-1

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

Gᴇᴏᴍᴇᴛᴇʀ
  • 915
  • 2
  • 14
  • 28
  • 2
    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
  • 1
    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?) –  Sep 30 '12 at 00:33
  • I would delete this post and start over if I could. Sorry. – Gᴇᴏᴍᴇᴛᴇʀ Sep 30 '12 at 00:34
  • 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 00:40

1 Answers1

4

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 $|a|<1$

Pedro
  • 122,002