How long is eternity?

First, we will lay out how much soul had returned to the Princess' body in the following days.

Day 1: $1/2$

Day 2: $1/2$ + ( $1/2$ )( $1/3$ ) = $1/(1\times2$ ) + ( $1/(2\times3$ ))

Day 3: $1/2$ + ( $1/2$ )( $1/3$ ) + ( $1/2$ )( $2/3$ )( $1/4$ ) = $1/(1\times2$ )+ ( $1/(2\times3$ )) + ( $1/(3\times4$ ))

As the series go on, we will see that on day n, the sum of series can be written as:

Day n: $\frac{1}{1\times2}$ + $\frac{1}{2\times3}$ + $\frac{1}{3\times4}$ + ... + $\frac{1}{n(n+1)}$

When we rearrange $\frac{1}{n(n+1)}$ = $\frac{1}{n}$ - $\frac{1}{(n+1)}$ , we can see that we can do the same with the other previous terms:

$\frac{1}{1\times2}$ + $\frac{1}{2\times3}$ + $\frac{1}{3\times4}$ + ... + $\frac{1}{n(n+1)}$ = $\frac{1}{1}$ - $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ + $\frac{1}{3}$ - $\frac{1}{4}$ + ... + $\frac{1}{n}$ - $\frac{1}{(n+1)}$ = 1 - $\frac{1}{(n+1)}$ .

As n approaches infinity, the sum of series converges to 1. (So the witch was mathematically correct; it took eternity to complete one soul.) So on day n, the soul was 99.99% or $\frac{9,999}{10,000}$ complete:

$\frac{9,999}{10,000}$ = 1 - $\frac{1}{(n+1)}$

$\frac{1}{(n+1)}$ = $\frac{1}{10,000}$

As a result, n = 9,999. Thus, on day 9999, the Princess woke up from her long sleep.

The prince brought back 1/3 of the half soul on day 2. So 1/3 of the full soul still remains in the underworld. Similarly on day 4, 1/4 soul remains in underworld and so on. We want to find the day when 1/10000soul remains in the underworld. So that is 9999 days.