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I2g-} find the sum n=I n=} Explain in YOur OWn words what is wrong with and how YOU might fix the following statement. ~The series 42X "+37 divergent by the Te...

Question

I2g-} find the sum n=I n=} Explain in YOur OWn words what is wrong with and how YOU might fix the following statement. ~The series 42X "+37 divergent by the Test for Divergence.

I2g-} find the sum n=I n=} Explain in YOur OWn words what is wrong with and how YOU might fix the following statement. ~The series 42X "+37 divergent by the Test for Divergence.



Answers

Use the $n$ th-Term Test for divergence to show that the series is divergent, or state that the test is inconclusive. $$\sum_{n=0}^{\infty} \frac{1}{n+4}$$

Hello, everybody. This is Kevin Truck with new Murad. Let's look at what this infinite Siri's might add up to. Well, there's a lot that's going on for this, but we do know that we're gonna be starting off with is is putting in a value of one and then indexing it every single term from there. So we've got one times one plus one. It's to over one plus two is three plus one times one plus three would be four. We're going to get something along. These lines will be adding that to the next index well behind right that we have two times three and then each of this is going to go up by one is also four times five, and we're gonna continue to add that through over and over and over again. Now, notice that we do have what looks like a denominator that is gonna be larger than the numerator. But instead of trying Teoh solve all this out, we can do what's called an end value on term test to find out whether or not this is going to decrease fast enough that I don't accidentally blow up to infinity. And so here's kind of what that looks like. Let's consider if at the end of all of this, at the end of all of this, I am going to be still adding terms that are not zero village, which there is no end. So really, what? We're talking about its limits here. So let's structure and thought little bit more properly. Let's look at what is the limit as n goes to infinity of end times and plus one over and plus two times and plus three. And really, what we're checking by doing this is we're saying, If I continue doing this forever, never, never, never will I still be adding things. Or will I theoretically eventually be adding zero at my forever term? So let's go ahead and work this on out. What? We can recognize that if I was just to kind of plug this and I'm gonna write this in brackets, because this is this is me being sloppy, but it's a nice little note to give yourself. I plugged in infinity here, which you can't do. It's not a number, but if we're going Teoh allow ourselves a little bit of error, we would see that we're gonna have infinity over infinity again. Not proper mathematics here, but we can recognize that we do have it's called indeterminate form. And so we might use what's called Loba tells rule to simplify out this This problem so low Beatles Rule says that if I have something which would have evaluated out to infinity over infinity, if I was incredibly sloppy, that I can simply do the derivative of the numerator and then the denominator. And that will be a much easier, much easier ratio to consider as I do the limits. Okay, so we're gonna do a product rule in top because we do see that we have end times n plus one. It's gonna be derivative of the first times. The second as is so it's gonna be endless one plus the first, as is times the driver of the second, which is gonna be one over. And then the bottom is going to the same thing driven or the first is gonna just simply be one times the second, as is, plus the derivative of the first times. The second as is now, I did forget some notation here. Remember that this Onley true love hotels rule on Lee works If you are still talking about limits, you're not allowed to step out of limit thought here. So all this is going to simplify out to sea Teoh. Let's see n plus one plus ends, too, and plus one over and then we have en plus two plus n plus three. So to end, plus five now we could actually do Loki tells Rule one more time. Gonna be a little bit simpler this time. Notice that there is a way to evaluate this by simply dividing by the largest and term over here. But I'll leave that as a as a exercise for another video. Lovato's real one, more time driven with the numerator is going to simply be too driven other than denominator is simply going to be, too. We're gonna include the fact that we're talking about the limits here, but the limit as N goes to infinity of 2/2 is just gonna be the limit as N goes to infinity of a constant. So it's just going to simply be that constant. So this is going to be ah, value of one, and what we can say is that because if we re contextualized all the math that we've done, the forever term, if you will, the forever term that I am adding into this series is one. So I'm adding a forever number off ones or things that are not zero is a very loose way of interpreting it. But essentially, what we can conclude is that because this serious does not decrease down to zero ever that I am going Teoh diverge. And this entire Syria's is going Teoh blow up to infinity. So we can kind of just put a little notation here and say that this thing, if I was gonna continue it forever, would approach infinity and grow in value. Okay, it's going Teoh diverge. So you could say maybe something like diverge to or towards infinity.

