Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Infinite series, the sum of an infinite sequence of numbers. So a sub 1 + a sub 2 + a sub three plus out to a sub NA sub n + 1 out to Infinity. So a sub N is the NTH term of the series, the sequence S sub N. Is the NTH partial sum adding the first N terms together, thus a finite series. If the sequence of partial sums converges to a limit L, we say that the series converges and its sum is L. So the infinite Series A sub 1 + a sub two plus... would equal the summation of N equal 1 to Infinity of a sub N which would equal L. If the sequence of partial sums of the series does not converge, we say it diverges. So our NTH term test if the limit is N goes to Infinity of a sub N fails to exist or is different from zero, then that series is going to diverge and the reason is if we can get the partial sum. This a sub one to a sub N to go to a limit and then every term after that is adding zero. Then that would go to the same limit as the a sub one to a sub N portion. If this a sub n + 1 is anything other than adding zero, you can see that it wouldn't be limit the L, it would be something other than L it'd be L plus whatever it was going to. So if a series. If two series converge, then the two series can be added together separately, the two series could be subtracted separately, and the two the one series could be multiplied by a constant, and the constant can come in and out of the summation if one series converges and one diverges. Then the sum of those and the difference of those is going to diverge because if I take a divergent series and I add a number to it, or I take a divergent series and I subtract a number, it's still going to diverge. So there's a caution here if the two converge, if the sum of the two converge. Then they both might independently diverge, and the example of that is if the series of a sub N is just 1 + 1 + 1 and the series of B sub N is -1 + -1 + -1, then those two added together as 0 + 0 + 0, which is a convergent series. So each of the original were diverging, but then the sum was converging. So that's a caution. We could look at things like doing our cosine and sine graphs. Also. If we took our cosine and our sine graphs and added them together, at certain locations, the sum may converge even though the two independently would diverge. Geometric series We're going to look at also, which is a + a * r + a * r ^2 + a * r ^3 plus... And eventually we could rewrite it as the summation of N equal 1 to Infinity of AR to the n -, 1, where the a is the first term and the ratio is R. Now the sum of a geometric series we can actually find. So we're going to say, let SN be the partial sum. So SN equal A+ AR out to a sub R to the n -, 1. And then we're going to multiply that by R on each side. So if I multiply R on the left and I multiply R on the right, then I'm going to subtract them. So these ARARS are going to cancel and the Arkansas squared, Arkansas squared is going to cancel. All these middle terms are going to cancel, leaving us SN minus RSN on the left, equaling a -, a RN on the right. So then I'm going to factor out the S sub N and I'm going to divide each side by 1 -, r Now R can't equal 1. If it did, it wouldn't be a geometric series, it would just be the same number added together over and over and over again. So if it did equal 1, it would diverge because it would be a + a + a + a N times. If R equal -1, then SMN would be equal a negative A, a negative a, and it would oscillate between 0 or A and hence it would also diverge. So let's look at what happens if the absolute value of R is less than one. If we look at the limit as N goes to Infinity of that series, we'd have a 1 -, 0 / 1 -, r or a. Over 1 -, r so it would converge because that's going to be an actual number. If the absolute value of R is greater than one, then the limit is N goes to Infinity of S sub N would equal A1 minus Infinity over 1 -, r and that's going to clearly diverge. It's going to go to Infinity or possibly negative Infinity depending on what our as and Rs are. So if N equal 1 to Infinity, a sub N converges, then you know that a sub N goes to 0. Caution, caution, caution. It doesn't say if a sub N goes to 0 then it converges. That's very very important. Does not say that. So if N equal 1 to Infinity of AF sub N or the series diverges. If the limit of N goes to Infinity of a sub N fails to exist or is different from zero. That's a repeat of the NTH term test, just a little different wording if we look at some examples now. If we look at a few examples, the summation of N equal 1 to Infinity of n ^2. So we have 1 + 4 + n + 16 plus... dot dot. Well, if we look at the NTH term, the limit is N goes to Infinity of the NTH term. We can see that that's going to be Infinity. Hence this series diverges. If we look at the summation of N goes to one of Infinity of n + 1 / n, we're going to see that. We could think of this as 1 + 1 / n So using the NTH term test, when we take the limit as N goes to Infinity, I get one that's not 0. Thus this