There are 2011 liters of kerosene. For the first time, he used 1 / 2 of his kerosene, for the second time, he used the remaining 1 / 3, for the third time, he used the remaining 1 / 4, and so on. Until 2010, he used the remaining 1 / 2011, how many liters are left?

There are 2011 liters of kerosene. For the first time, he used 1 / 2 of his kerosene, for the second time, he used the remaining 1 / 3, for the third time, he used the remaining 1 / 4, and so on. Until 2010, he used the remaining 1 / 2011, how many liters are left?

It's a good calculation
The first time is 2011 * (1-1 / 2)
The second time left 2011 * (1-1 / 2) * (1-1 / 3)
.
The rest of 2009
The rest of 2010
2011*(1-1/2)*(1-1/3)*(1-1/4)*...*(1-1/2011)=1
The result is one liter left in the end