# Help with describing a sequence

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1. Homework Statement

Describe the following sequence

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

2. Homework Equations

3. 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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Dick
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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 ...

tiny-tim
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Describe the following sequence

0.3,0.33,0.333,0.3333,0.33333,...
Hi rock.freak667!

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?

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Hi rock.freak667!

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?
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.

exk
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?

tiny-tim
Homework Helper
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!

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

So 3An = 1 - what ?

And so ∑An = … ?