00:02
For part a, if the even and odd parts go to the same number of l, then the whole sequence goes to l.
00:07
Think of the sequence a sub n as two interleaved lists.
00:12
We've got our evens, which would be a sub 2, a sub 4, a sub 6, and so on, and we've got our odds, which would be a sub 1, a sub 3, a sub 5, and so on.
00:27
Let's assume that the limit as n approaches infinity of a sub 2n is equal to l, and the limit as n approaches infinity of a sub 2n plus 1 also equals l, which means that far enough out, all even terms are as close to l as we want, and also all the odd terms are as close to l as we want.
00:52
But every term of the original sequence is either even index or odd index, so far enough out, every term is close to l.
01:00
So the limit as n approaches infinity of a sub n in general is equal to l.
01:12
We want to compute the first eight terms and find the limit.
01:16
So we're given that the first term is 1, and then a sub n plus 1.
01:21
The next term is 1 plus 1 over 1 plus a sub n.
01:25
So a sub 1 is 1.
01:27
A sub 2 would be 1 plus 1 over 1 plus 1 plus 1, which is 1 plus 1 over 1 plus 1 over 1 over 1 plus 1 over 2.
01:34
2, which is 1 .5.
01:37
A...