diverges by the NTH term test. Looking at some other ones, N equal 1 to Infinity of -1 N plus one. This is going to oscillate because it's. It's going to oscillate and thus it's going to diverge. It's an oscillating divergent. It's going to go one negative 1/1 negative 1/1 negative one. It's oscillating between zero and one depending on which term we're at. If we look at another example, N equal 1 to Infinity of negative n / 2 N +5. We could actually do the NTH term test here and use Lopa Tau's rule, so the NTH term test. They're both going to go to Infinity or negative Infinity over Infinity, which makes it an indeterminate form. So LOPA TAZ rule would take -1 / 2. We can see that that would equal negative 1/2, which isn't 0. So by the NTH term test, this one diverges if we look at 9 / 100 + 9 / 100 ^2 + 9 / 100 ^3 plus... This is going to be a geometric series because between each term we're actually multiplying something to get to the next term. So our R here is 1 / 100. We're multiplying each time 1 / 100. Our A sub 1 is going to be 9 / 100, or sometimes it's just referred to as our A as our first term. So 9 / 100 / 1 - 1 / 100 is going to be the sum because our R was less than the absolute value of R was less than one. So we know this does converge and we're going to actually find the sum of the convergence. So a / 1 -, r The easiest way to do these is to look at the denominator and multiply the top and bottom by the common denominator. So we'd get 9 / 100 - 1. Or 99 which would reduce to 111th. If we look at this next example, n = 0 to Infinity of cosine N π / 5 N So the numerator is going to oscillate back and forth. It's going to be 1 + -1 + 1 + -1 the denominator 1 five 5 ^2 5 ^3. So we can see that this is a geometric series also. Our first term is 1 and our R is -1 fifth because from one term to the next we're multiplying by -1 fifth. Thus the absolute value of R is less than one. Thus it's converges and we're going to find the sum of that convergence. So a / 1 -, r if I multiply by the common denominator here. Which I didn't do. There are lots of ways to do this, but if I multiply by the common denominator, I could see that that's going to turn into 5 / 5 + 1 or five sixths. You could also get the common denominator of the denominator, which is 6 fifths, and then take the reciprocal. This next one we're actually going to need to use our partial fraction concept. So we're going to look at 6 / 2 N -1 * 2 N plus one. A / 2 N +2 N -1 + b / 2 N plus one, multiply it out, combine our coefficients. So our 2A plus 2B is going to equal 0, our a -, b is going to equal 6. When we solve for A&B, we get A equal 3 and B equal -3. So we can rewrite this as. Our summation of 3 / 2 N -1 - 3 / 2 N plus one. Those threes are just constants, so by our constant rule we could pull the three out in front. We can then see that the summation is 1 / 2 * 1 - 1 + 1 / 2 * 2 - 1 plus... Doing the same thing down here if I list my terms. I'd have three 1 + 1/3 + 1/5 plus... and I'd have minus 1/3 + 1/5 + 1/7 plus... So what's going to happen is this positive 1/3 and this negative 1/3 are going to cancel. This positive 1/5 and -1 fifth is going to cancel. So we're going to eventually have the first term here and the last term here. So we could think of this as the limit as N goes to Infinity of three of 1 -, 1 / 2 N plus one. Now this actually isn't the NTH term because I didn't look at a single term. I expanded the series out and looked at all the terms. And this is actually called the telescoping series. When I look at this and put N equal Infinity, I actually found the limit of the series and the series is going to be 3. Thus the original series converges. Once again. This is called a telescoping series if we look at another example. If we look at the next example, 5 / 1 * 2, five over 2 * 3, five over 3 * 4 plus... dot .5 / n * n + 1 going to pull the five out because it's a constant. We're going to realize that this is another partial fraction, so 1 / n * n plus one a / n + b / n + 1, solving for my A and my B and substituting in. I'm going to get 5 S that five out in front. If I have A to the north plus b / n + 1. I could read or I could think of that as now that I've solved for A&B 1 / n -, 1 / n + 1. Instead of separating these as two different series, I'm going to leave them as one. And it might be a little easier to see. It's just a different way to look at it. Wanted to show both ways. So when I put in one, I get 1 / 1 -, 1 / 2. When I put in two, I get 1 / 2 -, 1 / 3. So all of these middle terms, as I expand out are going to go away and I'm going to end up with just the first term and the last term. So if we thought about this, this pairing here was one of that 1 / n -, 1 / n + 1. And then this next pairing here would be my next N term. And then here is my next N term, so the second term. And the first term are going to cancel of each of them. So the negative half and positive half, the negative 1/3, positive third, negative 1/4, positive 1/4. So down here, this 1 / n is going to cancel