1. ## Sigma Notation of Zeno's Pardadox

Achilles and the Tortoise Problem
Tsub1 = 10
Tsub2 = 1
Tsub3 = .1
Tsub4 = .01

Sigma with
infinity on top
i = 1 on bottom
tsubi on right

i got that it = 10/(10^0) + 10/(10^1) + 10/(10^2).....

so 0-10/9 so -(10/9)?

2. $\displaystyle \sum^{\infty}_{k=1} 10^{2-k}$

$\displaystyle \sum^{\infty}_{k=0} 10^{1-k}$

$\displaystyle \sum^{\infty}_{k=0} \frac{10}{10^k}$

$\displaystyle 10 \sum^{\infty}_{k=0} {\left( \frac{1}{10}\right)}^k$

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If $\displaystyle 0<r<1$, then $\displaystyle \sum^{\infty}_{k=0} r^k = \frac{1}{1-r}$
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So,
$\displaystyle 10 \sum^{\infty}_{k=0} {\left( \frac{1}{10}\right)}^k = 10\cdot \frac{1}{1-\frac{1}{10}}$

$\displaystyle 10\cdot\frac{1}{\frac{9}{10}} = \boxed{~\frac{100}{9}~}$

3. how did you get what you started off with?

$\displaystyle 10$
$\displaystyle 1$
$\displaystyle 0.1$
$\displaystyle 0.01$
...

You can write it as
$\displaystyle 10^1$
$\displaystyle 10^0$
$\displaystyle 10^{-1}$
$\displaystyle 10^{-2}$
...

Do you see how to formulize the sequence now?