Limits of Sequences and of Sums

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Take one over one times two, plus one over two times three, plus one over three times four, and keep going, forever. Every term is positive, so the total keeps growing. So, does it grow to infinity? Hmm, think about it. By the end of this video, you will know the exact answer. Today it is limits of sequences and of sums. They come up often in JEE, so let's begin.

0:29Four results to remember

So, here n is a counting number: one, two, three, and so on, and we ask where the terms settle as n tends to infinity. The arrow means tends to, and the sideways eight is infinity. Now, the root sign with a small n on it is the n-th root. In the n-th root of n, the number under the root keeps growing, but the root gets stronger too. So where does it settle? It settles at one! If you said one, very good! The n-th root of n tends to one. Okay? Next, take any fixed positive number a, say a hundred, or one half. The n-th root of a also tends to one. Note the condition in the bracket: a greater than zero. Clear? Now a tricky one: one plus one over n, to the power n. The base tends to one, and one to the power anything is one. So the answer is one, right? No, no, careful! This is the one to the power infinity form: the base tends to one, but the power runs off to infinity. The result: one plus a over n, to the power n, tends to e to the power a. So with a equal to one, the answer is e, not one. Got it? And the last one: a to the power n, over n factorial. The exclamation mark means factorial: one times two times three, up to n. For any fixed a, this tends to zero, because the factorial grows much faster. With me? Why? Take the course example with a equal to two. The dot means times, and the dots mean and so on. Two to the power n over n factorial is two over one, times two over two, times two over three, up to two over n. From the third factor on, each is at most two over three. So for n at least three, it is less than or equal to two, times two over three to the power n minus two, which tends to zero. Right?

2:26Telescoping: only the ends survive

Now the most useful trick for sums: telescoping. Write the k-th term as a difference: F of k, minus F of k plus one. Here k is the position of the term, and F is a function we have to spot. Okay? Now add the terms for k equal to one, two, three, up to n. Each pair of brackets is one term. Look at the first two brackets: minus F of two, and then plus F of two. What happens to them? Yes, they cancel! And F of three cancels the same way, all along the line. Like a telescope folding shut, only the ends survive: F of one, minus F of n plus one. Then let n tend to infinity. That's it!

3:10Sums of powers

Next, sums of powers. On top: one to the power p, plus two to the power p, up to n to the power p. At the bottom: n to the power p plus one. The word lim, with n tends to infinity below it, is the limit as n tends to infinity. This limit is one over p plus one, for p greater than zero. Got it? Why? Take p an integer. Expand k plus one, to the power p plus one, and subtract k to the power p plus one. The biggest term left is p plus one, times k to the power p; the dots stand for smaller powers of k. Clear? Now add this over k, from one to n. The left side telescopes, our trick again, and leaves roughly n to the power p plus one. The wavy equals sign means approximately, and the capital sigma means sum. So the sum of k to the power p is about n to the power p plus one, over p plus one. Right? Let's test it with a course example: p equal to three, the sum of cubes over n to the power four. What does the rule predict? One over three plus one: one quarter. Correct! And directly: the sum of cubes is n squared, times n plus one squared, over four. Divide by n to the power four, and it tends to one quarter. They match. With me?

4:36Example 1: limit of the sum of 1/(k(k+1)) by telescoping

Now, remember our puzzle from the start? Let's settle it. Find the limit, as n tends to infinity, of one over one times two, plus one over two times three, and so on, up to one over n times n plus one. Pause the video, and try it first. You already have the trick for it. Done? Okay, let's check. Split the k-th term: one over k, minus one over k plus one. Check it: over a common denominator, the top is k plus one minus k, just one. So F of k is one over k. Clear? So the whole sum telescopes. Every middle term cancels, and only the ends survive: F of one, minus F of n plus one. That is one, minus one over n plus one. Right? Now, as n tends to infinity, what happens to one over n plus one? It tends to zero, exactly! So the limit is one. And our puzzle is solved: infinitely many positive terms, yet the total never crosses one. Beautiful, isn't it?

5:37Example 2: limit of product (1 - 1/k squared), k = 2 to n

Now a harder one: a product. Find the limit, as n tends to infinity, of one minus one over two squared, times one minus one over three squared, and so on, up to one minus one over n squared. Pause the video and try it. Okay, first, a trap. Every factor is less than one. And when you multiply more and more numbers less than one, the product keeps shrinking. So the answer is zero, right? No, no, wait! Shrinking does not mean shrinking to zero. Let's compute it. One minus one over k squared is k squared minus one, over k squared. Factor the top: k minus one over k, times k plus one over k. Okay? Now the first pieces, for k from two to n: one half, times two thirds, times three quarters, up to n minus one over n. Each top cancels the bottom just before it, leaving one over n. Got it? Now the second pieces: three over two, times four over three, up to n plus one over n. This time each bottom cancels the top just before it, leaving n plus one, over two. Clear? So the product is one over n, times n plus one over two, that is, n plus one over two n. As n tends to infinity, this tends to one half. Not zero, one half! Telescoping works for products too.

7:03Quick revision: sequence limits, telescoping, sums of powers

So, quick revision. The n-th root of n, and of a for positive a, tend to one; one plus a over n, to the power n, tends to e to the power a; and a to the power n over n factorial tends to zero. Telescoping: write the k-th term as F of k minus F of k plus one; only the ends survive. And the sum of p-th powers, over n to the power p plus one, tends to one over p plus one. Now go and try the practice problems on this page. All the best!