# ALTERNATING SERIES ESTIMATION THEOREM SINX

So it’s going to be negative one over the square root of a million. I could show you in just right over here that this is going to be positive. So the smallest k for which this is true is going to be k is equal to , So how do we think about it? Notes Practice Problems Assignment Problems. So that tells us that the absolute value of our remainder can’t get any larger than one thousandth. How do I show that? Using the test points,. I’ll do that same pink color. This remainder is definitely going to be a negative value, but it can’t be any more negative than our first term. Let’s say it converges to some value S. So the numerator here just flips the sign. And then we can just subtract one from both sides and we have , is going to be less than or equal to k. So this is going to be positive.

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. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. 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.

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.

## Alternating Series estimation theorem vs taylor remainder

There are a couple of things to note about this test. Let’s say S sub k, this is the first k terms right over here and then you’re going to have a remainder, so plus the remainder after you’ve taken the first k terms.

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This one’s positive, this one’s negative. If you’re seeing this message, it means we’re having trouble loading external resources on our website. So, if you just pair all these terms up, you’re just going to have a whole series of positive terms. So, the divergence test requires us to compute the following limit. We’re going to start at n equals one, and go to infinity of negative one to the n snx one over n squared, which is going to be equal to So square root of k plus one so we can get this out of the denominator.

### Alternating Series estimation theorem vs taylor remainder | Physics Forums

The series can then be written as. And then plus, I’ll just do two more terms. So let’s just remind ourselves what R sub k looks like. This passes the alternating series test and so we know that this converges.

If we say divided bythat’s. Well once again I’m assuming you’ve had a go at sdries. The second condition requires some work however. And then we can just subtract one from both sides and we haveis going to be less than or equal to k. So, if we could figure out some bounds on this remainder, we will figure out the bounds on our actual sum. We’ll be theordm to figure out, “Well, how far is this away from this right over here? Example 3 Determine if the following series is convergent or divergent. So how do we think about it? I’m telling you sinnx far we’ve just kind of worked from the premise that it will be, but let’s actually look at that and feel good about it. Alright I’m assuming you’ve had a go at it. 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.

But the absolute value of this entire thing is going to be less than or equal to the absolute value of the first term, negative one to the k plus two over the square root of k plus one.

## Worked example: alternating series remainder

Now the first thing we could realize, if we just put parentheses like this, we see that this is going to be positive, this term is larger than this term, this term is larger than this term, and we could keep going. So this first term we already know, this right over here slternating this first term right over here is negative 0.

Due to the nature of the rheorem on this site it is best views in landscape mode. The first is easy enough to check. We have a smaller number being subtracted from a larger number.

### Alternating series remainder (video) | Khan Academy

If we take estimationn partial sum of the firstterms that this remainder is actually going to be less than this. So k just has to be greater than or equal toActually, I’ll just write it Let’s see, that isnegative 36 plus 16 is minus 20, so it’s minus nine, is Then you’re going to have a remainder, which hteorem going to be everything estijation.

So we know that the remainder can’t get more negative than this, but let’s also verify that it can’t get more than a positive thousandth either. Let me write that down. Using the test points. We’ve seen this before. If you’re seeing this message, it means we’re having trouble loading external resources on our website.