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### Course: AP®︎ Calculus BC (2017 edition) > Unit 12

Lesson 13: Optional videos- Formal definition for limit of a sequence
- Proving a sequence converges using the formal definition
- Finite geometric series formula
- Infinite geometric series formula intuition
- Proof of infinite geometric series as a limit
- Proof of p-series convergence criteria

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# Proof of p-series convergence criteria

A p-series converges for p>1 and diverges for 0.

## Want to join the conversation?

- The graphs are wrong! The right graph is integral from 1 to infinity of 1/x^p +1,kind of . But when Sal changed form left Reimann sum to right Reimann sum, the graph shrinks. If the left graph is the integral from 1 to infinity of 1/x^p, the right graph is not the left graph +1. Sal just transferred the left graph left 1 unit. That is confusing or least I don't understand. So I think we should make a new video or just explain to me......Please.(12 votes)
- It's a little confusing because there are only two graphs but there are THREE things in the inequality. The
**orange rectangles**in both graphs are exactly the same, Σ[1, ∞] 1/n^p, and**this is the infinite series we want to sandwich**. The idea is that in the left graph, this set of orange rectangles is exactly a Left Endpoint Riemann Sum, and is clearly greater that the integral from 1 to ∞.

Now, when he shifts the rectangles left one unit, they*almost*form a Right Endpoint Riemann Sum (RRS), BUT the first rectangle (area = 1), which he colors in reddish-purple, is**not part of the Right Riemann Sum**for the integral from 1 to ∞. So the (integral from 1 to ∞) > RRS, but our orange rectangles are 1 + RRS, and we want something bigger than the orange rectangles, so the right end of the inequality is 1 + the integral.

To sum up:

Integral from 1 to ∞ of 1/x^p = INT

Infinite series = orange rectangles (OR) = Left RS = 1 + Right RS**FIRST**

INT < Left RS

and thus

INT < OR**SECOND**

Right RS < INT

and thus

1 + Right RS < 1 + INT

and thus

OR < 1 + INT

And so finally (since OR = orange rectangles = infinite series):

INT < infinite series < 1 + INT(19 votes)

- It doesn't effect the outcome, but around7:20when Sal factors out the 1/(1-p) shouldn't it have made the rest [M^(1-p)]-1 ?(3 votes)
- At7:20, he factored 1/(1-p) out from the 1st term only.

For the second term, he simplified 1^(1-p) to just 1 in the numerator because one to any power (any p-value) is still just 1. Because the simplified second term 1/(1-p) is just a constant, and will be the same regardless of the value of m, it does not affect whether the limit converges or diverges. So to determine convergence/divergence, he only uses the first term. I hope this helps.(4 votes)

- I don't understand p-series. is 1/(ln3)^n n = 1 to infinity a p series, ? does it converge or diverge.?(2 votes)
- Variable as the exponent => Geometric series

Variable as the base => p-series(2 votes)

- why does f(x) decrease if p > 0? If p is a fraction, couldn't it increase?(2 votes)
- Why are Riemann sums being used to visualise? Aren't definite integrals the exact area under the curve, not approximations?(1 vote)
- The integral gives the exact area under the curve, but the p-series corresponds to the sum of the rectangles. So in this case it's not that Riemann sums are being used to approximate the area, but rather that the (exact) area is bounding the discrete sum. Interestingly, the integrals are giving the conditions under which the series converges, but they are not actually giving the
*value*to which it converges. (There's no general formula for the sum in terms of`p`

.)(2 votes)

- why p have to be greater than 0?(1 vote)
- Because we define p to be greater than 0.

We know that when p is zero, you're adding ∞ * 1.

We know that when p is negative, you're adding an infinite number of increasing positive terms, which obviously diverges.(1 vote)

- Okay, but you said in the integral test videos, that if the a function failed the divergence integral test, it doesn't mean for sure it converges, can someone explain ?(1 vote)
- What if the bounds were changed to 0 to infinity? How would the range of p change then? Thank you.(1 vote)
- The integrals diverge because you have an improper integral, and the series diverges because you always have a 1/0, and thus no value of p makes the series converge.(1 vote)

