Limit of the sequence .3, .33, .333

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OldMan

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Jan 14, 2009
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Hello,
Problem is to find the limit of the sequence .3, .33, .333
I have the solution but cannot understand one of the steps.
Un=3/10+3/10^2+...3/10^n = 3/10(1+1/10+...1/10^n-1)
If S=1+1/10+...1/1+10^n-1
then 1/10S=1/10+1/10^2+...1/10^n-1+1/10^n
subtracting gets 9/10S=1-1/10^n or S= 10/9(1-10^n)
Now here's the part I don't get:
Thus nth term=Un=1/3(1-1/10^n)
I don't understand how this follows from the previous (especially where the 1/3 comes from).
I apologize if I'm missing something obvious.
Any assistance is appreciated.
Old Man trying to learn some math.
Thank you
 
Do you want the sum of the infinite series or just the nth term?.

The infinite sum is:

\(\displaystyle \sum_{k=1}^{\infty}\frac{3}{{10}^{k}}\)

\(\displaystyle S=\frac{3}{10}+\frac{3}{100}+\frac{3}{1000}+...................\)

Factor out 3/10:

\(\displaystyle S=\frac{3}{10}\left(1+\underbrace{\frac{1}{10}+\frac{1}{100}+...........}_{\text{this is 1/3 S}}\right)\)

\(\displaystyle S=\frac{3}{10}\left(1+\frac{1}{3}S\right)\)

\(\displaystyle S=\frac{1}{3}\)

Since the total sum is 1/3, then the nth term is given by \(\displaystyle \frac{1}{3}\left(1-\frac{1}{10^{n}}\right)\)

Sometimes, that's how I do them for the nth term. Once we know the sum, we can find the nth term by \(\displaystyle \text{sum}\cdot\left(1-\frac{1}{10^{k}}\right)\)

Take \(\displaystyle \frac{7}{10}+\frac{7}{100}+\frac{7}{1000}+..........\)

The sum of the series is 7/9. Thus, the nth term is \(\displaystyle \frac{7}{9}\left(1-\frac{1}{10^{k}}\right)\)

You can also find the sum by the formula for the geometric series. \(\displaystyle \frac{\frac{3}{10}}{1-\frac{3}{10}}=\frac{1}{3}\)
 
Here is the way I would do it. Maybe is is easier to follow?
\(\displaystyle U_n = \sum\limits_{k = 1}^n {\frac{3}{{10^k }}} ,\quad \,10U_n = \sum\limits_{k = 1}^n {\frac{3}{{10^{k - 1} }}} ,\quad 9U_n = 3 - \frac{3}{{10^n }}\)

\(\displaystyle U_n = \frac{1}{3} - \frac{1}{{3 \cdot 10^n }},\quad \,\left( {U_n } \right) \to \frac{1}{3}\)
 
fasteddie65 said:
This is a geometric series, with a = 0.3 and r = 0.1
S = a/(1 - r)
Click to expand...

That is not correct. In the above case the sequence becomes:

0.3, 0.03, 0.003, 0.0003, .....
 
galactus said:
Do you want the sum of the infinite series or just the nth term?.

The sum is:

\(\displaystyle \sum_{k=1}^{\infty}\frac{3}{10}^{k}\)
Click to expand...

The nth term of the sequence is (expressed as a sum of a different sequence)

\(\displaystyle a_n \, = \, \sum_{k=1}^{n}\frac{3}{10^k}\)..... edited
 
Yep, thanks for the catch. I put infinity. That's a habit. :oops:
 
I think I was OK. I was finding the sum of the infinite series. Oh well.
 
galactus said:
I think I was OK. I was finding the sum of the infinite series. Oh well.
Click to expand...

Your work was correct - but the exponent of the equation should be on 10 (not on 3).
You wrote:

\(\displaystyle a_n \, = \, S_n \, = \, \frac{3}{10} \, + \, \frac{3}{100} \, + \, \frac{3}{1000} \, + \, \frac{3}{10000} \, + \, \frac{3}{100000} \, + ...\)

\(\displaystyle a_n \, = \, \frac{3}{10^1} \, + \, \frac{3}{10^2} \, + \, \frac{3}{10^3} \, + \, \frac{3}{10^4} \, + \, \frac{3}{10^5} \, + ... \, + \, \frac{3}{10^n} \,\)

\(\displaystyle a_n \, = \, \sum_{k=1}^{n}\frac{3}{10^k}\)

\(\displaystyle a_{\infty} \, = S_{\infty} \, = \, \sum_{k=1}^{\infty}\frac{3}{10^k}\)
That's all.....
 
Oh, I see, a flub up/typo. :oops:

I see why. I didn't put braces around my LaTex in one spot. That is why it displayed wacky.
 
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