00:01
We want to determine whether the sequence a is equal to x to the n, divided by 2n plus 1, raised to the 1 over n power.
00:09
When x is 3 to 0 converges or diverges, and if it converges what it converges.
00:16
So, first, let's do a little bit of algebra here.
00:20
So if we were to distribute the 1 over n to everything, in the numerator we just get x, and in the denominator we get 2n plus 1 to the 1 over n.
00:32
And so now this x is a constant with respect to n.
00:44
So when i take this limit, this x term doesn't really matter.
00:48
So whenever we take the limit, it can just be pulled out front.
00:52
So let's go ahead and take the limit now.
00:55
So the limit has an approach to infinity of a n.
01:01
Well, this is going to equal to where we pull that x out.
01:06
Or let me just write down black, pull that x out front of the limit by using the constant property for limits and then we would just be left with 1 over 2 n plus 1 to the 1 over n.
01:26
Now if we were to rewrite this, or at least if we were to look at the limit as an approach and venue for this right now, well 2 plus n is going to to go to infinity and then one over n goes to zero so this is one of the forms where it might be a good idea for us to apply lopatol rule or rewrite it so we can use opetal rule so let's see if we can do that so i'm going to go ahead and use the natural log and e so here now that one over in power will get pulled out and then i'll have the natural log of 1 over 2n plus 1, but i want to just write this as natural log of 2 and plus 1 and not divided, and by doing that, i need to move a negative out front here...