## Video transcript

- [Instructor] You might
recognize what we have here in yellow as the
general form of a p-Series, and what we're going
to do in this video is think about under which conditions, for what 'P's will this p-Series converge. And for it to be a p-Series, by definition P is going to be grater than zero. So I've set up some
visualizations to think about how we are going to understand when this p-Series converges. So over here you have the graph, this curve right here,
that's Y is equal to one over X to the P. And we're saying it in general terms, because P is greater
than zero, we know it's going to be a decreasing
function like this. Once again that's Y is equal to one over X to the P. And now what we've
shaded in ahead of time, underneath that curve, above the positive X axis, that's is the integral. From one to infinity, the improper integral
of one over X to the P, dX, so that's this area, that I've already shaded in, you see it in white in both of these graphs. And what we're going to hopefully see visually, is that there's a very close convergence or divergence
relationship between this p-Series, and this
integral, right over here. Because when we look at
this left hand graph, we see that this p-Series can be viewed as an upper Riemann
approximation of that area. What do I mean by that? Well think about the area of this first rectangle. The width is one, and its' height is one over one over one to the P. So this would be the first term, in this p-Series, this would just be an area of one. It's just the X and Y
scales are not the same. This one right over here; its' area would be one over two to the P. The area's one over three to the P. So the sum of the areas
of these rectangles, that is what this p-Series is, and you can see that each of these rectangles they are covering more than
the area under the curve. And so we know the area under the curve, that's gonna be greater than zero, this p-Series is going
to be greater than this integral, greater than
the area under the curve. But if we add one to the
area under the curve, now we're not just talking
about the white area, we're also talking about
this red area here, then our p-Series is going
to be less than that. Because the first term of
our p-Series is equal to one, and then all of the other terms, you can view it as a lower
Riemann approximation of the curve. And you can see, they fit under the curve, and they leave some area. So this is gonna be less
than that expression there. Now think about what happens; if we know that this right over here diverges, so if this improper integral diverges, it doesn't converge to a finite value, well the p-Series is greater than that, so if this diverges, then
that's going to diverge. Similarly, if this converges, the same integral right over here, if this converges, it
goes to a finite value, well one plus that is
still going to converge, so this, or p-Series must also converge, it must go to a finite value. All I'm talking about right here, this is really just the integral
test, when we think about tests of convergence and divergence, but I'm just making
sure that we have a nice conceptual understanding, and not just blindly applying the integral test. And you could go the other way too, if the p-Series converges,
then for sure this integral is going to converge, and if the p-Series diverges, then for sure this expression right
over here is going to diverge and the integral diverges. So we can say the p-Series converges if, and only if, this integral right over here converges. So, figuring out under what conditions for what P does the p-Series converge, it's boiling down to under what conditions does this integral converge? So let's scroll on down to
give us some real estate to think about what
has to be true for that integral to converge. So I'm gonna re-write it. So we've got the integral
from one to infinity, improper integral, of
one over X to the P, dX this is the same thing, this is the limit as, I'll use the variable
M since we're already using N, as M approaches infinity, and the integral from one to M of one over, and actually just let me
write that as X to the -P. X to the negative P, dX. And let me just focus on this, and we'll just remember that we're
gonna have to take the limit as M approaches infinity I don't wanna have to keep writing that over and over again. So let's think about what this is, so there's a couple of conditions; we know that P is greater than zero, but there's two situations
right over here: there's one situation
when P is equal to one. If P is equal to one,
then this is just the integral of one over X, and so this thing is going to be the integral of ln of X, and we're gonna go from one to M, and so this would be the natural log, of M minus the natural log of one, well E to the zero power is
one, so the natural log, I'll write it out, the natural log of one, but the natural log of one is just zero, so in this special case,
I guess we can say, when P equals one, this integral, from one to M, comes down to the natural log of M. Now let's think about the situation where P does not equal one, well there we're just
kind of reversing the power rule that we learned in basic differentiation. So we'd increment that exponent, so it'd be X to the negative P plus one, and we can even write that as X to the one minus P,
that's the same thing as negative P plus one, then we would divide by that. So one, minus P. We are gonna go from one to M, and so this is going to be equal to, we could write this as M to the one minus P, over one minus P minus, one to the one minus P, over one minus P. So now let's take the limits, so remember this integral,
we won't take the anti-derivative or the
definite integral here, but then we want to take the limit here as M approaches infinity. So what is the limit as
M approaches infinity, of natural log, of M? Well if M goes unbounded to infinity, well the natural log
of that is still going to go to infinity. So when P equals one, this
thing doesn't converge, this thing is just unbounded. So, P equals one, we diverge. So we know that. So now let's look over here, let's think about the limit as M approaches infinity, of this expression right over here. And the only part that's
really effected by the limit, is the part that has M, so we could even write
this as, we could take this one over one minus P out of this, we could say one over one minus P, times the limit as M approaches infinity, of M to the one minus P, and then separately we can subtract one to the one minus P, for any exponent that's just gonna be one, over one minus P. Is that right? Yeah no matter what exponent I put up here one to any power is going to be one. And so the interesting thing about whether it converges or not is,
this part of the expression. And it's all going to
depend on whether this exponent is positive or negative. If one minus P is greater than zero, if I'm going to infinity and I'm taking that thing to a positive exponent, then this is going to diverge. So in this situation we diverge, and one minus P is greater than zero we can add P to both sides, that's the situation, that's the same thing as one being greater than P, or P being less than one, we are going to diverge. So far we know that P is
going to be greater than zero, and we saw if P
is one, or if it's less than one, we're gonna diverge. But if this exponent right
over here is negative, if one minus P is less than zero, well think about it, then it's gonna be one over M to some positive exponent, is one way to think about it. So as M approaches infinity, this whole thing is going to approach zero. So this is actually
going to be a situation where we converge, where
we get to a finite value. And so we add P to both sides, we have one is less than P, we converge. So there you have it, we have established, this integral is going to converge only in the situation where P is greater than one. P>1 you're going to converge. And if zero is less than P
is less than or equal to one, you are going to diverge. And those are then the exact, cause this, our p-Series converges if and only if, this integral converges. And so these exact same constraints apply to our original p-Series. Our original p-Series converges only in the situation where P is greater than one, then we converge. And if zero is less than P
is less than or equal to one, we diverge, there you go.