 limit of a sum as definite integer. This is more important vis-a-vis the previous one. So in previous session what did we learn? We learned that we can write an integral of a function from a to b as limit of h tending to zero and n tending to infinity h times summation of f of a plus r minus 1 h r from 1 to n. This is what we learned. Now the same thing can be converted back to an integral if you are being asked a question like this. Now remember here, here you are actually dealing with something which we call as tending to zero tending to zero into infinity form tending to zero into infinity form. Okay, so I'll tell you the working rule. How do we convert a limit of a sum into a definite integral? Okay and it is mostly applied where your indeterminate form is tending to zero into infinity form. Okay, so I'll tell you a working rule. Please, please note this down because it is going to help you to solve all the questions related to this. Suppose you have been given any limit to evaluate. So they would give you something like this. Let me take an example, very simple example I'll take. Let's say you have been asked to evaluate this limit. So with this particular concept, we are going to touch upon our limits chapter also which was actually not completed fully. In the month of March when I did that, it was not done fully with you. With this concept it will be completed now. Okay, now let's say we have to evaluate this limit. How will you evaluate this? Let's say with your concept of limits, how will you evaluate this? By the way, most people will say, since n is infinity, this is tending to zero. This is also tending to zero. All these things are tending to zero. Okay, so answer will be zero. By the way, this is wrong if you are saying this. Why it is wrong? Because these quantities are not zero. They are actually tending to zero. If there were finitely many quantities like this, that means let's say there were 500 quantities like this, then you can say zero, no problem. But when such quantities are infinitely many number, it will actually become, this is actually tending to zero into infinity, which is an indeterminate form, which is an indeterminate form. That means it can take any value. Okay, it need not be zero. It can take any value. Okay, so a small quantity present in infinitely many in number can give you a non-zero quantity as well. So this method will not work. Okay, any other approach to solve this question? Any approach that you know of can help you solve this question? I would love to hear it. Okay, I don't think so. Any one of you will be able to find it out. Okay, now the working rule is we normally write down number one, the r-th term over here. Step number one is write down the r-th term. Write down the r-th term of the sum given to you. Okay, so in this example, okay, your r-th term would be 1 by n plus r. Yes or no? As you can see, this term is always in. Okay, only this term is varying from 1, 2, 3 till n. Okay, so your r-th term will be n plus r. That is step number one. Step number two. Step number two, take 1 by n common from, okay, t-r term. For example, in this case, if I take 1 by n out, it will become n by n plus r, right? 1 by n, n by n plus r. That's correct. Step number two clear. So take 1 by n common from t-r. If I take 1 by n common from t-r, I'll be left with 1 by n times n by n plus r. Correct. Okay, let me call this term as g n. Okay, let me not name it or let me name it because then I can refer to it. Okay, in fact, it is a function of both r and n, so I'll let it as function of r. Step number three, this is the most important step. Convert g r n as a function of, let's say, r by n. For example, in this case, in this case, you can write n by n plus r as 1 by 1 plus r by n. Okay, so this is basically converted as a function of r by n. Okay, so two things. Number one, take 1 by n out. Number two, the remaining part of the function, convert it as a function of r by n. Four step. Four step. Do these replacements. Do these following replacements. Number one replacement. Replace your 1 by n with r dx. Replace r by n with x and summation with an integral sign. Okay, and the limits also. And the limits also. Limits accordingly. Now I'll explain this through this example. So what I'm going to do is basically what I was trying to do, I was trying to sum up summation r. Okay, and the summation was going from n equal to r equal to one till n, where your n was very, very large number. Right, this is what you were doing. So what you would what you did now was you converted, you converted tr as 1 by n times 1 by 1 plus r by n. Okay, r equal to 1 till n. Now what you have to do is replace this guy with a dx. Replace this guy with an x. Replace summation symbol with an integration symbol. Now what about limit of integration? Now see, your x is r by n, right? You have put this as your x, right? Yes. So when your r is 1, x will be 0. So your lower limit will become 0. And when your r is n, your