BAM!!! Mr. Tarrou. In this Calculus lesson we are going to take a quick look at the convergence
or divergence of a p-series. We are going to be introduced to what a p-series is. We
are going to us the integral test which we just learned to figure out if the p-series...
um... from n going to 1 to infinity of 1/n. Basically use the integral test to prove whether
or not the harmonic series, which will be introduced in a second, either converges or
diverges. Then we will go through two last examples of just applying the rules of the
p-series. It is going to go very very quickly. The p-series is a series in the format of
1/n raised to some 'p' power, some constant power of p. And that is going to be, as you
let n start at one and go to infinity, we have 1 over 1 to the p power plus 1 over 2
to the p power plus 1 over 3 to the p power, so on and so on and so on, with the idea of
the series going on forever. And this p-series... this series here... is going to converge if
p is greater than 1 and it is going to diverge if p is greater than zero or less than or
equal to one. By the way, I mentioned the fact that our power can be a constant value
of 1. That means if p is equal to 1 we have this series of 1/n. Expanding that out 1/1
plus 1/2 plus 1/3 plus 1/4... so on and so on and so on. That is called the Harmonic
Series. It has a nice relationship with music. Ok. So we are going to use the Integral Test
to determine the convergence or divergence of this p-series... of the Harmonic Series.
And we are not going to waste our time doing the Integral Test if we can show with just
the n-th term test that this series diverges. In other words, we are going to take a little
minute here and we are going to find the limit as n approaches infinity of 'a' sub n which
is going to be 1/n. Well that is 1 over infinity which is equal to zero. So if the limit as
n approaches infinity of a sub n, if that is not equal to zero you are guaranteed the
series diverges so you can stop the problem right there. If the limit as n approaches
infinity of An is equal to zero, that opens the door... the possibility... the series
converges, it does not however guarantee it. And this is a nice... well when we get done...
keep this in mind, how we found the limit as n went to infinity of An. It came out to
be 0, kind of like... Yeah... Your adding by terms that are going to approach zero,
so it seems like you are effectively at some point by adding by zero. So it would kind...
approximately zero... it kind of makes sense that this is going to converge. But we have
it right here, if p is equal to 1 or less than 1 this actually will not converge. So
let's see... you know... basically if that is true. If we are going to use the Integral
Test to test whether or not this series converges or diverges, we need to make sure that we
meet the three requirements of the integral test. Those are first, if we are going to
work out some kind of integral we need a continuous function. So we are going to have... or let
f(x) equal 1/x. Sweet. Ok. Well now we have this function. Is f positive? Well, yes. If
you know... sometimes this function of 1/x would give negative values. But if we are
concerned about letting this function represent this A sub n term and have a function which
agrees this sequence everywhere we have a positive integer in the domain, then really
we care about what is happening on the positive side... basically where x is greater than
or equal to 1. This is continuous, really we can talk about x greater than or equal
to zero. So this is always going to be positive for those types of values. Is it continuous?
Well this function is undefined at zero, but again it is continuous for wherever we are
comparing... or using this function to represent or agree with this sequence A sub n is equal
to 1/n. So it is going to be continuous for all x values greater than zero. Really we
care about those x values greater than or equal to one. Is it decreasing? So we have
the positive taken care of, we have the continuous taken care of, is f(x) decreasing? Well let's
look at the derivative. So we have this is equal to x to the negative one power. So,
f'(x) is equal to... bringing the power down... -1x... reduce that power by one... We have
-1 over x to the negative two power. That means the first derivative is equal to -1/x^2.
Well yeah, if you take a number and you square it the denominator is always going to be positive.
The numerator of course being -1 is a constant negative value. This derivative is always
going to be negative. Thus, f(x) is a decreasing function. So
f prime... excuse me, yeah, f'(x) is less than zero for x greater than or equal to 1.
So we meet the requirements, the conditions, to apply the Integral Test.
So we are going to look at the definite integral from 1 to infinity of 1/x dx.
Awesome!
Since we have met the conditions for using the Integral Test, if this integral diverges
like if we get an answer of infinity or negative infinity, then this series diverges.
If this integral converges, then this series converges.
