00:01
For this problem, we have the sequence 3 to the power of k over 2 times 4 times 6 all the way up to 2k.
00:07
We are asked to use the ratio test to analyze the monotonicity of the sequence.
00:12
So we want to figure out if ak over ak plus 1 is greater than or less than 1.
00:18
So ak over ak plus 1 would be 3 to the power of k divided by 2 times 4 times 6 dot dot dot 2k, and ak plus 1 rather than dividing by a k plus 1, we can write this as being multiplied by 2 times 4 times 6 times dot dot dot up to 2 times k plus 1 divided by 3 to the power of k plus 1.
00:46
So we can recognize that 3 to the power of k is going to divide out with 3 to the power of k plus 1, leaving us with just a 1 over 3.
00:55
And we can see that our 2s will divide out and our 4s will divide out, and so our 6es leaving us.
01:02
With 2 times k plus 1 divided by 2k times 3.
01:10
So this would be equal to 2k plus 2 divided by 6k, which is then going to be equal to k plus 1 over 3k, dividing out the common factor of 2 here.
01:29
And we can see that if k is, so so if k is greater than 1, then we have that 3k will be greater than 3.
01:51
And actually, that's not the more significant thing here...