Limits of Sums

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take n over n squared plus one squared, plus n over n squared plus two squared, and so on, up to n over n squared plus n squared. Now let n grow without end. Only plain fractions, no circles anywhere. And yet the answer is pi over four! So where did pi come from? The secret is that this sum is really an area. Limits of sums come up often in JEE, so let's begin.

0:32A sum of thin strips is an area

So, look at the picture. The dark curve is y equals one over one plus x squared. We cut the x axis from zero to one into n equal strips, so each strip has width one over n. The marks read one over n, two over n, and so on, up to one. Okay? Strip number r has its right edge at x equals r over n, so its height is f of r over n. Its area is one over n, times f of r over n. Right? Add all n areas. Now make n bigger and bigger, so the strips get thinner and thinner. What happens to the total area of the rectangles? It becomes the area under the curve, yes, exactly! So the limit as n tends to infinity of one over n, times the sum of f of r over n, equals the integral from zero to one of f of x, d x. The word lim means limit, the arrow means tends to, and the sideways eight is infinity. The capital sigma means add up, from r equals one to n, and the long S is the integral sign. Clear? The same result holds when r runs from zero to n minus one. The two sums differ only in their end strips, each of width one over n, and that difference vanishes in the limit. With me?

1:53The method in four moves

Now, how do we use this in a question? Four moves. Move one: take one over n out of the whole sum. Okay? Move two: write every term using only r over n. Right? Move three: r over n becomes x, and one over n becomes d x. Here the arrow means: is replaced by. Got it? Move four: the limits come from the smallest and the largest values of r over n. So if r runs from one to n, what are the limits? Zero and one, correct! The smallest value is one over n, which tends to zero, and the largest is n over n, which is one.

2:34Trap: the upper limit is not always 1

Clear? Now the general form. If r runs from p times n up to q times n, then r over n runs from p to q, and the limit is the integral from p to q of f of x, d x. Okay? Here is one from the course's examples: r over n squared plus r squared, but now r runs from one to two n. Take out one over n, as before. So the integral runs from zero to one, right? No, no, careful here! This is the trap. The largest value of r over n is two n over n, which is two. So the upper limit is two. Got it? So we get the integral from zero to two of x over one plus x squared, d x, which is half natural log five. The l n on screen is the natural log. Always check where r stops!

3:20Products: take logarithms

Clear? Now, what if the question has a product? The capital pi means multiply all the terms, for r from one to n, and the whole product is raised to the power one over n. Call this limit capital L, okay? But our rule works only for sums. So what turns a product into a sum? Logarithms, yes, exactly! Take natural log. The log of a product is the sum of the logs, and the power one over n comes down in front. Right? This is our sum, so log L is the integral from zero to one of log f of x, d x. So L is e to the power of that integral; the e x p on screen means e to the power. Never skip this last step. With me?

4:04Example 1: sum of n/(n2+r2) is pi/4, of r/(n2+r2) is ln2/2

Now Example one; the first part is our puzzle. Evaluate the limit, as n tends to infinity, of the sum from r equals one to n of n over n squared plus r squared. Then the same sum with r on top in place of n. Pause the video and try it first. Done? Okay, let's check. Divide the top and the bottom by n squared: one over n, times one over, one plus the square of r over n. The dot means times. Okay? Now r over n becomes x, one over n becomes d x, and r runs from one to n. So what integral do we get? The sum tends to the integral from zero to one of d x over one plus x squared, yes! That is tan inverse x; the minus one means inverse, not a power. The square brackets with zero and one mean: put in one, then subtract the value at zero. So we get pi over four, the area under that very curve in the picture. That is where pi came from! Right? Now the second sum. Divide by n squared again: one over n, times r over n, over, one plus the square of r over n. Clear? So we need the integral from zero to one of x over one plus x squared, d x. The top is half the derivative of the bottom, so we get half natural log of one plus x squared. Got it? At one, that is half log two; at zero, half log one, which is zero. So the answer is half natural log two. That's it!

5:39Example 2: nth root of n!/n to the n tends to 1/e, by log

Now Example two, a harder one. Evaluate the limit, as n tends to infinity, of n factorial over n to the power n, all raised to the power one over n. The exclamation mark is factorial: one times two times three, up to n. Pause the video and try it first. Done? Okay, let's check. First, is this a sum, or a product? It is a product, exactly! Give each factor of n factorial its own n: one over n, times two over n, and so on, up to n over n. The dots mean and so on. Okay? So call the limit capital L, and take natural log. The power one over n comes down in front, and the log of the product becomes a sum. So log L is the limit of one over n, times the sum of log of r over n. Right? That is our standard sum: the integral from zero to one of log x, d x. By parts, that is x log x minus x, from zero to one. With me? At one, we get zero minus one. At zero, x log x tends to zero, so that end gives zero. So log L is minus one. Clear? So the answer is minus one, right? No, no, careful here! Minus one is log L, not L itself. So L is e to the power minus one, which is one over e. Students often forget this step, so always come back from the log. Very good!

7:08Quick revision: take out 1/n, r/n to x, products by log

So, quick revision. Take out one over n, and write every term through r over n. Then r over n becomes x, one over n becomes d x, and the limits come from the smallest and largest values of r over n. For a product, take log, integrate, then raise e to that power. Now go and try the practice problems on this page. All the best!