August 8th, 2017, 02:41 PM  #1 
Newbie Joined: Aug 2017 From: Costa Rica Posts: 2 Thanks: 0  Prove that this sum is irrational
I've been trying to solve this problem using the proof that e is irrational, but I don't know what I'm doing wrong! Here's the problem: 1. Prove that 1/7+1/7^2 +1/7^6 +⋯+1/7^n! +⋯ is irrational. 2. Enunciate possible extensions for part 1, propose solutions. Last edited by skipjack; August 9th, 2017 at 03:59 PM. 
August 8th, 2017, 03:10 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,600 Thanks: 942 Math Focus: Elementary mathematics and beyond  It's not irrational! Any term in the sum is a rational number and the sum of two or more rational numbers is rational.
Last edited by skipjack; August 9th, 2017 at 04:00 PM. 
August 8th, 2017, 04:02 PM  #3  
Member Joined: Aug 2017 From: United Kingdom Posts: 97 Thanks: 28  Quote:
The issue is that an infinite sum $\sum_{n=1}^{\infty} a_k$ is NOT just obtained by repeatedly adding numbers. It is in fact the limit $\lim_{n \to \infty} \sum_{k=1}^{n} a_k$, so there is no reason that it couldn't be irrational even if all the $a_n$ are rational. Last edited by cjem; August 8th, 2017 at 04:04 PM.  
August 8th, 2017, 04:04 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,600 Thanks: 942 Math Focus: Elementary mathematics and beyond 
Ah, I see. Thanks. There are many examples of that; for instance $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ . I was thinking of finite $n$. Last edited by greg1313; August 8th, 2017 at 04:15 PM. 
August 8th, 2017, 04:18 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,344 Thanks: 534 
The expression seems to be $\displaystyle e^{\frac{1}{7}}$. Since e is irrational, the expression can't be rational. If it were, the 7th power (=e) would be rational.

August 8th, 2017, 04:23 PM  #6 
Senior Member Joined: Aug 2012 Posts: 1,575 Thanks: 380  Its base7 expression clearly has no repeating block. You can add as much detail as the assignment expects. Can you assume that rationals eventually have repeating blocks in any base? Likewise $\frac{1}{7} + \frac{1}{7^3} + \frac{1}{7^5} + \dots + \frac{1}{7^{2n + 1}} + \dots$ is rational. Do you see why? Last edited by skipjack; August 9th, 2017 at 04:03 PM. 
August 8th, 2017, 04:30 PM  #7 
Member Joined: Aug 2017 From: United Kingdom Posts: 97 Thanks: 28 
@OP If you're familiar with Liouville's approximation theorem, you could actually go further and show this is transcendental  the partial sums provide suitably good rational approximations to the number that (by the theorem) it cannot be algebraic of any degree. If you work through the details, it will suggest a whole family of very similar numbers that can also be shown to be transcendental with the same argument.

August 8th, 2017, 04:43 PM  #8 
Member Joined: Aug 2017 From: United Kingdom Posts: 97 Thanks: 28  $e^{\frac{1}{7}} = \sum_{n=0}^{\infty} \frac{1}{7^{n}n!} \neq \sum_{n=1}^{\infty} \frac{1}{7^{n!}}$

August 9th, 2017, 08:59 AM  #9 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24  
August 9th, 2017, 02:44 PM  #10 
Member Joined: Aug 2017 From: United Kingdom Posts: 97 Thanks: 28  

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irrational, prove, sum 
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