# Thread: Two cauchy sequence questions

1. ## Two cauchy sequence questions

1. prove that the infinite decimal .123456789101112....
is not eventually periodic, i.e. it represents an irrational number.

2. Let x be a real number having the infinite decimal expansion
x = 1.01001000100001... without the use of a calculator find the first four digits of x^(1/3)

Thanks!

2. Setting $\displaystyle x=1+\xi$ is $\displaystyle \xi= .01001000100001\dots$ and now we remember that is...

$\displaystyle (1+\xi)^{\frac{1}{3}} = 1 + \frac{\xi}{3} - \frac{\xi^{2}}{9} + \dots$ (1)

... so that with four decimal digit accuracy is...

$\displaystyle x^{\frac{1}{3}} = 1.0033\dots$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

i want to answer the question in a way that references Cauchy sequences however. I know that the decimal 1.010010001... is a Cauchy sequence because you can make
a_1 = 1.01
a_2 = 1.01001
a_3 = 1.010010001
...

and you can make the difference between two successive terms of this sequence arbitrarily small by going out far enough

so is there some way to evaluate a Cauchy sequence to a certain degree of accuracy? I had trouble finding a sequence that represents this decimal.

my shot at #1 (would appreciate it if someone can tell me yes it's valid/no it's not and why):

1. prove that the infinite decimal .123456789101112....
is not eventually periodic, i.e. it represents an irrational number.

define a_n = n

then if this decimal is periodic, for some m and k,

a_m = a_(m+k)
a_(m+1) = a_(m+1+k)
...
a_(m+k-1)=a_(m+2k-1)

but a_n is strictly increasing, so a_m <> a_(m+k) for k > 0

so the decimal cannot be periodic.