In this video we're investigating the convergence of the sum then goes from one
to infinity of natural log n over n quantity squared and we're given this weird hint of
well you should do a comparison to this series one over n to the 3 halves and most of this video
is going to be spent trying to motivate why we would even think to compare to something like that
and so when i look at this my first thought
is well this is kind of like 1 over n squared but i have a natural log in the numerator
and that's growing as n becomes large so maybe this thing is not converging as aggressively as 1
over n squared i'm going to compare it to 1 over n and that is the divergent harmonic series so we're
going to do a limit comparison here and see what happens okay so those are the terms of the series
that we're working on and then we divide by the terms of the series we're limit comparing two
that's the same as multiplying by the reciprocal and i can clean things up a little bit
i get natural log of n all squared divided by n squared but one of those is going to cancel
with the one in the numerator so i end up with a divided by n okay that's like an infinity over
infinity indeterminate form so we're going to use l'hopital's rule for this and i want to be
careful not to say that l'hopital's rule applies to a discrete sequence that's being evaluated at
integer values of n so i'm going to instead look at the limit as x goes to infinity
natural log x all squared over x and we know that if the limit of a function of real numbers
settles down to some value then the same is true for the discrete sequence
corresponding to those integer values of x so i use l'hopital's rule once
and i get 2 natural log of x times the derivative of natural log of x there's the
chain rule in action and then the denominator is 1 and i'll rewrite this real quick
as the limit as x goes to infinity 2 natural log
x over x that's also an infinity over infinity in determinant form so i've got to go again
and i end up with a 2 times derivative of natural log of x that's 1 over x over 1 which gives me a
2 over x and the limit as x goes to infinity which is going to go to 0. now unfortunately
we can't conclude anything from this because we took the ratio of the terms of our series
to the terms in a divergent sequence and we found out in the large n limit that goes to zero this is
essentially saying that the terms in our series are smaller than the terms in a divergent series
that doesn't tell us anything because being smaller than infinity doesn't necessarily mean
it's a finite result so this is an inconclusive limit comparison test the next thought that occurs
to me is maybe we should compare to something that's converging a little more aggressively
and the natural choice for that is to compare to the convergent p series 1 over n squared
so let's give that a try when i divide by 1 over n squared that gives n squared over 1.
and out of this one the n squares are going to cancel and i just get the limit as n goes
to infinity of natural log n all squared which clearly goes to infinity and so i've just shown
that my terms are essentially bigger than the terms of a convergent series that's another way of
being inconclusive being larger than a convergence series doesn't show divergence or convergence
all right and then i think to myself and this is where the motivation finally comes in for
the hints i compared to the divergent harmonic series i compared to a series that was converging
more aggressively to that in each case i got these inconclusive results for kind of opposite reasons
why don't we try a power of n that's just right in between those two
so that's the motivation and i set up the limit comparison
again dividing by one over n to the three halves is the same as multiplying by its reciprocal and
i have an n squared in the denominator n to the three halves in the numerator which is
going to leave me with a square root of n in the denominator and that's an infinity over infinity
in determinate form so again i'm going to talk about a function of the real numbers in terms of
x's now just because that makes it so i'm allowed to use differential calculus and so i'm going to
re-express the thing in terms of x then i'm going to take the derivative of the top and bottom again
using l'hopital's rule the square root of x is x to the one half so that differentiates
to one half x to the negative one half which is the same as saying one over two root x
all right i'm going to multiply by the reciprocal
there and i get the limit as x goes to infinity of 4
natural log of x and then when i multiply root x times 1 over x that gives me 1 over
square root of x so i'm just going to throw that in the denominator in the same step
and that's an infinity over infinity in determinate form so i have to differentiate
again using l'hopital's rule it gives me 4 over x in the numerator and again 1 over 2
root x in the denominator and i multiply by the reciprocal of the denominator and i end up with 8
root x over x all right cancel that root x out of the numerator and end up with 8 over the square
root of x as x goes to infinity which is finally an unambiguous thing and that gives me 0. and
again the result of the limit on the function of real numbers is going to be the same as the result
of the limit on the function of integers over here so we know that n to the one over n to
three halves is a convergent p series because p is bigger than one and we just showed basically
that the terms in our sequence are substantially less than the terms in this convergent p series
in the large n limit so in this case we can conclude now that our series actually converges
you