How many liters of water remain in the first urn after 1977th pouring? (Full Question in description)

Question

One urn contains 1 liter of water, while a second urn is empty. after 1/2 of the water in the first urn is emptied into the second urn, 1/3 of the water in the second urn is returned to the first urn. Then, 1/4 of the contents of the first urn is poured into the second urn, followed by the return of 1/5 of the contents of the second urn. At each successive pouring from alternate urns, the denominator of the fractional part poured increases by 1. How many liters of water remain in the first urn after 1977th pouring? (Please be specific as to how you come up with each step, thank you)

Answer 1

$$\begin{array}{|c|c|c|c|c|c|c|} \text{Pouring #} & \text{Amount poured} & \text{Into Urn} & \text{Urn 1 before} & \text{Urn 2 before} & \text{Urn 1 after} & \text{Urn 2 after}\\ \hline \text{0} & 1 & 1 & 0 & 0 & 1 & 0 \\ \hline \text{1} & 1/2 & 2 & 1 & 0 & 1/2 & 1/2 \\ \hline \text{2} & 1/3 & 1 & 1/2 & 1/2 & 2/3 & 1/3 \\ \hline \text{3} & 1/4 & 2 & 2/3 & 1/3 & 1/2 & 1/2 \\ \hline \text{4} & 1/5 & 1 & 1/2 & 1/2 & 3/5 & 2/5 \\ \hline \text{5} & 1/6 & 2 & 3/5 & 2/5 & 1/2 & 1/2 \\ \hline \text{6} & 1/7 & 1 & 1/2 & 1/2 & 4/7 & 3/7 \\ \hline \text{7} & 1/8 & 2 & 4/7 & 3/7 & 1/2 & 1/2 \\ \hline \text{...} & ... & ... & ... & ... & ... & ... \\ \hline \end{array}$$

We can note that for all odd pouring number $n \ge 1$, the "after" is perfectly even - $1/2$ to $1/2$ - and that for all even pouring number $n \ge 2$, the "after" is $\frac{n+2}{2n+2}$ to $\frac{n}{2n+2}$. This means that for pouring number $n = 1976$, where $1/1977$ of the contents from urn 2 will be poured to urn 1, the contents in urn 1 will be $$\fbox{(1976+2)/(2*1976+2) = 989/1977}$$

Answer 2

Let $\,a_n\,$ be the liters of water in the first urn after $\,n\,$ pourings. Since the total amount of water is constant, the liters in the second urn after $\,n\,$ pourings will be $\,1-a_n\,$.

The initial condition is $\,a_0=1\,$, and the rules of pouring give the following recursive relations:

$$ \begin{align} a_{2n-1} &= a_{2n-2} - \frac{1}{2n}a_{2n-2} = \frac{2n-1}{2n}a_{2n-2} \tag{1} \\[5px] a_{2n} &= a_{2n-1} + \frac{1}{2n+1}(1 -a_{2n-1}) = \frac{1}{2n+1} + \frac{2n}{2n+1}a_{2n-1} \tag{2} \end{align} $$

Writing $\,(1)\,$ for $\,2n+1\,$ and using $\,(2)\,$:

$$ a_{2n+1} = \frac{2n+1}{2n+2}a_{2n} = \frac{2n+1}{2n+2}\left(\frac{1}{2n+1} + \frac{2n}{2n+1}a_{2n-1}\right) = \frac{1}{2n+2}+\frac{2n}{2n+2}a_{2n-1} \tag{3} $$

Relation $\,(3)\,$ implies that if $\require{cancel}\,a_{2n-1}=\frac{1}{2}\,$ then $\,a_{2n+1}=\frac{1}{2n+2}+\frac{\bcancel{2}n}{2n+2}\frac{1}{\bcancel{2}}=\frac{n+1}{2n+2}=\frac{1}{2}\,$ and, since $\,a_1=\frac{1}{2}\,$, it follows by induction that all odd-index $\,a_{2n+1} = \frac{1}{2}\,$. Therefore $\,a_{1977}=\frac{1}{2}\,$.