00:01
For this problem, we are told to let ak be the sequence a1 equals 3, a2 equals 3 .1, a3 equals 3 .14, and so on, such that each term ak contains the first k decimal digits of pi.
00:14
In part a, we're asked to explain why ak is a rational number for each positive integer k.
00:19
So the reason why is that we have, obviously, a1 is equal to 3, which is just equal to 3 over 1, which is a rational number.
00:29
Then for a 2 can see that that's 3 .1 which is equal to 31 over 10 again integer over integer so that's a rational number a 3 .14 equals 314 over 100 which is now a rational number and so on so we can see that the pattern would just continue on each one as long as long as you as we have a finite number of decimal places, we can represent it as a rational number.
01:06
Then, for part b, we're asked to explain why the sequence ak is increasing.
01:12
Well, we can think of it as ak plus 1 minus ak would be equal to, and this is a sort of informal way of writing it, it would be equal to every digit of pi after k, which therefore would be some value greater than zero.
01:44
So we, by definition, then, have a monotonically increasing sequence.
01:50
For part c, we're asked to provide an upper bound for the sequence ak.
01:54
Well, for instance, 4 would be a valid upper bound for the sequence.
02:00
Then, for part d, we are asked, what is the least upper bound of the sequence, ak? well, clearly, the least upper bound would be pi...