that if you add 0 up often enough you get 2. And the next problem is, if you define the sum to equal 1 instead of 2 you end up with 1=2. So obviously there must be a mistake here.

Is there any reason why this approach can be used for infinite powers but not sums and products?

Answer

The problem is when you make those kinds of substitutions you are assuming that the associated sequence is convergent. That is true for the first case but it is not for the second.

In the second case you have:

xxx...=2

It is equivalent to

lim

But if you have |x|>1 then we have two cases: if x>1 then the limit will be \infty but if x<-1 then x^n will oscillate for +\infty or -\infty and then we don't have limit is this case:

\lim_{n\to \infty}x^n=\infty \text{ or } \nexists \lim_{n\to \infty}x^n \quad (1)

which means that it is not convergent. Otherwise, if |x|<1 then

\lim_{n\to \infty}x^n=0 \quad (2)

For, if x=1 we get

\lim_{n\to \infty}x^n=1 \quad (3)

and, finally, for x=-1 we have that x^n will oscillate. It will be 1 when n is even and -1 when n is odd so \lim_{n\to \infty}x^n doesn't exist. (4)

From (1), (2), (3) and (4) we conclude that you will never have

x\cdot x\cdot x ...=2

Attribution
Source : Link , Question Author : zockDoc , Answer Author : Arnaldo

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