Infinite decimal expansion

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  • #1
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"An infinite decimal expansion has the following meaning:
0.a1 a2 a3 a4... =

∑ ak / 10k =
k=1
sup{∑ ak / 10k : n E N}
where ∑ is the sum from k=1 to k=n."
======================================

Why is the second equality (about the supremum) true? (or why does it make sense?)

Thanks.
 
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Answers and Replies

  • #2
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Here each a_k is positive, so the sequence of partial sums is increasing, and its supremum and limit are equal.
 
  • #3
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This is a consequence of the monotone convergence theorem, which states that if we have a monotonically increasing sequence of numbers (i.e. a sequence [itex](a_n)[/itex] for which [itex]a_k \leq a_{k+1}[/itex] for all k, then the sequence has a finite limit if and only if the sequence is bounded.

Remember an infinite series is nothing more than a limit of partial sums and these partial sums can be treated as a sequence. In this case, the partial sums are

[tex]s_n = \sum_{k=1}^{n} \frac{a_k}{10^k}.[/tex]

The limit [itex]\lim_{n\rightarrow \infty} s_n[/tex] if it exists, is denoted [itex]\sum_{k=1}^{\infty} \frac{a_k}{10^k}.[/itex]

Now clearly [itex](s_n)[/itex] is monotonically increasing (Why?). If it is unbounded, then the equality we are seeking is trivial. If the sequence is bounded, the monotone convergence theorem tells us that the sequence not only has a limit, but in fact this limit is the least upper bound of the terms of the sequence, which is exactly what we want.

Check out

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

for more info.
 
  • #4
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This is a consequence of the monotone convergence theorem, which states that if we have a monotonically increasing sequence of numbers (i.e. a sequence [itex](a_n)[/itex] for which [itex]a_k \leq a_{k+1}[/itex] for all k, then the sequence has a finite limit if and only if the sequence is bounded.

Remember an infinite series is nothing more than a limit of partial sums and these partial sums can be treated as a sequence. In this case, the partial sums are

[tex]s_n = \sum_{k=1}^{n} \frac{a_k}{10^k}.[/tex]

The limit [itex]\lim_{n\rightarrow \infty} s_n[/tex] if it exists, is denoted [itex]\sum_{k=1}^{\infty} \frac{a_k}{10^k}.[/itex]

Now clearly [itex](s_n)[/itex] is monotonically increasing (Why?). If it is unbounded, then the equality we are seeking is trivial. If the sequence is bounded, the monotone convergence theorem tells us that the sequence not only has a limit, but in fact this limit is the least upper bound of the terms of the sequence, which is exactly what we want.

Check out

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

for more info.
The theorem in my textbook is as follows:
if a_n is nondecreasing and bounded, then lim a_n converges to sup a_n.

But in our case, I can see that s_n is nondecreasing, but how can we prove that s_n =
n
∑ ak / 10k
k=1
is bounded?

Thanks!
 
  • #5
HallsofIvy
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Homework Helper
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The theorem in my textbook is as follows:
if a_n is nondecreasing and bounded, then lim a_n converges to sup a_n.

But in our case, I can see that s_n is nondecreasing, but how can we prove that s_n =
n
∑ ak / 10k
k=1
is bounded?

Thanks!
It is less than [itex](a_1+1)/10[/itex].
 
  • #6
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It is less than [itex](a_1+1)/10[/itex].
Why?? (this is not so obvious to me...)
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
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because [itex].0a_2a_3a_4.... < .1[/itex].

Each digit is less than 10 so [itex]a_210^{-2}\le .09[/itex], [itex]a_310^{-3}\le .009[/itex], etc.

[itex]a_2 10^{-2}+ a_3 10^{-3}+ ...\le .09+ .009+ ...< .1[/itex].
 

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