Sum of an infinite series

  • #1

Main Question or Discussion Point

i have to prove that the sum of the follwing infinite series is irrational.

[tex]\frac{1}{2^3} + \frac{1}{2^9} + \frac{1}{2^{27}} + \frac{1}{2^{81}} + \cdots[/tex]

i have no idea where to begin. thanks in advance.

note: this is not a homewrok problem.
 

Answers and Replies

  • #2
StatusX
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What's true about the decimal expansion of rational numbers?
 
  • #3
the decimal expansion of a rational number either terminates or repeats a pattern. but how does that help?
 
  • #4
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the decimal expansion of a rational number either terminates or repeats a pattern. but how does that help?
Is that fact unique to a decimal expansion?
 
  • #5
HallsofIvy
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the decimal expansion of a rational number either terminates or repeats a pattern. but how does that help?
That doesn't necessarily help you FIND the sum, but your problem is to prove the number is irrational. What happens if you think of this as a number written in binary?
 
  • #6
That doesn't necessarily help you FIND the sum, but your problem is to prove the number is irrational. What happens if you think of this as a number written in binary?
it becomes 0.001000001000000000000000001.......
clearly it doesn't terminate and no pattern repeats. but does this fact of irrational numbers hold true for any base?
 
  • #7
Gib Z
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Of course it does. If a number, in a rational base, has a terminating decimal expansion, then that is true of any other rational base. Obviously you can get a terminating decimal for pi in base pi...
 
  • #8
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Fine recipe for proving that real numbers represented by sums of murshid's kind are irrational or even trancendent is http://mathworld.wolfram.com/LiouvillesApproximationTheorem.html" [Broken]

:smile:
 
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  • #9
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The "general" sketch for proving irrionality is to express the sum as S = a/b

and then prove that if that was the case, then

x<a<x+1

where x is some integer and therefore there is contradiction; an integer cannot lie between two integers.
 
  • #10
CRGreathouse
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The "general" sketch for proving irrionality is to express the sum as S = a/b

and then prove that if that was the case, then

x<a<x+1

where x is some integer and therefore there is contradiction; an integer cannot lie between two integers.
The most common form I see is expressing it as a/b in lowest terms and then showing that a/b = c/d with 0 < c < a.
 
  • #11
HallsofIvy
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The "general" sketch for proving irrionality is to express the sum as S = a/b

and then prove that if that was the case, then

x<a<x+1

where x is some integer and therefore there is contradiction; an integer cannot lie between two integers.
The most common form I see is expressing it as a/b in lowest terms and then showing that a/b = c/d with 0 < c < a.
But, in this particular case that would be much more difficult that just observing that the number, in base 2, is neither a terminating nor a repeating "decimal".
 
  • #12
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Werg22,CRGreathouse:

Of course,"Reductio ad absurdum" is ,pretty often, a logical way of organization of the proofs.Unfortunately,as HallsofIvy indicates, it doesn't tell anything on how to construct proofs from case to case!
It may appear you two sound like suggesting there is elementar proof in this case,something like [itex]\sqrt{2}[/itex] is irrational?
Please,show your proofs ,I'm not aware this can be done at elementar level.
Proofs that real numbers :
[tex]\sum_{n=1}^{\infty}\frac{1}{2^{(n!)}}[/tex]
Or
[tex]\sum_{n=1}^{\infty}\frac{1}{2^{(m^n)}};m\geq 2[/tex]

are irrational and transcendent can be constructed by employment of joined consequences of Dirichlet's Theorem and mentined above Liouville's Theorem.I don't think I can classify Lioville's Tm by elementar label...
 
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  • #13
HallsofIvy
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What do you mean by "elementary proof"?

Euclid's proof seems pretty elementary to me: If [itex]\sqrt{2}[/itex] is rational then [itex]\sqrt{2}= \frac{m}{n}[/itex], reduced to lowest terms. Then [itex]2= \frac{m^2}{n^2}[/itex] so [itex]2n^2= m^2[/itex]. But the square of any odd number is odd ((2n+1)2= 4n2+ 4n+ 1= 2(2n2+ 2n)+ 1). Since m2 is even, m must be even: m= 2k for some integer k. Then [itex]2n^2= (2k)^2= 4k^2[/itex] and so [itex]n^2= 2k^2[/itex]- that is, n is also even, contradicting the fact that m and n are relatively prime.

A direct proof, somewhat less "elementary" is this: [itex]\sqrt{2}[/itex] obviously satisfies the equation x2- 2= 0. By the "rational root theorem", any rational root of that equation must have numerator that evenly divides the constant term, 2, and denominator that evenly divides the leading coefficient 1. The only possible rational roots, then, are 2 and -2, neither of which satisfies the equation. x2- 2= 0 has no rational roots so [itex]\sqrt{2}[/itex].

That, of course, has little to do with the original question, but I like to show off!:approve:
 
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  • #14
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What do you mean by "elementary proof"?
The one that doesn't rely on theorems from math analysis (complex analysis in particular.)
Read my post more carefully.I'm not questioning if the irrationality of [itex]\sqrt{2}[/itex] can be prooven by elementar math methods,but if OP's problem can be solved in similar way.
I think it can't (but you can never know for sure :smile:).
If you can solve problems ,and prove irrationality (let alone transcendence) of the numbers represented by the sums given above,in "Euclid's" fashion ,post your solution.We would like to know.
 

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