This is a problem on sequences and series involving describing a repeating decimal as an infinite series in order to express it as a ratio of two integers; for some reason, I'm off by a factor of ten, and I can't figure why. Posted below in a minute.

The repeating decimal is just 0.777- \[0.777... = \frac{ 7 }{ 10 }+\frac{ 7 }{ 100 }+ \frac{ 7 }{ 1000 } ... \sum_{n = 1}^{\infty}\frac{ 7 }{ 10^{n} }\]

ok... so the 1st term is 7/10 and the common ratio is 1/10

the series is geometric and converges... since -1 < r < 1

Since it's obviously a geometric series, \[\sum_{n = 1}^{\infty}\frac{ 7 }{ 10^{n} }, a = 7, r = \frac{ 1 }{ 10 }, ar ^{n-1}\] The infinite sum for a geometric series is \[\frac{ a }{ 1-r }\] \[\frac{ 7 }{ 1-(\frac{ 1 }{ 10 }) } = \frac{ 7 }{ \frac{ 9 }{ 10 } } = \frac{ 70 }{ 9 }\]

The answer is 7/9ths. I can't figure out why I'm off by a factor of ten. But yeah, full aware of the common ratio, and convergence is a given. I'm just looking for a simple algebraic error or something.

so the way to express it as an rational number is let x = 0.7777..... multiply it by 10 so 10x = 7.777777777 so now subtract the 2nd equation from the 1st 10x = 7.7777777777777.... - x = 0.7777777777777..... -------------------------- 9x = 7

so now solve for x

I can do that, too, but I'd like to use the formula, rather than just plug and chug. I know how to do it otherwise, but I'm trying to do it with the formula.

well its not plug and chug, its a valid method... so what you need to do is find the limiting sum of the geometric series \[S_{\infty} = \frac{a}{1 - r}\]

thats if you want to solve it as a series

Yeah, got that. I know it's a-that's exactly what I'm trying to do, and why I wrote the formula.

so you have \[S_{\infty} = \frac{\frac{7}{10}}{1 - \frac{1}{10}}\]

a is now 7/10.

Thanks, that's what I was looking for; It was just the number. I understand the process completely, but I couldn't figure out what a should be. That's what I needed.

a is 7/10 1st decimal place

Got it.

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