00:01
Hello, so here we're giving the series, we have n going from 1 to infinity of n times e to the negative n squared.
00:08
So we then have our function, f of x, be equal to x times e to the negative x squared.
00:15
And we have that this is going to be greater than 0 for all x greater than equal to 1.
00:20
So we're positive.
00:21
And we have that both x and e to the negative x squared are going to be continuous.
00:28
So therefore the composition of these two functions is going to be continuous and if we take the derivative we find that the derivative is going to be negative so we are decreasing so we're positive continuous and decreasing so therefore we then use the integral test and we then consider the integral going from one to infinity of well x times e to negative x squared and then take the limit as x our our let's say as r approaches infinity of the integral from one to r of x times e to the negative x squared we're they going to use here a u substitution we can let u be equal to x squared that gives us that d u is equal to two x d x and then we have that one half d u is then going to be equal to x d x so, and then we have our endpoints.
01:29
When x is equal to 1, we get that u is equal to 1, 1 squared.
01:34
And when x is equal to r, we have u is equal to r squared.
01:38
So therefore we then have our, we take the limit again, as r tends to infinity.
01:44
And now we have the integral going from 1 to r squared of just e to the, well, to the negative u times 1 half d u...