Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. A series from an equal 1 to Infinity of the terms a sub N of non negative terms converges if and only if it's partial sums are bounded from above. Now in reality this index could be anything. It doesn't have to start at 1:00 because we could adjust it because we're going out to Infinity. But the big important piece here is that it's got to be bounded from above. If we look at the harmonic series, N equal 1 to Infinity of one. Over north, we're going to have 1 + 1/2 + 1/3 + 1/4, et cetera. Now we're going to group these, and we're going to group them in terms of halves. So 1 is just going to be 2 / 2, 1/2 is 1/2, and we're going to do a comparison of 1/3 + 1/4, and we're going to group those together as 1/4 + 1/4. We know that 1/3 is bigger than 1/4. So if we can show that this series below is going to diverge, since 1/3 was bigger than 1/4, we would know that the top one would also diverge. So 1/5 and 1/6 and 1/7 and 1/8. One eighth is smaller than 1/5 and smaller than 1/6 and smaller than 1/7. So these four are going to add together to give me a half, and then we're going to do one 9th, 10th, 11th, 12th, 1314, fifteenth, and 16th. And if we think about that, that's 8 terms, 12345678, where each of those terms are in comparison to the 116th. These are each going to be bigger than the 116th or equal to. So eight of those would give me one half, and then the next one, instead of 1248, it would be 16 of them, and 16 of them would add up to give me 1/2 in comparison. So now when we look at this, we can see that we have 1 + 1/2 + 1/2 + 1/2 + 1/2 plus... Dot halves and halves and halves, so it's going to diverge. And this bottom sequence, this bottom series is going to be smaller than the one right above it. And so by the sandwich theorem, we know actually not by the sandwich theorem, by the theorem that says that if the smaller one diverges, than the bigger one must also diverge. This is a really important one that we're going to look at a lot. The harmonic series is going to diverge. If we look at this next one, we're going to have an integral test and we're going to let the Series A sub N be a sequence of positive terms. Suppose that a sub N equal F of N where F is a continuous positive decreasing function of X for all X that's greater than or equal to some N where N is a positive integer. This piece is going to be important because it doesn't have to be true everywhere, it just has to be true at some location where N is a positive integer. So it might not be true up until X equal 10 or X equal 50. Then the series N equal that capital N, whatever it was out to Infinity of a sub N and the integral N to Infinity of F of XDX both converge or both diverge. So we're going to look at a sequence of terms. We're going to see if the function is going to be continuous positive and decreasing. If so, we're going to try to evaluate the integral, and if the integral goes to a number, it converges and so does the series. If it diverges, then so does the series. So if we look at an example here, N equal 1 to Infinity of 5 / n + 1 first thing F of X = 5 / X + 1. We can see that that's positive for all the values one to Infinity, and it's continuous. The place it's discontinuous is at -1 when that denominator is 0, and that's not part of our values here. To figure out if it's decreasing or not, we're going to look at the first derivative, and the first derivative is going to be -5 / X + 1 ^2. So we know that this function is decreasing. So now we're going to take the limit as N goes to Infinity of one to N of 5 / X + 1 DX. We know that that's just the limit as N goes to Infinity of five natural log of X + 1 from 1 to N Putting in our bounds, we get limit as N goes to Infinity of five natural log of n + 1 -, l and 2. So we would get Infinity minus a constant or Infinity. Thus they both have to diverge by the integral test. So important pieces on the integral test, the original function has to be positive and continuous. The derivative of the original function has to be negative to show it's decreasing. We have to be able to evaluate the integral. If it goes to Infinity or if it oscillates, or if it does not exist, it's going to diverge. If it goes to a number, it's going to converge. They're both going to diverge or both going to converge. Let's look at another one. If I have N equal 1 to Infinity of 1 / n times the quantity one plus lane squared of NF of X = 1 over. So it's positive and continuous. If we look, it's going to be discontinuous at zero and then the one plus lane squared X is never going to be 0. So looking at the first derivative, derivative of top is 0 times the bottom minus the derivative of the bottom using the product rule times the top, which was one all over the bottom squared. When we simplify this, we can see all the terms in the numerator are negative and the bottom is both squared. So those are going to be positive. So it's got to be decreasing. So we're going to evaluate the limit as N goes to Infinity of one over N12 north of 1 / X times the quantity one plus lane X ^2 DX. We're