Mistake converting $2.3151515...$ to a fraction?

Question

So I think I made a mistake but I just want to confirm I'm getting this right:

In the example, we had the number $2.3151515...$ and were supposed to convert this number into a fraction. It was then shown to be equal to the sequence $2.3 + \frac{15}{10^3}+\frac{15}{10^5}+\frac{15}{10^7}$ etc. As such the common ratio is $r = \frac{1}{100}$

Thus we get the sum of the sequence equalling $(2.3)\frac{1}{1-\frac{1}{100}} = \frac{230}{99}$ The problem is, when I checked my solution $\frac{230}{99}$, it's rougly $2.32323232...$ yet the right solution would actually be $\frac{382}{165}$

Does the problem lie within the simplification of the sequence or am I doing something wrong?

Thanks in advance.

Answer 1

The $2.3$ is not be part of your series. The first term in the series is actually $\dfrac{15}{10^3}$ with the common ratio of $\dfrac1{10^2}$.

Your sum should be $$2.3+\frac{\frac{15}{10^3}}{1-\frac1{10^2}}=\frac{382}{165}$$ as desired.

Answer 2

The correct answer is $2.3+\dfrac{\dfrac{15}{10^3}}{{1-\dfrac1{100}}}=\dfrac{382}{165}$

Answer 3

Here is the standard technique. We have $$2.3\overline{15}=\frac{2315-23}{990}=\frac{2292}{990}$$ The overline / bar denotes the group of repeating digits. Just form the integer without any decimals to get $2315$ and subtract the number formed by non-repeating digits $23$ and you get numerator $2315-23=2292$.

For denominator just check the digits after decimal point. It consists of as many 9's as there are repeating digits followed by as many zeroes as there are non-repeating digits after decimal point. Thus we get $990$ as denominator.

The above law can be proved using the standard formula for sum of an infinite geometric progression.

Your mistake is in ignoring the first term $a$ in formula for sum of infinite GP $$a+ar+ar^2+\dots=\frac{a} {1-r}$$ and treating the sum as just $1/(1-r)$.