00:01
Hello, so here we have our sum is equal to the sum of 1 over n to the n.
00:07
So that's going to be equal to 1, and then plus 1 over 4, 1 over 2 squared, plus 1 over 3 cubed, and so on, up to plus 1 over n to plus 1, and we continue, plus 1m plus 1 plus 1 plus 1 plus 1 plus 2 to the m plus 2, and so on.
00:22
So the partial sum then, s sub n, is going to be equal to just 1 over 1 to the 1 to the 1.
00:30
Plus one over two to the two, plus one over three to the three, and so on, up to plus one over n to the n.
00:40
So if we subtract, for any positive series, we have that s minus s of n is greater than zero.
00:45
So if we do s minus s of n, here we obtain an infinite geometric series, where we have that our initial term a is one, and then we have our common ratio is going to be equal to one over, n plus 1.
01:02
So then we know that the sum then is going to be equal to just the one over the first term, which is 1 divided by 1 minus the common ratio.
01:12
So 1 minus 1 over n plus 1.
01:16
So that's going to give us just n plus 1 divided by n.
01:20
This is equal to just n plus 1 over n...