going to do AU sub here next. So if we let U equal lane X, then DU is one over XDX. So here is my one over XDX. And that's all going to get replaced by the DU. So then we end up with 1 / 1 + U ^2. Well this is just really a formula. It's our formula for tangent inverse of U. If we change our bounds, our original bounds were one to N. We know the natural log of one is 0, so our new bottom bound is 0. Putting an N, our natural log of N turns into our top bound. So the limit as N goes to Infinity. Because if we put Infinity in here, natural log of Infinity still going to be Infinity tangent inverse of UL and N over or zero to L&N. The limit is N goes to Infinity of tangent inverse of natural log of N minus tangent inverse of 0. Well, when we stick in Infinity, what angle gives us out tangent of Infinity? And that would be π halves. What angle gives us out tangent of 0? And that's zero. So this one's going to go to π halves. Thus it converges, which means that the original series also converges. They both converge. If we look at this next one, we're looking at the series of N equal 1 to Infinity of 2 / 1 + e to the north. Let F of X equal 2 / 1 + e to the X that's continuous everywhere. It's never undefined, and it's positive everywhere from 1:00 to Infinity. If we look at our first derivative, we get -2 E to the X over the quantity 1 + e to the X ^2. That's going to be negative everywhere, so we're going to use the integral test limit as that goes to Infinity of one to north of 2 / 1 + e to the X DX. Here I'm going to let U equal 1 + e to the X. If U = 1 + e to the X, then DU is going to be E to the X DX. So we could think about dividing E to the X on each side because we don't have an extra E to the X in this integral. So if I divide each side by E to the XE to the X is really the same thing as U - 1. So if U equal 1 + e to the XE to the X is U -, 1. So there's a little bit of tricky. We have to see what's going on kind of stuff happening at this point. So the two, the 1 + e to the X is going to be U, but I needed this DX to be DU over U -, 1. We're going to change our bounds so that we're going to put one in for the X and we're going to get 1 + e to the 1st, and then we're going to put N in and we're going to get 1 + E to the north. Now this turns into a partial fraction, so 2 / U * U - 1 equal a / U + b / U - 1, getting a common denominator 2 equal AU -1 plus BU if I let U equal 1, I can see that B is 2, and if U = 0, then A equal -2. We could also do it just by looking at our coefficients. So we know that A+B has to equal 0 and negative a has to equal 2. So that's coefficients of the U's have an equal coefficients of the U's. That would give us a equal -2 and B equal 2 again. So lots of ways to do that. So now if we substitute, we get limit as N goes to Infinity of 1 + e to the 1 + e to the north of -2 / U + 2 / U -, 1 DU. Evaluating that integral limited as N goes to Infinity of -2 natural log of the absolute value of U + 2 natural log of U -, 1, putting in the upper bound minus the lower bound. So we get limited as N goes to Infinity of -2 natural log 1 + E to the north +2 natural log 1 + E to the north -1 and then it's a minus. But it was a negative. So minus a negative would make it 2 natural log of 1 + E -, 2 natural log of 1 + E - 1. So 1 + E - 1 is really just E. If we simplify these terms up, we're going to get the limit as N goes to Infinity of the natural log of E to the N squared over 1 + e to the north quantity squared plus 1 + e / e all squared. And that's going to really give us these E to the north over 2. The leading coefficient on top and bottom of the same power would be 1. So the natural log of one is just zero. Because when N goes really, really big, we're going to have E to something really, really big over E to something really, really big. So it's going to be the leading coefficients because those will cancel each other out. You could also think, if you wanted, about dividing each term by E to the north squared. Down here there was no N, so N going to Infinity doesn't affect this last term. So 0 plus the natural log of 1 + e / e ^2. So it converges by the integral test. Thus they both converge because the integral converged P series. If P is greater than one, IEP is a positive value greater than one, then F of X = 1 / X to the P is a positive decreasing function of X. It's also going to be continuous. Remember we have to look at the first derivative to make sure it's decreasing, and the first derivative would give us a negative P / X to the P -, 1. But due to the fact that P is greater than one, we can see that that's clearly less than 0. So now we're going to use the integral test because it obeys the three things that have to be true, which is positive and continuous original function and decreasing. So limit as N goes to Infinity of one to N of 1 / X PDX. Using the power rule, add 1 to the exponent and divide by the new exponent. We'd get the limit as N approaches Infinity of X to the negative P + 1 / -P + 1 from 1 to N. Putting in the upper bound minus the lower bound, we can see that the denominators