I talk about the relationship between these two in the beginning of my... my lesson involving or introducing
the Integral Test for convergence and divergences of series.
I will put a link to that in the description.
At any rate, the integral of the... of 1/x or x to the negative one power is going to be the natural log.
So we have the natural log absolute value of x which really is redundant here because
we are just looking at lower and upper limits which are positive.So we will just go ahead and drop that.
We get the natural log of infinity minus the natural log of 1.
The ln(1) is equal to zero and the natural log of infinity is infinity. So this is going to diverge giving us an answer
of infinity.
So since the... umm...
Since the integral tests shows that we have divergence for that definite integral then therefore
this given series diverges as well.
So, it is certainly not a proof of the rules that we have here for the convergence or divergence rules for
p series. But at least we have done an example showing when p is equal to 1 showing that the Harmonic Series
is a divergent p-series.
Let's get on to two more examples and finish this lesson up.
nananananana...
Alrighty. So we are going to stumble on the paper shredder.
And we are going to determine the convergence or divergence of the p series.
Now here we just have
uh...
a series of additions going on. It is addition, we are adding for infinity.
We don't have a sigma notation here. That means we don't have an A sub n.
And...
We don't really know what type of series this is.
But looking at the n values of
1, 2, 3, 4, and so on.
Well clearly it kind of seems like the numerators are staying a constant value of one.
The denominator has a cube root in it.... and...
Whatever n is, that value keeps showing up in the denominator so it would appear
that A sub n is 1 over
Let's see here. These are matching so 1 over n.
We have the cube root of... and then these match as well.
So it appears that we have 1 over n times the cube root of n which means that our series notation will be
the summation from n starting at 1 and going to infinity
of 1 over again n times the cube root of n.
Well that does not look like a p series. It is not 1 over n to some power.
But, it is not hard to make that happen. We have the summation
from n going to from 1 to infinity of 1 over n times the n to the...
now this power over root, so n to the 1/3. When you multiply like bases you add the exponents.
So we have the summation
from infinity... or from 1 going to infinity of 1 over... 1 plus 1/3 of course is 4/3.
Ah... OK. So now we have it in that exact format of 1 over n to the P.
So this is a p series. This is a p series
with p equaling 4/3.
Well... um.. you know...
4/3 is greater than 1, so therefore
the given
what?!
If p is greater than zero and less than or equal to one.
If p is greater, then the given p series
converges. Alright, one last example.
If you were trying to do these examples before I went over them, maybe you had the same pause that I did.
Because my intention was to take this A sub n term and split it up and talk about how
we have... well... you are going to hear that discussion in a second>
And I realized that I am about to split this up into two separate summation symbols.
This is really just a copy error. I don't know why I included it in the lesson.
Maybe just for a little bit of learning experience.
Ah... Yeah... If you have two terms in the denominator, you are not just going to just separate that fraction
into two separate fractions. It would work if that was in the numerator.
This is a copy error from my notes from when I was getting the on the board.
I apologizes. It should be
1/n^3 plus 1/n^4. Ah... Ok.
Let's try this again.
So we have the series of 1/n^3 plus 1/n^4
And... you know...
Does this converge or diverge? It seems like I have a p series, a couple of p series
in my A sub n... my nth term...
expression... my A sub n expression that is not exactly the same kind of format that we needed
or that we were given in our rules for determining convergence or divergence of p series.
Well that is simple enough.
We just pull it apart. We give each of these terms their own summation symbol just to get
this problem to exact match those rules that we were given.
or the p series rules.
Now...
Great...
We have a p series
exactly in that same format that we were given
with a p
which is equal to 3. That means that 3 is greater than 1.
Therefore this
this term... this series converges.
With the same idea here we have a p series
so I have a p series with a p value of 4. 4 is greater than 1, so that p series converges.
So individually we have 2 series which converge, and now we have a series of the sum of those two nth terms.
That means that also converges.
So since both of these converge
Since both terms in this nth term... both of these
p series
converge the entire... or the given series converges.
... as well. That is the end of my last example.
I am Mr. Tarrou. BAM!!! Go Do Your Homework!