All right, everybody, welcome. This is Kevin Truck with New Murad. Here we have an infinite Siris of what looks like end over end squared plus three. Now there's a means of evaluating whether or not this is going to converge or diverge. Called the term test. Essentially, what it's asking is it was going to do this salvation forever. Would I eventually be adding nothing mawr to my total or what I'd be adding something indefinitely more to my total. And really, what we're gonna test is what is it that I'm adding as I approach my forever term, if you will grant me a little bit of loose speak right there. No, there's, ah way to test this where we just simply consider that forever term. So we're really looking at the limit is an approaches infinity of this expression and over and squared plus three now lumpy tiles rule. If you're familiar with it, would provide a nice way of evaluating this because I do get an indeterminant form. But I'm also gonna in this video, I'm going to look at just simply comparing the magnitudes of each of these terms. Somebody is a little bit a technique where we just simply divide through by the largest value so noticed that it is fair in limits to divide by, ah, cards to divide by variables. Normally, I wouldn't be allowed to do this in algebra because that variable could be a zero. But limits don't care about moments of dis continuity. They care about overall structures. So essentially what I can say is I can divide the top by Let's propose M squared and I will divide everything of the bottom by in squared two. So let's do end squared over and squared, plus three over X squared. This would be then the limit as n goes to infinity of one over end over one plus three over and scored. So essentially all we've done is we've just taken away the magnitude of the largest term, and we're considering how everything else is changing in relation to that. What we do know that one over n is n goes to infinity is gonna be a zero and that one plus three over and square. It is gonna be one plus zero as any goes to inferior. So I've got 1/1 plus zero, and so this says that I do have AH value of zero. So I am going to eventually be adding It looks like numbers that are of the zeroth power or zeroth size. But that doesn't tell me that I'm going to get there fast enough so I can kind of write a little note to myself here. That says, the term test, the inter term test for divergence. And that's the important part. This test four divergence is inconclusive, and it's inconclusive because I was able Teoh, too. I wasn't able to confirm that. I did always add values that were larger than one. But it doesn't necessarily say anything about how this structure as a whole if it goes down to adding values of zero mega.

You were given following Siri's and you're asked to use a term test to determine just seriousness time. Urgent. First thing that you do is you take the limit. It was an your purchase, infinity facia, and which in this case, is on over one over which is equal to the limit. This other approaches infinity. Both I think it out. You can shock throughout the negatives those one that is needed limit as any purchase and plenty of how earned, which is just simply equal to negative infinity. So you know that was the limit. Has a new approaches infinity from eight to end. This is equal to zero. Then if this is not true than this, Siri's is divergent. So you just found that the limit of Ellen of a one over and is night of Vivendi and this is not equal to zero. So you know that the Siri's diverges based off the end test test for divert, gets

Welcome back, everybody. This is Kevin Chirac with new Murad. Let's take a look at this infinite Siri's. Now we do have, Ah, a structure here that is a little bit difficult to determine whether or not we're going to be converging, or perhaps whether not all of this effort would be, ah, a little bit wasted. Um, and we can just kind of see that again. Let's let's take a look at what the first couple terms are. So eat of the zeros just could be one. And so I've got 1/1 plus zero, so it's gonna be a value of one, and then I'm going to be adding that to e over e plus one, and it's gonna be difficult to determine whether or not all of this is is converging. And so maybe one thing we would want to dio is to test briefly whether or not this diverges. So say whether or not it converges Teoh. Let's use s for some. So if we can confirm that this diverges will have saved ourselves lots of times, we're going to use the term test for divergence, which is going to simply help us to disqualify that Syria's. But if it doesn't work, we we will not have been able to say that we found any value or that it does in fact work, were just essentially doing a really quick filter on everything. So what we can see is that, uh, by the end term test, we're really gonna be testing to figure out whether or not at the D forever value. I'm adding something of size. If I'm always adding something of size than my Siri's is going to continue to grow, and if it continues to grow, then it's obviously not going to convert. Now, if you are too loosely, imagine plugging in an equal to infinity. In all of this structure, you would see that we're gonna get a indeterminant form of infinity over infinity. Or if you want to use more proper Matt speak, if you consider end to grow largely without bounds, then you're going to notice that we're gonna have the numerator grow without bounds, and then the denominator is gonna grow without bounce. So you can very loosely and put it in brackets because this is this is error notation. But if you'll forgive me a little bit of slapping this here. It helps to get the point across that the top is going to be approaching infinity on the bottom is going to be approaching infinity. And so we have an indeterminant form of this infinity over infinity that we would be approaching loopholes. Rule is gonna be a very useful tool for us here because it allows us to say that the limit is of the of ah of a quotient. Structure like this is going to be equal to the limit of the question structure made up of the derivative of the numerator over the drift of the denominator. So if we do each of the and we're going, consider the derivative of each of the end. That would just simply be you to the what about the bottom? We've got E to the end. Plus end is going to be the driven. It would be e to the end plus one. Okay, now notice again that if we're to plug an infinity, refer to consider if end were to grow without bounds in this structure that that would still get me an indeterminant form of the numerator approaching infinity, the nominator approaching infinity. So money is low. Patel's rule one more time wonderful little trick will be tells Rule, and that's going to get me that the limit could further be equated Teoh and growing to infinity of driven eat of the end would be e to the end driven Here the n plus one would just be eat of the n. And so that shows me that I have a value of one. So this entire limit is going to be a value of one. And what that says that we re contextualize is that even as you continue down this sequence, your eventually it's still going to be adding values that are not zero. So they're gonna be adding values. They continue to get closer and closer to one and are definitely not zero. So we get this summation, the summation is going. Teoh diverged. So we're going to say S does not exist. And it's because we have tested for divergence and has passed. And that's it. So we can say because of divergence


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