in this video we're asked to test the convergence of the series n squared e to the negative n as n
goes from zero to infinity and we're going to use the integral test for this i'm not saying it's
the only way to do it but i wanted to look at the integral test to approach this thing so to perform
the integral test we look at the integral of the related function x squared e to the negative x on
the interval one to infinity and if this integral converges then we're guaranteed that the entire
infinite series converges in other words it's got to add up to something finite well this is
a classic integration by parts integral and we're going to let u equal x squared and that means d u
is 2x dx dv is the rest of the integral that was e to the negative x dx and that means v is negative
e to the negative x now we apply the integration by parts formula and that starts with a u times v
so negative x squared e to the negative x that's evaluated from one to infinity minus
the integral of v d u and v is carrying a minus sign here so we're going to pull that in front
of the integral and cancel that minus sign and i'm left with the integral of 2x e to the negative x
dx and things have gotten simpler here the integral that we're left with only has
x to the first power in it because we chose that original substitution u equals x squared
so that integral's a little bit better because we knocked down the power of x
we are going to have to apply integration by parts one more time to knock that power of x
down again before we do that i want to evaluate this first piece of the expression and the lower
limit is not a big deal there if i sub in x equals one i get a negative one out in front and
an e to the negative one so not a big deal but subbing in the infinity is a little bit tricky
and the issue there is that i end up with an indeterminate form if i sub in infinity for x i
get an infinity squared e to the negative infinity well e to the negative infinity is 1 over e to the
infinity or 0. so this is an infinity times zero indeterminate form so we need to be a little more
formal about it and we're going to re-express this as a limit the limit as x goes to infinity
negative x squared e to the negative x again that's an infinity times zero indeterminant
form and what we want to do is we express this as a fraction and then apply l'hopital's rule so
i write it as negative x squared over e to the x and now this thing is an infinity divided by
infinity indeterminate form i'm allowed to apply l'hopital's world to this and with a
little bit of experience you see what's coming here i differentiate the numerator and i get a
negative 2x differentiate the denominator it just returns an e to the x to me and what i mean by
seeing what's coming is that that denominator is never going to change even as the power in
the numerator keeps going down so this is still an infinity over infinity indeterminate form i
apply l'hopital's rule for the second time and i get a negative 2 over e to the x now this is
unambiguous if i sub in x equals infinity i have a constant divided by e to the infinity
a constant divided by infinity is always going to give you 0. so when i sub in the upper limit
to this first piece i'm getting zero and then i subtract what i get when i sub in the one
for the lower limit subtracting that lower limit kills the minus sign that was out in front and
i just get a 1 times e to the negative 1 or a nicer way to say that is 1 over e i'm going to go
ahead and pull the 2 out of the integral while i'm writing this step and there we go we have 1 over e
plus twice the integral from 1 to infinity of x e to the negative x dx we have to apply integration
by parts for the second time and i'm going to let u equal x this time that means d u is equal to dx
dv is the rest of the integral which was the e to the negative x dx which means v is negative
e to the negative x now don't forget to copy down that one over e that was a spectator
here and then we have twice this integral and we're going to apply the integration by parts
and u times v is a negative x e to the negative x we're evaluating this from one to infinity and
then i have minus the integral of v d u and again the v is carrying a minus sign and i can pull that
in front and now i just have the integral from one to infinity e to the negative x dx and i'll
go ahead and just compute that integral before we worry about evaluating across the limits of
integration all i do is guess the anti-derivative which is negative e to the negative x
and i'm evaluating from one to infinity now when i sub in the infinity in the second term the result
is unambiguous that's e to the negative infinity or one over e to the infinity i'm going to get
0 out of that but the upper limit in the first term when i sub in that infinity i'm stuck in the
same situation where i get an indeterminate form infinity times 0. so i need to do a little side
note to compute that using l'hopital's rule i'm going to go ahead and streamline it a little bit
this time and immediately write it as a negative x over e to the x that's an infinity over infinity
in determinant form i apply l'hopital's rule and get a negative 1 over e to the x
this unambiguously goes to 0 as x goes to infinity so that upper limit vanishes so getting back to
our expression for the original integral i have a 1 over e plus a 2 and then in my first chunk here
my upper limit vanished and then i subtract what i get when i sub in the lower limit well the
subtraction is going to kill the minus sign that was out in front so i get an x e to the negative
x where x is one that's a one e to the negative one or a one over e in the second piece again the
upper limit vanished i subtract what i get in the lower limit giving me a positive sign and what i
get in the lower limit is just e to the negative 1 or 1 over e now in those parentheses i have a 1
over e plus 1 over e which is 2 over e and then i'm multiplying it by 2 which gives me 4 over e
i add that to 1 over e out in front and i get a 5 over e and the whole point with the integral
test again is that if this integral converges to some finite number it guarantees the series itself
converges to some finite number so this is really the end of the argument i showed that the integral
converged therefore the series itself converges as well and we're done if you enjoyed this video
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