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August 8th, 2017, 02:41 PM   #1
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Thumbs up 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.
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August 8th, 2017, 03:10 PM   #2
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Quote:
Originally Posted by edugonzaa View Post
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.
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August 8th, 2017, 04:02 PM   #3
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Quote:
Originally Posted by greg1313 View Post
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.
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Last edited by cjem; August 8th, 2017 at 04:04 PM.
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August 8th, 2017, 04:04 PM   #4
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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.
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August 8th, 2017, 04:18 PM   #5
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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.
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August 8th, 2017, 04:23 PM   #6
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Quote:
Originally Posted by edugonzaa View Post

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?
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Last edited by skipjack; August 9th, 2017 at 04:03 PM.
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August 8th, 2017, 04:30 PM   #7
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@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.
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August 8th, 2017, 04:43 PM   #8
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Quote:
Originally Posted by mathman View Post
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.
$e^{\frac{1}{7}} = \sum_{n=0}^{\infty} \frac{1}{7^{n}n!} \neq \sum_{n=1}^{\infty} \frac{1}{7^{n!}}$
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August 9th, 2017, 08:59 AM   #9
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Quote:
Originally Posted by cjem View Post
$e^{\frac{1}{7}} = \sum_{n=0}^{\infty} \frac{1}{7^{n}n!} \neq \sum_{n=1}^{\infty} \frac{1}{7^{n!}}$

He said $(e^{1/7})^7$
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August 9th, 2017, 02:44 PM   #10
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Originally Posted by complicatemodulus View Post
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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