upper limit will become n by n, which is 1. So it will be integrated from 0 to 1. So this summation of let me write 1 by n plus r, r from 1 to n, where n is a very, very large number will be converted to an integral like this. And once it is converted to an integral like this, you know how to evaluate it. So it is limit mod 1 plus x, 0 to 1. So that will give you ln 2. So this series that you see is actually ln 2, which is obvious also because if you look at ln 1 plus x, Maclaurin series expansion, it is given by x minus x square by 2 minus x cube by 3 and so on till infinity, correct? If you put a 1, what do you get? 1 minus half, sorry this is plus, plus 1 third, minus 1 fourth and so on till infinity, correct? Does it match with this or is there any connection between these two? Hello, somebody is trying to say something? It is very feeble. Who is this? It is Siddhartha sir. Siddhartha, yeah Siddhartha. Hello, Siddhartha, can you hear me? Yeah, yeah, yeah. Why can't you put 1 in this series? Its radius of convergence is from minus 1 to 1, 1 inclusive. This is the radius of convergence, got it? Okay, so I leave this as an exercise for you to figure out that this sum and this sum will converge to the same, okay? Anyways, so you got an idea about what is the process involved when you are trying to convert the limit of a sum to a definite integral. Step number 1, I'll repeat once again, write your R at term. If you make a mistake in this, then no point going forward. Everything will become wrong from there on. Step number 2, take one man out. Always, one man will always be taken out and that will play the role of your DX term. Step number 3, whatever is the term remaining after taking one by an out. Okay, this term, write it as a function of R band. So here I will write it down. Your aim should be to write it as a function of R band because ultimately we have to substitute R band as an X, okay? And step number 4, make these replacements change one man with DX, change R band with an X, change your summation symbol to integration symbol and depending upon your R is going from what value to what value, your limit of integration will accordingly change. Is that clear? Let us take some examples on it that will make it even more clearer. Let's take these four questions. Let me know the answer for the first one. First one, I think you do not need definite integrals to evaluate it. Your I think sigma method of summation will work. But still I would like you to solve it by your limit of a sum as definite integral. Done first one. One second sir. Is that in somewhere in the middle of a one by N comes that's directly zero no? No, no, no. Convert it to DX. One by N you just take out from the TR term that one by you reserve for a DX. Okay. And the rest of the functions should definitely be a function of R band somehow. No sir, I meant one by N is taken out and then you get R by N minus one by N. So that one by N minus one. No, no, no, no. You'll never get a case like that. Then some mistake is happening. See, R term here would be R by N square. Correct? No. So when you take a one by N out, you'll be left with R band. So this will be a straight of a X. Okay, okay. Okay. Yes, I got it. So when you're doing a summation of this, some R equal to one till N, in fact, N minus one. Okay, doesn't make a lot of a difference to us. So what will happen? What will happen? This will play the role of DX. This will play the role of DX. This will play the role of X. Okay, so you're indicating X DX. Now what is the limit of integration? Remember, when R is one, when R is one, X will be zero. And when R is N minus one, your X will be N minus one by N. And if you take a limit of this, it'll actually be a one. Okay, so it'll be zero to one. That means it'll be just X square by two from zero to one. That's nothing but a half. Anyhow, this was an easy problem which you could have done it in this way also. One plus two till N minus one by N square limit N tending to infinity. So it will be a clear cut case of infinity by infinity. Okay, so this was nothing but N minus one into N by two. Okay. And if you cancel this out and divide by N, you will get half limit N tending to infinity one minus one by N. And one by N as N into infinity is zero. So it'll be half again. Okay, so another method to do it. So we have just basically confirmed that our answer is correct. Whether we go by this approach or whether we go by this approach. Second one we have already done. So you can skip the second one. You can go to the third one directly. Third one. Sir, could you go down slightly? Is that okay? Yeah. Done, five by four. Okay, good. Okay, sir. Second. Done. Okay. Again, quickly. I'm directly writing it like this. Okay, so that don't waste all of time. Correct. So now just one shot it'll take to convert it to a function of R by N. So this is a function of R by N. I hope you understand the meaning of function of R by N. That means R by N should be the only composite