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What would the equation to sum (shown below) of $ 3\cdot 10^{-n-1}$? Just like $ 2^x-1$ is the sum of $ 2^{x-1}$.

$$\sum_{n=0}^{\infty} 3\cdot 10^{-n-1}$$

This would be 0.3+0.03+0.003. . .

this would help me greatly in finding a limit for something.

UPDATE: I found (via calculator) it's $0.3021339806 \times 0.099570245^{x}$ but I'd like something smoother

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The sum of that series is actually $\frac{100}3$, not $\frac13$, and it’s equal to $\frac{100}3$, not approaching it. But it’s not clear what you’re asking. –  Brian M. Scott Sep 29 '12 at 23:34
    
thanks for the input, currently editing –  WebMaster Sep 29 '12 at 23:36
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You keep changing the question. I think you mean $\sum 3\cdot 10^{-n}$. I also think you need to get more of a grip on geometric series to do this. The sum you are looking for is equal to 1/3, not $0.3021339806\times 0.099570245^x$ (what is $x$ supposed to be?) –  user39572 Sep 30 '12 at 0:33
    
I would delete this post and start over if I could. Sorry. –  WebMaster Sep 30 '12 at 0:34
    
Don't worry about changing the question, even if you realize more than once that you've misstated something. What you currently have in the sum still doesn't match up with $0.3+0.03+...$ as you've written below. If the latter is what the sum should be, you should leave everything the same except for starting the sum at $n=0$, so the first term is $\frac{3}{10}$. In that case, as you noted in an earlier version, the sum is $\frac{1}{3}$; if you leave the summation starting at $n=1$ as now, the sum is $\frac{1}{30}$. –  Kevin Carlson Sep 30 '12 at 0:40
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1 Answer 1

You have that

$$0.3 + 0.03 + 0.003 + \cdots = \sum\limits_{n = 1}^\infty {\frac{3}{{{{10}^n}}}} $$

Now use that $$\sum\limits_{n = 1}^\infty {{a^n}} = \frac{a}{{1 - a}}$$

whenever $|a|<1$

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sorry i was unclear, the question was updated –  WebMaster Sep 29 '12 at 23:39
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@WebMaster You want to find $$\sum\limits_{k = 0}^n {{a^k}} = {s_n(a)}$$ for $n$ and $a$ arbitrary? –  Pedro Tamaroff Sep 29 '12 at 23:41
    
Or you want to find $s_n$ such that $s_n-s_{n-1}=10^{2-n}$? –  Pedro Tamaroff Sep 29 '12 at 23:43
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I want to find 0.3+0.03+0.003. So I think the first one –  WebMaster Sep 30 '12 at 0:05
    
@WebMaster I edited. I guess you can work it out. –  Pedro Tamaroff Sep 30 '12 at 1:16
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