Telescoping Sums

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. One over one times two is one half. Add one over two times three, and the total is two thirds. Add one over three times four: three quarters. Add one over four times five: four fifths! Messy fractions, but such neat totals. Why? Hmm, there is a beautiful trick hiding here: telescoping sums. Sums like these come up often in JEE, so let's begin!

0:32The idea: neighbours cancel

So, here is the whole idea. Say the k th term of a series is T k, written as T with a small k below it. Suppose it is a difference: f of k, minus f of k minus one, two neighbouring values of some function f. Okay? Now we add these terms, for k from one up to n. What do you think will happen? Let's write them out. The big Greek letter is capital sigma, and it means add up, from k equals one to n. In square brackets: f of one minus f of zero, then f of two minus f of one, and the three dots mean and so on, up to f of n minus f of n minus one. Now look: minus f of one, plus f of one. They cancel! With me? Then f of two cancels, and so on: everything in the middle cancels in pairs! So the sum is just f of n, minus f of zero. Like an old pocket telescope: the middle tubes slide inside, and only the two ends remain. Hence the name, telescoping. Clear? So the real work in every question is just one thing: write T k as a difference of neighbours. Then the sum falls out in one line. Okay?

1:45Split the term into a difference

Let's start with the most common one. One over k times k plus one: two neighbouring numbers multiplied in the bottom. How can we write this as a difference? One over k, minus one over k plus one. Yes, exactly! Two neighbours again, so this is the telescoping form. Right? Now, never trust a split without checking it. Over the common denominator, the top is k plus one, minus k, which is just one. So we get the term back. Perfect! Got it? Now the general case. In the bottom, two neighbouring terms of an A P: a plus k d, and a plus, k plus one, times d. Here a is a fixed number, and d is the common difference. Split it the same way, but with one over d outside the big round brackets. Okay? Now, why do we need that one over d? Check it, just like before! The top becomes the difference of the factors, which is d, not one. So we divide by d. Careful, this is a classic trap: students forget the one over d, and the answer is off. So always subtract the factors first. Clear?

2:52More splittings to remember

Now, one over k times k plus one times k plus two: three factors in a row. Split it into two pieces with two factors each: one over k times k plus one, minus one over k plus one times k plus two. The outer factors, k and k plus two, differ by two, so a half goes outside. Same rule as before, okay? Next, factorials. The exclamation mark means factorial: k factorial is one times two times three, up to k. So tell me, what is k plus one factorial, minus k factorial? Look, k plus one factorial is k plus one, times k factorial. Now take away k factorial: k plus one, minus one, leaves k. So the difference is k times k factorial, where the dot means times. Very good! With me? Now roots. The tick-shaped sign is the square root. Take one over a sum: in the bottom, root k plus root k plus one. Multiply the top and the bottom by root k plus one, minus root k. On top we get root k plus one, minus root k, and the bottom becomes k plus one, minus k. Okay? And k plus one, minus k, is just one! So the term is simply root k plus one, minus root k. Beautiful, right?

4:15Example 1: sum of 1/k(k+1) and its limit

Now, Example one, and it finishes our puzzle. Find the sum of one over k times k plus one, for k from one to n, and its limit. Pause the video, and try it first. Done? Okay, let's check. Split every term: one over k, minus one over k plus one. Written out: one minus a half, plus a half minus a third, and so on, up to one over n, minus one over n plus one. The half cancels, the third cancels, and so on. Okay? So which two pieces survive? The very first and the very last. Correct! One minus one over n plus one, which is n over n plus one. Got it? Now, back to our puzzle. Put n equal to one, two, three and four: one half, two thirds, three quarters, four fifths! Mystery solved. And a topper's habit: always test your answer at n equals one. Here it gives one half, the first term. Clear? Last, the limit. The arrow means tends to, and the sideways eight is infinity. As n grows without bound, one over n plus one tends to zero. So the sum tends to one. Simple!

5:32Example 2: sum of k/(k^4 + k^2 + 1)

Now, Example two, a harder one, JEE style. Find the sum of k over k to the power four, plus k squared, plus one, for k from one to n. The small four up top means the fourth power. Pause and try it. Hint: factorise the bottom first. Done? Okay, let's check. The bottom does not look friendly. So how does k to the four, plus k squared, plus one factorise? Here is the trick: write k squared as two k squared, minus k squared. Then the bottom is k squared plus one, the whole squared, minus k squared. That is a difference of two squares, so it is k squared minus k plus one, times k squared plus k plus one. Lovely, right? Now our rule: subtract the two factors. The bigger one minus the smaller one is exactly two k. And the top of our term is k, half of that. With me? So T k equals one half, times one over k squared minus k plus one, minus one over k squared plus k plus one. Careful with that half! The factors differ by two k, but the top is only k. Clear? But wait. Are these two pieces really neighbours? Will they cancel? Yes, they will! Put k plus one in place of k in the first factor. It simplifies to k squared plus k plus one. So the second piece of each term cancels the first piece of the next term. Got it? So only the ends survive. At k equals one, the first factor is one, so the first piece is one; the last piece is one over n squared plus n plus one. So the sum is one half, times one minus one over n squared plus n plus one, which simplifies to n times n plus one, over two times n squared plus n plus one. Okay? And the topper's check at n equals one: one times two, over two times three, is one third. T one, the first term, is one over one plus one plus one, also one third. Perfect!

7:40Quick revision: split the term, only the ends survive

So, quick revision. If each term is f of k minus f of k minus one, the sum is f of n minus f of zero. Split the term into a difference, and when the factors differ by d, put one over d outside. Know the factorial and the root splittings. And always check your answer at n equals one. Now go and try the practice problems on this page. All the best!