00:01
We want to determine whether the sequence an is equal to 4n plus 1 plus 3n over 4n converges or diverges, and if it converges, what it converges.
00:13
And we want to give a reason for our steps.
00:20
All right, so first before we do anything, let's just do a little bit of algebra to rewrite this.
00:25
So i'm going to divide the 4n into what we have in the numerator.
00:29
So 4n plus 1 divided by 4n, just give us 4.
00:33
And then 3n divided by 4n.
00:36
Well, since they have the same power, we can write that as 3 .4 to the n.
00:42
Now, if we go ahead and take the limit as in approaches infinity of a .n is equal to the limit as in approach to infinity of 4 plus 3 over 4n.
01:03
Well we can possibly distribute that limit so i'm going to put a little question mark here because i don't quite know if we can do it yet because we still have to show that the limit of each of these things i'm adding exists so let's just go ahead and write this out first so the limit as an approach is infinity of three over four to the end now on the next line let's just go ahead and write what we think both of these limits should be if they exist will the limit of a constant always be itself? so that's 4 plus...