term that is to be seen apart from constants. They should not be any stray R or stray N terms. Okay. So now if you're summing this up from R equal to one till N, it is like integrating one by one plus x squared dx. Okay. Now remember your limit of integration will be from from R is going from one to N. That means your limit of integration will be from zero to one. And that is nothing but tan inverse x from zero to one, which is five by four. Fourth one, please try out fourth one. Sir, can you show the question? Thank you, sir. I'll write it down. Limit ending to infinity. One to the power P to the power P to the power P is a positive power. So this is very easy. Again, our term would be what our term would be R to the power P by N to the power P plus one. If you take out a one by an out, it'll be having a balanced power of power and P both in the numerator and denominator. Okay. So summing this up, summing this up from R equal to one till N is like integrating x to the power P dx from zero to one. So it'll be x to the power P plus one by P plus one, zero to one. So that'll be one by P plus one. Easy. Okay. So let's move on to the next set of questions. Okay. Do you have access to pages, Venkat? I have another one. Okay. Very good. So if you take a N out, you will automatically see under root of four minus R by N the whole square. Okay. So clearly, your summation of TR from R equal to one till N minus one, sorry, zero to N minus one. Okay. Won't make a difference where your N is a very, very large quantity. We'll convert to one by integral of four minus x square dx. Remember when R is zero, x will be zero. And when R is N minus one, x will tend to one. Okay. So this will give you nothing but sine inverse of x by two, zero to one, which is five by six. Can I go on to the next page? Yes. Evaluate limit and tending to infinity. Just give me a second. Let me just write down this. N factorial by N to the power N, whole less to the power one by N. So please copy that, then I'll go to the previous slide. None copied. Okay. Somebody wanted to have a look at the previous slide. Okay. Here you go. Thank you. Any idea? See, first of all, you have to take a log. That is for sure. Okay. Take a log to the base E on both the sides. Remember log can be applied to this function directly. So it'll be one by N, LN of N factorial. Now N factorial, you write it as one, two, three till N. And this N to the power N, you distribute between these terms like this. That's a hint for you. Now you can proceed from here on. So think as if this is a new limit problem that you have been given. How will you evaluate it? Okay. Let me give you a further hint on this. So this will be one by N, LN one by N. Okay. One by N, LN two by N and so on till one by N, LN, N by N. Correct. So if you pick out any rth term from this, your rth term here will be one by N, LN, R by N. Okay. So automatically one by N is being separated out. So your summation of TR from R equal to one till N will be nothing but integral of LNX dx from zero to one, if I'm not wrong. This is something which you already know XLNX minus X from zero to one. Correct. And that's going to give you, when you put a one here, everything will become a zero other than a minus one here or here. And when you put a zero, everything will become a zero. So finally log of the limit to the base E is equal to minus one. That means the limit value is one by N. Is the idea clear? So don't be, don't shy away from taking a log. Okay. After taking a log that gets converted to a series, write down the rth term of a series or take out the one by N term treated as DX, the remaining term, write it as a function of R by N, replace your R by N with an X and integrate it, replace your summation symbol with an integration symbol. And depending upon from where to where you're integrating, okay, your limits of integration will change. Is that clear? Yes, sir. Okay. Try this one out. This one would be a relatively easier one for you to crack now. Done. Very good. What is the answer, Venkat? The log three. Okay. There are two types of answers I'm getting. One is log three and log three by two. Okay. Never mind. Let's check. Okay. Now, if you take a one by N common from here, you will end up getting, correct me if I'm wrong, two plus R by N. R by N, yes. So your summation is happening from R equal to one till four N my dear. Yes, sir. Limit N tending to infinity. Okay. That means your integration will be one by X plus two DX from zero to four. So this will give you ln mod X plus two from zero to four. That is ln six minus ln two, which is ln three. Is that fine? So this problem I gave you because I wanted you to be careful about from where to where you are summing it up. You're not summing it to three N. That means your R is not going from one to N. It is going from one to four N. Okay. So your limit of integration will become zero to four. So be careful about that. Next question. Let me put the poll on for this. Let's