47. In one vessel, there are $a$ liters of water, and the other is empty. From the first vessel, half of the water in it is poured into the second, then from the second, $\frac{1}{3}$ of the water in it is poured back into the first, then from the first, $\frac{1}{4}$ of the water in it is poured into the second, and so on. How many liters of water will be in the first vessel after the 1987th pouring?
47. After pouring from the second vessel into the first, the first will contain $\frac{1}{2} a+\frac{1}{3} \cdot \frac{1}{2} a=\frac{4}{6} a=\frac{2}{3} a$, and the second will contain $\frac{1}{2} a-\frac{1}{3} \cdot \frac{1}{2} a=\frac{2}{6} a=\frac{1}{3} a$. Reasoning this way, we get the table

| pouring | 1 | 2 | 3 | 4 | 5 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 1st vessel | $\frac{1}{2} a$ | $\frac{2}{3} a$ | $\frac{1}{2} a$ | $\frac{3}{5} a$ | $\frac{1}{2} a$ |
| 2nd vessel | $\frac{1}{2} a$ | $\frac{1}{3} a$ | $\frac{1}{2} a$ | $\frac{2}{5} a$ | $\frac{1}{2} a$ |

This table can be extended as far as we like. We notice that after each odd number of pourings, the vessels contain the same amount of water $-\frac{1}{2} a$ liters. Therefore, after the 1987th pouring, the first vessel will contain $\frac{1}{2} a$ liters of water. (This can be proven.)