Geometric Series Derivation for Given Identities

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

<br /> \begin{array}{l}<br /> \frac{1}{{0.99}} = 1.0101010101... \; \; \; {\rm{ (1)}} \\<br /><br /> \frac{1}{{0.98}} = 1.0204081632... \; \; \; {\rm{ (2)}} \\ <br /> \end{array}<br />

Here is my answer for (1),

<br /> \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{100}}} \right)} ^n + 1<br />

Here is my answer for (2),

<br /> \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{50}}} \right)} ^n + 1<br />

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?
 
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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.
 
You can get rid of adding the ones by changing the summation to start from zero instad of one.
 
Ok, I see how changing the lower limit to 0 solves the problem of having
to add 1 to the series.

Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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