Help with describing a sequence

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The sequence 0.3, 0.33, 0.333, 0.3333, 0.33333,... approaches the limit of 1/3 as the number of decimal places increases. The differences between each term and 1/3 form a geometric progression with a first term of 0.0333... and a common ratio of 0.1. The sum of this series does not converge to 1/3 as initially thought, but rather to 1/27. The task is to find a formula for the nth term of the sequence, which remains to be explicitly defined. Ultimately, the sequence can be described in relation to its convergence towards 1/3.
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Homework Statement



Describe the following sequence

0.3,0.33,0.333,0.3333,0.33333,...


Homework Equations





The Attempt at a Solution



I tried to put the sequence as

\frac{3}{10},\frac{3}{10}+\frac{3}{100},\frac{3}{10}+\frac{3}{100}+\frac{3}{1000}+...

But I didn't see anything happening there. Also I would hope the answer is convergent to \frac{1}{3} since it looks like the number of 3's tend to infinity which would be the number of 3's in \frac{1}{3}
 
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It's 3 times the geometric series (1/10)^n starting at n=1, isn't it? The series sums to 1/9, doesn't it?
 
.3,.33,.333,.3333...
> 3 [.1+.11+.111+.1111...]
Solve in factors of 10 now ...
go ahead...
 
rock.freak667 said:
Describe the following sequence

0.3,0.33,0.333,0.3333,0.33333,...

Hi rock.freak667! :smile:

You're only asked to describe the sequence.

In other words: find a formula A_n for the nth term.

For example, if the sequence is 1 8 27 64 …, then you describe it as {n³}.

Hint: As you point out, they're obviously getting closer to 1/3. So what is the difference between each term and 1/3? :smile:
 
tiny-tim said:
Hi rock.freak667! :smile:

You're only asked to describe the sequence.

In other words: find a formula A_n for the nth term.

For example, if the sequence is 1 8 27 64 …, then you describe it as {n³}.

Hint: As you point out, they're obviously getting closer to 1/3. So what is the difference between each term and 1/3? :smile:

well the differences are like this

1/3-0.3=0.0333...
1/3-0.33=0.00333...
1/3-0.333=0.000333...

So the differences run in a GP of first term a=0.0333... and common ratio r=0.1

But the sum to infinity = 0.0333../0.9 = 0.333...*1/9 = 1/3 * 1/9 = 1/27..Not 1/3 as I wanted.
 
rock.freak667 remember that \sum\frac{1}{10^{n}} = \frac{1}{1-\frac{1}{10}} only when n goes from 0 to infinity. Is that the case in your example?
 
rock.freak667 said:
well the differences are like this

1/3-0.3=0.0333...
1/3-0.33=0.00333...
1/3-0.333=0.000333...

Hi rock.freak667! :smile:

(remember, the question asks you to describe the sequence, and you haven't specifically done that yet.)

So 3An = 1 - what ?

And so ∑An = … ? :smile:
 
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