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 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
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Quote:
 Originally Posted by edugonzaa 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!
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
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 Originally Posted by greg1313 It's not irrational! Any term in the sum is a rational number and the sum of two or more rational numbers is rational.
How about $\pi = 3 + 0.1 + 0.04 + 0.005 + ...$?

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: 7,842 Thanks: 1068 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: 6,555 Thanks: 600 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
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Quote:
 Originally Posted by edugonzaa 1.- Prove that 1/7+1/7^2 +1/7^6 +⋯+1/7^n! +⋯ is irrational.
Its base-7 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 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 211 Thanks: 63 Math Focus: Algebraic Number Theory, Arithmetic Geometry @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
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 Originally Posted by mathman 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.
$e^{\frac{1}{7}} = \sum_{n=0}^{\infty} \frac{1}{7^{n}n!} \neq \sum_{n=1}^{\infty} \frac{1}{7^{n!}}$

August 9th, 2017, 08:59 AM   #9
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Quote:
 Originally Posted by cjem $e^{\frac{1}{7}} = \sum_{n=0}^{\infty} \frac{1}{7^{n}n!} \neq \sum_{n=1}^{\infty} \frac{1}{7^{n!}}$

He said $(e^{1/7})^7$

August 9th, 2017, 02:44 PM   #10
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Quote:
 Originally Posted by complicatemodulus He said $(e^{1/7})^7$
He said the expression in the original post is $e^{\frac{1}{7}}$; I was just pointing out that this is not correct.

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