see. Two minutes gone. One minute. Okay. All right. Last 15 seconds. Just four of you have responded so far. I would like all of you to put your response on the poll. Okay. Five, four, three, two, one, go. Almost four and a half minutes gone in this question. So most of you have gone with option A. The second highest vote has gone to option C. Okay. Of course, the first thing that would come to your mind is if let's say I call this as your L, you take log to the base here on both the sides, isn't it? So this would convert to limit and tending to infinity. This will become one by N log of sine pi by two N plus log of sine two pi by two N. Okay. I'm interested in what is going to be my last term that it shows like this. So basically it is nothing but log of sine off. Correct me if I'm wrong. It's two times N minus one pi by two N, isn't it? That's the only way I can get up. N minus one pi by N over here. Right. Now, basically it tells us that your rth term was actually one by N ln off sine r pi by two N. Okay. And you are trying to sum up the series from r equal to one till two times N minus one. Am I right? Isn't it? See, look here carefully. Your r is varying from one to two times N minus one. Okay. So this will get converted to this will get converted to let me just make the convergence and tending to infinity. Let's write everything that makes this look correct. Yeah. So this will become ln sine. This will become our x term. So this will become sine pi x by two. Okay. Limit of integration will become from zero to two my dear. Please note that. Please note that it will be from zero to two. Correct. Now let us write pi x by two as a t. So dx will be two by pi dt. So this will be two by pi ln sine t dt. And your limits of integration if I'm not wrong will become from zero to pi. This is your ln ln ln. Okay. Now this term seems to satisfy this property that f of t is f of pi minus t. Isn't it? That means I can use half the limit property. So it'll be two times zero to pi by two ln of sine t dt. This integral must be known to you by now. Pi by two log half. Pi by two log half. Absolutely. So it'll be two ln half that will come out from here because pi by two and two by pi will get cancelled. Two will be left off. Okay. So this guy will get cancelled with this. And this will be ln L. Which clearly means L is one-fourth. L is one-fourth. So which option? Option number C. Janta was wrong. Okay. Sir, one minute, sir. Just to show the last one. Yeah. Yeah. Yeah. Thank you sir. Okay. All right. Let's take up one last problem on this and then we can probably you know, take a break and start with differential equations. So let's have one more problem. Evaluate. Evaluate. Limit n tending to infinity. Summation of root r from r equal to one till n. Summation of one by root r from r equal to one by one to n by summation of r from one to n. So this has been multiplied. Yes. Any breakthroughs? Sir, can a bunch of terms be considered as kth term or something? There should be a series. No, a series should come out. But these are, there are three series operating here. This is a series. This is a series. This is a series. Okay, sir. Okay. So in the present form, whatever you're trying to apply that might not fit in. Okay, sir. Okay. Let's have a look at this. What I'll do here is this term, I will write it as summation under root of r by n. That means n have introduced. Remember, summation is applied to r, not on n. So n is like a constant. You can introduce it to any term. Okay. At the same time, I'll put under root n over here. Okay. Similarly, what I'm going to do is I'm going to make this as under root of n by r. Correct? So since I've multiplied with root n, I have to divide with the root n. So this root n is basically non-existent for me. This term will go off. Okay. Now, this term I'm going to do here one by n and r by n. That means, that means I've introduced one by n square in the denominator, right? So I'll give one by n to this and I'll give one by n to this in the numerator. Okay. And limit is basically following this algebra, so we can apply limit on this separately. Correct? So I can see this as a three definite integral problems. So here I can see this as a dx. Correct? This has under root of x. So you're integrating under root of x dx from zero to one into this I'm seeing as under root of one by x or you can say one by root x, zero to one. And this, of course, is integration of x dx from zero to one. Okay. So this will be your desired limit. So let us evaluate this. So this comes out to be, if I'm not mistaken, two by three. Okay. This is two root x. So that will give you a two and this will give you a half. So your answer is eight upon three. Correct? Is this fine? Any questions, any concerns? Okay. Now, we'll take a break. Since it is a three and a half hours class, I'll give you a 10 minutes break. Okay. So right now, as per my watch, it is 646. Let's meet at 656. Okay. On the other side of the break, we'll start with