The harmonic series again.
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We've already seen that the harmonic series diverges.
So we've shown that the harmonic series diverges by grouping together the terms
and then showing that each of the groups is bigger than one half.
So summing the harmonic series is even worse then summing 1 plus a half,
plus a half, plus a half, plus a half, plus a half, and so on.
We can also check this with the integral test,
so let's say a sub n equal 1 over n and I am trying to analyze the convergence of
the series n goes from one to infinity of a sub n, that's just the harmonic series.
And to help me I'll look at the function f of x which is 1 over x.
And note that a sub n is f of n.
I want to make sure that f is positive and decreasing.
Right, so I need the function f to be positive.
And decreasing the interval that I care about,
which is the interval from one to infinity.
Well, I know that f is positive because if x is at least one, even if x is at
least zero, alright, f of x, which 1 over x,
is positive, I mean, the reciprocal of a positive number
is positive.
I need to check if f is decreasing, and what that means is that if a is bigger
than b, then I need that f of a, is less than f of b.
On the interval I'm thinking about here.
Well, that's true because if I've got a
and b, say they're both positive numbers even.
Then how does 1 over a compared to 1 over b if a is bigger than b, or if
I got a larger positive number but I take its reciprocal now its smaller.
So this is telling me that F is a
decreasing function, now I will check that the integral diverges.
Well let's compute the integral from one to infinity of f of x dx.
by definition this integral is the limit as big N approaches infinity
of the integral from 1 to big N of f of x dx, now in this case f
is just the one over function, right it's the function that sends x to one over x.
So this is the limit as big N approaches infinity, of
the integral from 1 to big N, just 1 over x dx.
Now, to compute this definite integral, I'm
going to use the fundamental theorem of calculus and
recall that an anti-derivative for the 1 over x function is the natural logarithm.
So, by the fundamental theorem of calculus, this is
the limit as N approaches infinity of an anti-derivative,
which is the natural log, at the one endpoint,
minus the anti-derivative natural log at the other endpoint.
And a log of 1
is 0, so this is just asking.
What's the limit as N approaches infinity, of the natural log of N, right?
What happens when you take the natural log of an enormous number?
Well, admittedly, it's less enormous, but it's still as large as you'd like, right.
By choosing big N, big enough, I can make log N as large as you'd like.
Which is to say that the limit of log N
as N approaches infinity is infinity. So this integral diverges.
So the same is true of the series.
Right. So, by the integral
test. The sum n goes 1 to infinity of' 1 over n,
the harmonic series, diverges as well.
So now you get two proofs that the harmonic series diverges.
Any time in mathematics that you've got two different proofs of
the same theorem, it's worth thinking about how those proofs relate.
In this case which proof do you think is easier?
I mean in a certain sense, the interval test is easier.
Requires less creativity to just integrate one
over x than to do that complicated grouping.
And yet the grouping argument
predates integration.
Alright, the grouping argument is from the 1300s,
and integration, right, wasn't really around until Newton, say.
So yes, although the integral test seems like the easier way to show the harmonic
series diverges, the integral test is dependent on the huge edifice of calculus.
It is requiring more technology. And a grouping argument,
how well somehow requiring more creativity.
Doesn't
require so much theory, to be built up.