-1
$\begingroup$

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

  • 0
    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 $\frac{1}{30}$.2012-09-30

1 Answers 1

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 $|a|<1$

  • 0
    @WebMaster I edited. I guess you can work it out.2012-09-30