When the ratio is small enough, an endless sum still settles on a number

0:00Add a hundred numbers and you get a number. Add endlessly many, and it is not clear you get anything at all. That is not a rhetorical worry. Add one, then one, then one, and keep going: the running totals climb past every number you could name. So an endless addition is not automatically a number. An endless sum is not an instruction to add forever. Nobody adds forever. It is a name for what the finite totals do. Add the first term. Add the first two. The first three. That gives a run of ordinary numbers, every one of them computable.

0:42If that run closes in on a single number and stays there, that number is what the endless addition means. If it does not, the addition means nothing. Take the run that opens at one and multiplies by two thirds at every step. One, two thirds, four ninths, eight twenty-sevenths. The terms get smaller, they never stop, and there are endlessly many of them. Add the first four and you get sixty-five twenty-sevenths, just under two and a half.

1:17You already know how to total the first n terms of a run like this. Put the opening term and the multiplier in, and the total comes out as three, times one minus two thirds multiplied out n times. Look at what that says: three, times one minus a quantity that is getting small. Draw the totals as a staircase and you can watch it happen. The steps climb, they climb by less each time, and they flatten off.

1:48Where they flatten is the question. So track that quantity on its own: two thirds multiplied out n times, at n of one, five, ten and twenty. The first is two thirds. The last has three zeros after the point before it says anything at all. They are exact fractions underneath: thirty-two over two hundred and forty-three at five, and a thousand and twenty-four over fifty-nine thousand and forty-nine at ten.

2:22And notice they are not even quoted to the same precision. Four figures, then ten, then ten, then eight. Three different precisions in one row. That is a display, not four measurements. But the trend is unmistakable. The quantity is being squeezed towards nothing. And if it does go to nothing, the total goes to three times one minus nothing. Three. Stop. That was four numbers. Four numbers make a claim believable. They do not establish it, and everything downstream depends on that quantity really going to nothing. So the checker behind this video builds a second quantity on purpose.

3:09At n equal to one, five, ten and twenty, it agrees with the real one exactly. All four. Not close — equal. Put it in the table and you could not tell the two apart. Now go to thirty. The real quantity has shrunk to about five millionths. The impostor is a hundred and forty-five larger than that, and climbing. Four agreeing values told you nothing at all about where a quantity was going.

3:40The table is evidence. The argument is a different thing, and it is one rearrangement long. Take the total of the first n terms of any such run, opening term a, multiplier r. It is a, times one minus r multiplied out n times, all over one minus r. Now split it. Multiply out the bracket and keep the two pieces apart. a over one minus r, minus a times r-to-the-n over one minus r.

4:16The first piece has no n in it anywhere. It is a fixed number, decided by the opening term and the multiplier and by nothing else. The second piece carries every appearance of n there is. It is the whole of the total's dependence on how far you went. That split is exact. It is not an approximation, and it holds at every single n — which the checker confirmed across three hundred and twenty finite totals, with no disagreements.

4:49And the question has changed shape: it is no longer what the total does, but what the second piece does. The second piece is a fixed number, times r multiplied out n times. So everything turns on that quantity, and everything about it turns on one thing: the size of r. If the size of r is under one, multiplying by it makes things smaller every single time, and repeated shrinking gets under any bound you care to name.

5:24Not eventually, in a vague way. At a countable step. For two thirds: six steps to get under a tenth. Eighteen to get under a thousandth. Thirty-five to get under a millionth. Fifty-two to get under a thousand-millionth. And notice that it is the size. Minus a half is the same size as a half, and takes the same twenty steps to get under a millionth. That is the condition. Not a safety rail bolted on at the end — precisely what the argument spends.

6:01So the total is a fixed piece minus a shrinking piece, the shrinking piece goes to nothing, and the totals close in on the fixed piece and nothing else. The endless total is a over one minus r. That is not a second formula to learn. It is the finite one with the shrinking part deleted, and the size condition is what makes the deletion legal. One more thing. The distance still to go — the gap between a finite total and the answer — is exactly that second piece.

