# Deriving a geometric series

1. Feb 26, 2006

### opticaltempest

I am trying to derive the geometric series for the following given
identities,

$$\begin{array}{l} \frac{1}{{0.99}} = 1.0101010101... \; \; \; {\rm{ (1)}} \\$$

$$\frac{1}{{0.98}} = 1.0204081632... \; \; \; {\rm{ (2)}} \\ \end{array}$$

Here is my answer for (1),

$$\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{100}}} \right)} ^n + 1$$

Here is my answer for (2),

$$\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{50}}} \right)} ^n + 1$$

Are my answers correct? The only way I can get the correct answer is by
adding 1 onto the series. Is this the correct way represent the series?

2. Feb 26, 2006

### 0rthodontist

Yes, the answers are correct. The reason you are adding 1 onto the series is because this formula for the infinite sum of a geometric series
1/(1-r)
holds when the first term is 1. 1/.99 can be written as 1/(1-.01) so you have ratio r and first term 1, and the sequence you wrote for the first one reflects that only when you add 1 to it. You could sum from 0 to infinity and remove the 1, instead of from 1 to infinity and adding in the 1 afterwards, if you think it looks neater.

3. Feb 26, 2006

### d_leet

You can get rid of adding the ones by changing the summation to start from zero instad of one.

4. Feb 26, 2006

### opticaltempest

Ok, I see how changing the lower limit to 0 solves the problem of having
to add 1 to the series.

Thanks!