From Andrew Love comes a puzzle about planning ahead and resolving to get things done:
It’s the morning of Jan. 1, 2023, and you have a task you want to complete before the end of the year, on Dec. 31. The task is easily divisible into portions, so you always know exactly what fraction of the task is left.
Your initial plan is to do exactly 1/365 of the task every day. But then you think to yourself, “If I do 2/365 today, I can do a little less every day from now on.” And so, on Jan. 1, you would complete 2/365 of the task. On Jan. 2, 363/365 of the task remains with 364 days left to complete it, so you would do another 2/364 × 363/365 of the task.
If you continue in this fashion every day, dividing the remaining work by the remaining number of days in the year and doing twice that amount, when would you finish?
Let $W(d)$ represent the amount of work remaining on day $d$.
On day $d$, the amount of work you do is $\dfrac{2}{366-d} \cdot W(d-1)$ and the amount remaining is $\dfrac{364-d}{366-d}\cdot W(d-1)$.
There will be no remaining work on day 364.
Note that:
$$W(d) = \dfrac{365-(d+1)}{365-(d-1)}\cdot W(d-1)$$
$$W(d) = \dfrac{365-(d+1)}{365-(d-1)}\cdot\dfrac{365-(d+0)}{365-(d-2)}\cdot W(d-2)$$
$$W(d) = \dfrac{365-(d+1)}{365-(d-1)}\cdot\dfrac{365-(d+0)}{365-(d-2)} \cdots \dfrac{365 - 3}{365 - 1} \cdot \dfrac{365 - 2}{365 - 0}$$
$$W(d) = \dfrac{(364-d)(365-d)}{364\cdot 365}$$
Also note that:
$$\dfrac{1}{365} = \dfrac{2}{366-d} \cdot W(d-1)$$
$$\dfrac{1}{365} = \dfrac{2}{366-d} \cdot \dfrac{(365-d)(366-d)}{364\cdot 365}$$
$$364 = 2(365-d)$$
$$d = 183$$
On day 183, you still end up doing 1/365 of the original task.