Use the formula

Take a table of length L (Figure 8). At Stage 1, remove the section of length L/4 centered at the midpoint. Two sections remain, each with length less than L/2. At Stage 2, remove sections of length L/4² from each of these two sections (this stage removes L/8 of the table). Now four sections remain, each of length less than L/4. At Stage 3, remove the four central sections of length L/4³, and so on.
(a) Show that at the Nth stage, each remaining section has length less than L/2N and that the total amount of table removed is

(b) Show that in the limit as N →∞, precisely one-half of the table remains.
This result is intriguing, because there are no nonzero intervals of table left (at each stage, the remaining sections have a length less than L/2^N). So, the table has “disappeared.” However, we can place any object longer than L/4 on the table. The object will not fall through because it will not fit through any of the removed
sections.

Transcribed Image Text:

1+r+²++N-1 1-N 1-r