6:36So the gap is itself a geometric run, carrying the same multiplier as the terms do. Over three hundred and twelve consecutive pairs the checker found no exception. The distance to the answer shrinks by the same factor the terms do. That is the staircase flattening. Now a question students ask and rarely get a straight answer to. Is the sum really three, or is it just very close to three?

7:08Every finite total is three, minus something above nothing, so every one of them is under three. The checker looked at three hundred and sixty finite totals, across the six runs here with a multiplier above nothing. The number equal to the answer: none. The number past it: none. So no finite total is ever the answer. And the answer is still exactly three, not approximately three. Because the endless total was never defined as one of the finite totals. It is the number they close in on, and being closed in on is not the same as being reached.

7:53Two cases, to watch the condition work. One, a half, a quarter, an eighth, onward: opening term one, multiplier a half, and one over a half is two. The totals climb. One, then one and a half, then one and three quarters. Always under two, always closer to it. Now turn the sign over. One, minus a half, a quarter, minus an eighth. The multiplier is minus a half. Its size is still a half, so the condition holds, and the answer is one over one plus a half, which is two thirds.

8:31But look at the staircase. It does not climb. It closes in from both sides, overshooting and undershooting by less each time. Of twenty finite totals, ten sit above two thirds and ten sit below. None of them sits on it. The condition was about size, and it always was. Now the two multipliers the argument refuses, for genuinely different reasons. Multiplier one. Five, five, five, on forever. The totals are five, ten, fifteen, twenty.

9:08They get past a hundred at the twenty-first term, past a thousand at the two hundred and first, and past a million at the two hundred thousand and first. Name any bound and they get past it. Multiplier minus one. Five, minus five, five, minus five. The totals go five, nothing, five, nothing, and across sixty of them they take exactly two different values. These do not grow — they never get past six — and they still settle on nothing at all, because settling means closing in, and swinging between two values forever is not closing in on either.

9:51Failing to grow is not the same as settling. That is the case worth remembering. And here is why the formula has to refuse rather than compute. Put minus one into a over one minus r and the arithmetic cheerfully hands back five halves. Nothing whatsoever settles on five halves. With that in hand the computations are short. Find the multiplier, check its size, divide. One, a third, a ninth. The multiplier is a third, and one over two thirds is three halves.

10:30Six, one point two, nought point two four. The multiplier is a fifth, and six over four fifths is fifteen halves. The other two are on the board, and the fourth is the one to look at: its multiplier is minus a quarter, whose size is a quarter, so the condition holds and the answer comes out below nothing. Every one of those four was found here by searching two thousand two hundred and thirteen candidate numbers for the one the totals close in on, and only then checked against the formula. There were no disagreements.

11:11Two of the questions are not computations at all. The first: three to the half, times three to the quarter, times three to the eighth, and on forever. Show the product is three. Multiplying powers of the same number adds the exponents, so the product is three, raised to the total of a half, a quarter, an eighth, and onward. Those exponents are themselves an endless geometric run, opening at a half and multiplying by a half. Its endless total is one, so the product is three to the first power. Three.

11:51And the same structure shows up inside it: after four factors the exponent total is fifteen sixteenths, and of the first sixty, none reaches one. So every finite product falls short of three. The second is the best of them. Let x be the endless total of one, a, a squared and onward, and y the same for b, both sizes under one. Find the endless total for the powers of a b, in terms of x and y.

12:25x is one over one minus a, so a comes back out as x minus one, over x, and the same for b. Then one minus a b works out to x plus y minus one, over x y, and the answer is x y over x plus y minus one. Which is worth hearing: an endless total is a number like any other, and you can solve for it.

12:54So what was bought here. Not a formula to memorise alongside the finite one. It is the finite formula with the shrinking part removed, and the removal is legal exactly when the multiplier's size is under one. Not a new kind of arithmetic either. Nobody added forever, and nobody put infinity in for n — you cannot put infinity in for anything. One piece of an expression was driven towards nothing and the rest was left alone.

13:28And not a proof from a table. Four values made it believable. The split made it true. The thing to carry away is the shape of the question. An endless addition has a number when its finite totals close in on one — and for a geometric run they do exactly when the multiplier is small enough that multiplying by it destroys the tail. That is the whole of it.