with whatever was right in front of it. So we're still going to have the very first and the very last to look at. Now this is a telescoping series. So what happens when I look at this limit? As N goes to Infinity, I have 5 of 1 -, 1 / n + 1. This 1 / n + 1 is going to go to 0, so we're going to get 5 and hence that one converges. If we look at this next example, 7 / 4 + 7 / 4 ^2 + 7 / 4 ^3 + 7 / 4 to the 4th. This is multiplying by a fourth each time. So it's a geometric series. Our R is 1/4, so it's going to converge because the absolute value of R is less than one. Our S sub N is going to equal our first term divided by 1 minus our R so 7 fourths over 1 -, 1/4 here I * 4 / 4, so I got 7 on the top and 4 -, 1 on the bottom, or three. Looking at this next one, it's a more complex partial fraction, so we're going to have a / n + b / n ^2 + C / n + 1 + d / n + 1 ^2. When I multiply that all out, I can do some shortcuts if I see them. So if I thought about if N equaled 0 this term and this term and this term would all cancel and I'd end up with my B. N + 1 ^2 equaling 2N plus one. If I put in zero there my I would end up with B equaling one. Now I can see that there's an n + 1 and three of the terms. So if I said well let's let N equal -1, those 3 terms would cancel. I have to put in -1, so I'd have 2 * -1 + 1 equaling d * -1 ^2, so D would be -1. Now, if we look back here and I actually put in my B and my D, so my B was one, my D is -1. I still have that A / N and C / N + 1. I don't know what those are quite yet, but if I look at the B&D portion and combine them getting a common denominator, I can actually see that the numerator here is the numerator of what I wanted. So then I know that A and C are both going to be 0. So I could actually ignore this piece down here. Now, if you're not comfortable with that, we can do it the way that's a little more traditional, which is to multiply it all out, getting your common denominators, leading coefficients, equaling leading coefficients. So the N cubes equal N cubes, N squareds equal N squareds, etcetera, and solve for ABC and D that way. So what this is going to equal is it's going to equal N equal 1 to Infinity of 1 / n ^2 - 1 / n + 1 ^2. This is called a telescoping series. If I put in one, I get 1 -, 1/4 and the next term is plus 1/4 -, a ninth and the next term is plus a 9th minus a 16th plus... So at this point. We can see that the negative 4th and positive 4th and -1 ninth positive 1/9 are all going to cancel all the way down to that -1 / n ^2. So I have my first term left and my last term left. So the limit is N goes to Infinity is going to equal 1 and it converges. This is the actual sum of the series. It's a telescoping series. We're not using the NTH term test. Really trying to emphasize the difference here. If we look at this next, 1 N goes to 0 to Infinity of negative 1/2 to the NX -3 to the N. If I expand this out, I can see that this is really a geometric series and each time we're multiplying by X -, 3 negative X -, 3 / 2. So the question now becomes what does X have to be in order for this to converge? Well, if I know my ratio, because it's what I multiply to get from one term to the next, one term to the next, I'm going to make that ratio be less than one the the absolute value that less than one. So I know that the absolute value that's got to be in between -1 and one. If I multiply everything through by two and add 3, I can say that this geometric series is going to converge when my X is in between 1:00 and 5:00. I can also find what it's going to converge to. So my S sub NA over 1 -, r My first term back here was 1, so 1 / 1 minus the negative X -, 3 / 2. Just simplifying it out, I'd get 2 / X -, 1. So it's going to converge if my X is in between 1:00 and 5:00 and it's going to converge two 2 / X - 1. For this next example, we're going to do the summation of N goes to 0 to Infinity of e / π to the N So we have 1 + e / π + e / π ^2 plus... dot dot. Hopefully at this point we can see that these are geometric series. When I have a term and I multiply it by something to get to the next term, term multiply it by something to get to the next term. So here our first term is one, our a is 1, and our R is e / π E over π is going to be less than one. The absolute value of e / π is going to be positive, so it is less than one. So we know it's going to converge. To find what it converges to, we're going to use our S of N formula equaling a / 1 -, r Our first term 1 / 1 -, e / π. So multiply top and bottom by π and I get π / π -, e Another example is going to be a telescoping series again. So we have the summation of N equal 1 to Infinity of one over in the natural log of n + 2 - 1 over the natural log of n + 1. If we put in the first few terms to see what's happening when N is one, we get one over natural log of 3 -, 1 over natural log of two. When we put in two, we get one over natural log of 4 -, 1 over natural log of three. When we put in three, we get one over natural log of 5 minus the natural log one over natural log of four. If