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Current time:0:00Total duration:9:18

AP.CALC:

LIM‑7 (EU)

, LIM‑7.B (LO)

, LIM‑7.B.1 (EK)

- [Voiceover] Let's explore
the infinite series. We're going to start at n equals one, and go to infinity of
negative one to the n plus one over n squared, which is
going to be equal to ... Let's see, when n is one,
this is going to be positive. It's going to be one. This, you go minus one over two squares, is minus 1/4 plus 1/9 minus 1/16 plus 1/25 ... I'm actually going to go pretty far ... Minus 1/36, plus 1/49, minus 1/64. Yeah, that's pretty good. I'll stop there. Of course, we keep going on and on and on, and it's an alternating
series, plus, minus, just keeps going on and
on and on and on forever. Now, we know from previous tests, in fact, the alternating series test, that this satisfies the constraints of the alternating series test, and we're able to show that it converges. What we're doing now is,
actually trying to estimate what things converge to. We want to estimate
what this value, S, is. We're going to do that by doing a finite number of calculations, by not having to add this
entire thing together. Let's estimate it by taking, let's say, the partial sum of the first four terms. Let's take these four
terms right over here. Let's call that, that's
going to be S sub four. Then you're going to have a remainder, which is going to be everything else. All of this other stuff, I don't want even the brackets to end. That's going to be your remainder, the remainder, to get
to your actually sum, or whatever's left over when you just take the first four terms. This is from the fifth term
all the way to infinity. We've seen this before. The actual sum is going to
be equal to this partial sum plus this remainder. Well, we can calculate this. This is going to be, let's see ... Common denominator here, see, nine times 16 is 144. That's going to be 144, and then that's going
to be 144 minus 36/144, plus 16/144, minus 9/144. Let's see, that is 144,
negative 36 plus 16 is minus 20, so it's
124 minus nine, is 115. This is all going to be equal to 115/144. I didn't even need a
calculator to figure that out. Plus some remainder. Plus some remainder. So, if we could figure out
some bounds on this remainder, we will figure out the
bounds on our actual sum. We'll be able to figure out, "Well, how far is this away
from this right over here?" There's two ways to think about it. Let's look at it. The first thing I want to see is, I want to show you that this
remainder right over here is definitely going to be positive. I actually encourage
you to pause the video and see if you can prove to yourself that this remainder over here is definitely going to be positive. I'm assuming you've had a go at it. Let's write the remainder down. Actually, I'll just write it ... Actually, I'll write it up here. R sub four is 1/25. Actually, I don't even have
to write it separately. I could show you in just right over here that this is going to be positive. How do I show that? Well, we just pair ... Let's just put some parentheses in here, and just pair these terms like this. 1/25 minus 1/36. 1/36th is less than 1/25. This one's positive, this one's negative. So this is positive. Then you have a positive term. Subtracting from that,
a smaller negative term. So this is going to be positive. So, if you just pair all these terms up, you're just going to have a whole series of positive terms. Just like that, we have established that R sub four, or R
four, we could call it, is going to be greater than zero. R four is going to be greater than zero. Now, the other thing I
want to prove is that this remainder is going to be less than the first term
that we haven't calculated, that the remainder is
going to be less than 1/25. Once again, I encourage
you to pause the video and see if you can put
some parentheses here in a certain way that will convince you that this entire infinite sum here, this remainder, is going to sum up to something that's less
than this first term. Once again, I'm assuming
you've had a go at it, so let's just write it down. I'll do that same pink color. Our remainder, when we
take the partial sum of the first four terms, it's 1/25. The way I'm going to write it, instead of writing minus 1/36, I'm going to write minus, I'm going to put the parentheses now around the second and third terms. This is going to be 1/36 minus 1/49. Then we're going to have minus 1/64 minus ... Actually, the next terms is going to be one over nine squared, 1/81. Then minus, and we keep going like that, on and on and on, on
and on and on, forever. Now, notice what happens. This, this term right over here is positive. We have a smaller number being subtracted from a larger number. This term right over here is positive. We're staring with 1/25, and then we're subtracting a bunch of positive things from it. This thing has to be less than 1/25. R sub four is going to be less than 1/25. Or, we could even write that as R sub four is less than 0.04. 0.04, same things as 1/25. Actually, this logic right over here is the basis for the proof of
the alternating series test. This should make you feel pretty good, that, "Hey, look, this
thing is going to be "greater than zero," and it's increasing, the more
terms that you add to it. But it's bounded from above. It's bounded from above at 1/25, which is a pretty good sense that hey, this thing is going to converge. But that's not what we're going to concern ourselves with here. Here, we just care about this range. The sum is the sum of these two things. So the entire sum is going to be less than 115/144 plus the upper bound on R four. Plus 0.04, and it's
going to be greater than, it's going to be greater than, it's going to be greater than
our partial sum plus zero, because this remainder is
definitely greater than zero. You could just say, it's going to be greater
than our partial sum. And just like that, just doing a calculation that
I was able to do with hand, we're able to get pretty nice bounds around this infinite series. Infinite series. Let's now get the calculator out, just to get a little bit
better sense of things. If we say 115 divided by 144, that's .79861 repeating. This is 0.79861 repeating, is less than S, which is less than this thing plus .04. Let me write that down. Plus .04 gets us to .83861 repeating, 83861 repeating. Actually, I could have
done that in my head. I don't know why I
resorted to a calculator. 0.83861 repeating. And just like that, just a calculation we're
able to do by hand, we were able to come up with a pretty good approximation for S. And the big takeaway from here ... We're going to build on this, but this was really to
give you the intuition with a very concrete example, is when you have an
alternating series like this, the type of alternating
series that satisfies the alternating series test, where you can write it
as negative one to the n, or negative one to the n plus one, times a series of positive terms that are decreasing and
whose limits go to zeros and approaches infinity, not only do those things, not only do those things converge, but you can estimate your error based on the first term
that you're not including. Now, this was one example. It's going to be different depending on whether the first term
is negative or positive, and we're going to have to introduce the idea of absolute value
there, the magnitude. But the big takeaway here is that the magnitude of
your error is going to be no more than the magnitude
of the first term that you're not including
in your partial sum.