both have a negative P + 1. So I'm going to factor that out in front. Negative P + 1 is really the same thing as 1 -, P in the denominator. This negative P in the exponent, If I wanted it to turn into a positive P, I'm going to put it in the denominator, which is basically going to change the signs of the exponent. So we get 1 / n to the P -, 1. One to any power is really just going to be 1. So now when N goes to Infinity, remember P is a positive value greater than one. So Infinity to something greater than one when it's in the denominator is going to make it go to 0. So 1 / 1 - P * 0 - 1 is going to equal 1 / P - 1. That's some value. That P is some value. Now what this says is it's convergent, and if the integral converges, then the original series converges. It does not say anything about the sum. The sum is not 1 / P -, 1. We just know it converges, not what value it converges to. Now let's look at what happens if P is less than one. If P is less than one, then we know that P -, 1 is less than 0. Literally all we did was subtract the one. So if we multiply through by a -1, we can see that 1 -, P is going to be greater than 0. Using the same steps that we did just a moment ago, we'd get the limit as N goes to Infinity of 1 / 1 -, P and to the negative P + 1 - 1 to the negative P + 1. So the limit as N goes to Infinity of 1 / 1 minus PP is now less 1 -, P or P - 1. P - 1 here is less than 0, so it's a negative value. If P - 1 is a negative value, if we put it in the numerator, it would then be a positive value. So Infinity to a positive value is going to be really, really big. So Infinity -1 is still going to be something really, really big times a constant still going to be something really, really big. So if the P is less than one, it's going to diverge. Now we looked at if P = 1 already. If P equal 1, we get N equal 1 to Infinity 1 / n which is the harmonic series. And the harmonic series we showed diverges and as a refresher it was 1 + 1/2 + 1/3 + 1/4 etcetera. This 1/3 and 1/4 is bigger than 1/2. This one fifth one sixth one 7th 1/8 is bigger than 1/2. So we're going to have 1 + 1/2 + 1/2 + 1/2. So the smaller function diverges, which means the bigger function also has to diverge. So the P test, the P series test, basically says if P is greater than one, it's going to converge. If P is less than or equal to one, it's going to diverge. So if P is greater than one, it's going to converge. If P is less than or equal to one, it diverges. So when we look at an example N equal 1 to Infinity of 1 / n ^2, we could do the integral test. We could show that F of X = 1 / X ^2 F prime of X = -2 / X ^3. So it's continuous, it's positive, and it's decreasing. Then we could do our integral of 1 / X ^2 DX as the limit as N goes to Infinity OF1TO N. Now we can do this every time. But the P series we just proved and the P series says that if we have 1 / n to some P, we just have to look at the P to decide whether it converges or diverges. In this case the P is 2. If P equal 2 it converges. So if we look at some others, N equal 1 to Infinity of -2 / n square roots of N we can bring out the -2 N times. Square root of N is N to the three halves. Our P is 3 halves. If the P is 3 halves, we know that it converges because our P is greater than one. For convergent, if we look at a different example, N equal 1 to Infinity 3 over the cube root of N. Pull the three out in front. If I pull the three out in front, I get the summation N equal 1 to Infinity of 1 / n to the one third. This is AP series where P is 1/3. One third is not greater than 1, so therefore it has to diverge. Summary The P series N equal 1 to Infinity of 1 / n to the PP is greater than one, it converges. P is less than or equal to one. It diverges the integral test N equals some capital N that's a positive integer. So we need a positive integer to start with going out to Infinity of a sub N and the limit as N goes to Infinity of N to NF of XDX. Then both of these converge or both diverge. The geometric series says the limit as N goes to Infinity of a partial sum S of N If the ratio R is less than one it converges to and. This is actually knowing what the sum does converge to a / 1 -, r and if the ratio the absolute value is greater than or equal to one it diverges. The NTH term test says if limit is N goes to Infinity of a sub N doesn't equal 0 or does not exist, it diverges If the limit is N goes to Infinity of a sub n = 0. It's inconclusive. There's no information given. We have to use a different test. Telescoping series N equal 1 to Infinity of a sub N minus the series N equal 1 to Infinity of B sub NA sub 1 -, b sub 1 + a sub 2 -, b sub 2 + a sub 3 - b sub three. Somehow or another, we're going to have combinations of those, and we're going to get an A sub 1 -, b sub N, or it could be a A sub n -, b sub one, depending on how these terms correspond. And it may converge or diverge. And it's going to converge if we actually get a value. If we get a sum, then it's going to converge. If it comes up with anything other than a single sum, it will diverge. Thank you and have a wonderful day.