N, log N.
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The harmonic series diverges, but the sum of the reciprocals
of the squares converges. P series with P equals 2 converges.
Alright. So this series converges.
The harmonic series diverges, so this series diverges.
But what's something in between, right?
Where's the boundary between convergence and divergence?
Well here's an example of a series that sits in between.
Its the sum N goes from 4 to infinity of 1 over N log N.
Now, 1 over N log N is less than 1 over
N, because N times log N is bigger than N, right?
A bigger denominator makes the fraction smaller.
And because N squared is bigger than N log N, 1
over N squared is less than 1 over N log N.
So, this thing is bigger
than something that converges, smaller than something that
diverges, so it's, it's really a question, right?
Does this series converge or diverge?
We can use Cauchy Condensation.
So let's analyze this series using Cauchy Condensation.
So I'll write down the condensed series.
It's the sum N goes from 2 to infinity of 2 to the N times the 2 to the Nth term.
So 2 to the Nth term
mean I put a 2 to the N times. Log 2 to the N in the denominator.
Okay, so this is the condensed series, this is the original series.
Cauchy Condensation tells me that these two series share the same fate.
They either both converge or both diverge.
So it's enough to figure out what's going on with this series.
Oh and look!
I've got a 2 to the N in the numerator and a 2 to
the N in the denominator, so we cancel those and we can rewrite this as
the sum N goes from 2 to infinity, which is 1 over log 2 to the N.
Now I could use property of logs to simplify this.
This is the sum N goes from 2 to infinity of 1 over N times log 2.
That's just because log 2 to the N is N times log 2.
And then, well look at this.
This is
1 over log 2 times the sum N goes from 2 infinity of 1 over N.
Oh, but this is bad or really great news depending on your attitude.
This thing here, is the tail of a harmonic series.
So that means that the condensed series diverges.
And because the condensed series diverges, this original series must also diverge.
If you like integrals, you can use an integral test.
Okay.
So let's do this with the integral test.
So I could look at this integral, the integral from
4 to infinity of 1 over X times log X DX.
That's a suitable function to consider.
By definition, this integral from 4 to infinity just means the limit as N
approaches infinity of the integral from 4 to big N of 1 over
X times log X DX.
But now, how do I evaluate that definite integral?
Well, I happen to know this.
The derivative of log of log of X is what?
Well by the chain rule, this is the
derivative log, which is 1 over, evaluate the
inside, log of X, times the derivative of the inside function, which is 1 over X.
Well, here I've got 1 over
X times log X. Here, I've got some other function.
Here I've got an anti-derivative for this integrand.
So, that's enough for me to be able to
calculate this definite integral using the fundamental theorem of calculus.
So this is the limit as N approaches infinity of what is this?
This is telling me an anti-derivative, so log log X evaluated
at 4 and N.
And then I could plug in N and plug in 4 and take the difference.
This is the limit N goes to infinity of log log N minus log log 4.
What is this limit?
Well, it's growing very slowly but it is, in fact, running off to infinity, right.
By choosing N big enough, I can make log N as large as I like,
and that means I can also make log log N as large as I like.
So this limit is in fact, infinity.
That means that this integral diverges.
That means that the original series diverges as well.
So that diverges, but maybe if we mess around
with it a bit we can make it converge.
So our original question was whether or not this series
converged or diverged.
And now we've seen, both by using condensation
and the integral test, that this series diverges.
We're going to go back to our original story, right?
We were thinking of this series as somehow sitting in between these two series.
But now where's the boundary between convergence and divergence, right?
This series converges, these two series diverge.
So let me try to fit another
series in between these two series. Here's an example.
Its the sum N goes from 4 to infinity of 1
over N times log N, and that log N is squared.
Does that series converge or diverge?
We can use condensation. So lets write down the condensed series.
In this case, this is the sum N goes from 2 to infinity of 2
to the N times the 2 to the Nth term.
So, 2 to the N times log 2 to the N squared.
Now, how do I evaluate this series, right?
If I can determine the convergence of this
series, I've determined the convergence of the original series.
Good news.
This 2 to the N and this 2 to the N cancel.
So now I'm left with the sum N goes from 2 to
infinity of 1 over log 2 to
the N squared. What else can I do there?
Oh, I can use properties of logs again. So this is the sum N goes
from 2 to infinity of 1 over N times log 2 squared.
I could factor out the 1 over log 2 squared.
So this is 1 over log 2 squared times the sum N goes from
2 to infinity of 1 over N squared. this is really great.
Right?
because what do I know? This is a P series, where P equals 2.
This series converges.
Well, that means that this condense series converges.
That means the original series converges. So, we've got a convergence
series and a divergence series. It's worth comparing these two series.
In this series, which we were asking
about, we now know converges by Cauchy condensation.
And we've already seen a little while ago that this series diverges.
So again, we could try to play the same game, there's
just something that we try to fit in between these two.
Well, here's our question.
Does this series, which sort of fits
in between these two series, does this series converge or diverge?
This is the series N goes from 4 to infinity
over 1 over N times log N times log log N.
Well, I'll leave it to you to analyze this series.
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