A math challenge?

Question

Determine, with proof or disproof, if the number determined by the following convergent series is rational or irrational:

1/10 + (1/10)^(2!) + (1/10)^(3!) + (1/10)^(4!) + ...

You may take it for granted that the series does actually converge (although this should be fairly obvious).

This is a math challenge for all the self-proclaimed gurus around here. I love math challenges, and I solved this one myself a month ago or so. It's not too difficult. My solution was a bit lengthy, so I'm also anxious to see if someone could come up with a shorter, more elegant solution.

I'll give the correct answer (without my proof/disproof because it's too long) if nobody gets it.

Answer 1

This number exists and is real and positive because the series is bounded by convergent geometric series 1 + 1/10 + 1/100 + .... and is clearly greater than 0.



[intuitive proof]

Since the decimal expansion of this number is infinite and non-repeating, we know the number must be irrational (the decimal expansion is 0.110001.... with a 1 in position m where m = n! for some integer n >=1).



[rigorous proof]

If this number converges to a positive rational number, it can be written in the form p/q with p and q positive integers. Multiplying both sides of the equation,



p/q = 1/10 + (1/10)^(2!) + (1/10)^(3!) + (1/10)^(4!) + ...



by q10^(q!) we get



p10^(q!) = [q10^(q! - 1) + q10^(q! - 2!) + ...q10(q! - q!)] +

q10^(q! - (q + 1)!) + q10^(q! - (q + 2)!) + .....



The left hand side is an integer as is the sum of terms in the brackets. This must imply that the remaining terms also sum to an integer. However, it is easy to see that the remaining sum is bounded from above by



q/10^(qq!) + (1/10) q/10^(qq!) + (1/100) q/10^(qq!) + ....

which sums to 10/9 q/10^qq!.



For all q >= 1, we have 10^qq! >= 10^q.



Since we know that e^x >= 1 + x, we have

q = 1 + (q - 1) <= e^(q - 1) <= 10^(q - 1) for q >= 1.



Therefore, 10^q = 10(10^(q-1)) >= 10q and combinining this with our earlier inequalities, we have



10/9 (q/10^qq!) <= 1/9 which means that the remaining terms cannot sum to a positive integer which is a contradiction.



Therefore, the original sum must converge to a positive irrational number.

Answer 2

It is irrational.



A number has a recurring decimal expansion if and only if it is rational.



So to prove it is irrational, all you have to do is to prove the decimal expansion is non recurring.



This is fairly simple. Say it starts recurring from a certain point ie the first period starts at 10^-([n-1]!).



Then there must be at least one period later on in the expansion where there are two 1s that are only n! - ([n-1]!) away. However it won't be as the number of zeroes in the expansion between 10^-m! - 10^-([m-1]!) is greater than the number of zeroes between the decimal point and 10^-([m-1]!) for m >= 4 certainly. So if m>n>=4 f zeroes in the expansion between 10^-m! - 10^-([m-1]!) is strictly greater than the number of zeroes between 10^-n! and 10^-([n-1]!). This contradicts the assumption that the decimal expansion repeats.



So the decimal expansion is not repeating and the number is irational.



This is a bit brute force and not too elegant, but it is fairly short.



Edit: This is a rigourous method. It fairly easy to prove that number has a recurring decimal expansion if and only if it is rational.



If a number x has a recurring decimal expansion then let the period be k. wlog assume |x| <1 (consider x - floor(x)): 10^k x = x + r where r<10^k. Then x = r/ (10^k-1)



If a number is rational then say x = p/q where q has k digits. By considering the remainder of the fraction, we can assume p


The second part is proof by contradiction.

Answer 3

This is simple. Irrational. The sequence (n + 1)! - n! is strictly increasing. Hence, the decimal expansion of the limit is something like



0.10..10.......10...



Only 0s and 1s and the "distance" between the 1s is increasing. Hence, the limit's decimal representation is infinite and non periodic, so that the limit is irrational.



About a year ago I posted a similar problen, actually a bit more difficult, and gave my solution.



https://answersrip.com/question/index?qid=20070805063138AAjnvG4



Addendum:

The proof based on the decimal expansion is rigorous. It's based on a known fact. Of course, the proof of transcendentality is more advanced and much more interesting.



Edit.



A cave asked me to post this link about Liouville numbers



http://en.wikipedia.org/wiki/Liouville_number



In his opinion, this answers all the questions about the subject we're discussing.

Answer 4

Let N = 1/10 + (1/10)^(2!) + (1/10)^(3!) + (1/10)^(4!) + ...



Expanding each term of the series and summing them we have:

0.1

0.01

0.000001

0.00000...(23 zeros)...1

0.00000...(95 zeros)...1

.

.

.

__________________

0.11000100000...(17 zeros)...100000...





Therefore N = 0.11000100000...



If N is a rational number, then we should observe a recurring set of it's digits to the right of the decimal point. However, as we observe, there is no uniform recurring set of digits after the decimal point. Therefore N has to be an irrational number.



Thus the sum of the series is IRRATIONAL