we go out far enough, we're going to have this one over natural log of n + 2 -, 1 over natural log of n + 1, and then it's going to keep going because it goes out to Infinity. We're going to look at the partial sum. So if we look at the pattern here, we can see that this negative natural log of three and the positive natural log of three are going to cancel. And we can see that this negative natural log of four and the +1 over natural log of four are going to cancel. And if we keep looking this one over natural log of five and the -1 over natural log of five, if we had the next term out, would cancel. So in this first term, it's the last term that still exists. It didn't cancel. So on this very, very far end, it's going to be the first term in that term that didn't cancel. So we're going to get the limit as N goes to Infinity of that -1 over natural log of 2 + 1 over the natural log of n + 2. One over something really, really big is really, really small. So this limits going to go to -1 over natural log of two. And if we look at a term, a single term out far enough, we can see that that's actually going to be 0. So it makes sense that our sum of the partial fraction, our S of N is just going to equal this. So our next thing, we're going to look at repeating decimals. So point D bar over it means that it's going to repeat point DDDDDDDDDDD. So we could think of these place values as d / 10 and D / 100, D over 1000, D over 10,000. And what we're going to do is we're actually going to add up all these pieces. So we'd have d / 10 + d / 10 * 1/10 + d / 10 * 1 / 10 ^2. So to get from one place to the other, we're really multiplying by 110th. To get from one place to the other, we're really multiplying by 110th. So this is a geometric series again, we're going to have our summation from n = 0 to Infinity of DD over 10 of 1 / 10 to the north. Now, we could have used some other summations. That just happens to be the one that I saw quickly. So our a is going to be d / 10 and our R is going to be 110th. And it looks like some of my writing has disappeared. Sorry about that. So what we're going to do is we're going to realize that our R is less than 1, so it's going to converge because it's a geometric series with the R less than one. So we can actually find the sum S of N equal a / 1 - r So in this case S sub N equal d / 10 / 1 - 1 / 10. If I multiply the top and bottom through by 10, I get d / 10 - 1 or 9. So d / 9's the final summation there that the sum goes to and it does converge. If we look at another 1.06 repeating, we could think of this as 6 / 100 and 6 / 1006 / 10,000, etcetera. So our first term is going to be 6 / 100, and then to get to each of the terms after that, we're going to multiply by 1 / 10 to the north. So 6 / 106 / 100 times. 110th, 6 / 100 * 1 / 10 ^2. It's a geometric series. Again, our R is 110th, so it's going to converge again. And when we look at the convergence, S of N equal a / 1 -, r six over 100 / 1 minus 110th 6 / 100 -, 10, six 90th, 115th. Now, the power of these two examples really is coming from the fact that now I can take a repeating decimal and find out what it's equivalent to in its fractional form. So .06 is really the same thing as 115th. If I have point D, that's really the same thing as d / 9. Point D repeating is really the same thing as d / 9. We can use this even when the repeating is not part of the beginning. So if I have 1.24123123123123, I'm just going to take this 1.24 and I'm going to add it in to the summation. So now we have 123 over 12345 zeros, 12345. And then from this place to the next, we're multiplying by 1 / 1000 because there are three decimal places to the end power. So when we look at this piece here, it's a geometric series. Our ratio is 1 / 1000, so it converges. So now we know that our first term, our A is going to be that 123 / 100,000. So we have our original 1.24 that wasn't part of the repeat plus 123 / 100,000 / 1 / 1000. So when we simplify that all up, plug and chug, I multiply by 100,000 / 100,000 here. So 123 over 99900. Those are both divisible by three, so 41 over 333001.24 turned to 124 / 100, getting common denominator, adding them together. That's my final answer. Looking at this next example, if we have n = 0 to Infinity of 5 / 2 N plus 1 / 3 N, If we look at this expanded out, we get 5 + 1 + 5 has plus 1/3 + 5 fourths plus a 9th, etcetera. So now if we split these up into two separate summations, we can see that 5 + 5 has plus 5 fourths plus... This is a geometric series because we're multiplying by 1/2 each time, so our R is 1/2. That's less than one, thus it converges. When we look at the second piece, 11, third one, ninth, etcetera, our R is 1/3. Here it's less than one, so it converges. So we can find the sum. Now remember, if either of them diverge, then the whole thing diverges and we couldn't find a sum. So the limit is N goes to Infinity of the partial sum. SN equals the a term divided by 1 -, R Plus the other a term divided by 1 -, r So we get 10 + 3 halves or 23 halves. Thank you and have a.