does this apply only for terms that are less than 1 eg 1 + 1/2 + 1/4…. or does this apply to all ap/gp. I remember studying this, but it’s been so long I remember only the gist.
In: 248
Take a whole (1) and cut it in half (1/2 + 1/2). Take a half and cut it in half (1/2 + 1/4 + 1/4)… You can repeat this infinitely many times without ever exceeding the value of the original whole.
They don’t always, but there may be a limit to what they can equal, as eventually those fractions/decimals get so small that they become insignificant.
These are called geometric series, and if I recall it’s a Calculus I topic.
For your example, it does actually equal ~1 (well ~2, as you added a 1 to it) and there is a Wikipedia article with illustration:
https://en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/16_%2B_%E2%8B%AF
Notice the language that the partial sum *tends* to 1, so not exactly; as yes, you end up adding something like 0.0000000000000000004 and next 0.0000000000000000002 and so on; but when we say that we go to infinity, then it’s 1.
It’s hard to ELI5 without losing some mathematical rigor, but what matters is that the amount you add with each “step” of the sum is less than the difference between the current sum and the value that sum converges to.
Let’s take a variant of Zeno’s paradox. To walk to the flag at the end of a race, I must first cover half the distance of the way there. But before I can finish the race I must first cover half of the distance remaining. But then before I can finish the race I must cover half of *that* distance remaining, and so on.
Do this an infinite number of times, and you reach the flagpole. “Eventually” over an infinite number of increasingly smaller divisions the distance between you and the flag becomes zero, and thus you have completed the race. However you don’t *cross* the flagpole,
This is exactly what happens when you add up 1 + 1/2 + 1/4 + 1/8…. Notice each step takes you half of the distance towards two. Of course it doesn’t have to be half, just small enough that you never cross the limit.
As for your other question, there is nothing that says it applies to terms that are less than one. For example 1000 + 500 + 250 + 125 + 62.5… will eventually converge on 2000. This is the same principle as before, just with larger numbers.
Ok so it often doesn’t, the sun of infinite terms only turns out to be a finite value in very specific circumstances
So I’ll give two explanations, the first is a bit more ELI5 than the other
So really, in your example, it’s a sum of infinite terms 1/2^n with n starting at 0 and going to infinity. So we have 1 + 1/2 + 1/4 + 1/8 + …
For now, let’s ignore that 1, we’ll just talk about every one past that then add the 1 at the end.
So then we have 1/2 + 1/4 + 1/8 + …. Well each time, we’re adding enough to get us to the half way point between where we are and 1. So the first term gets us half way from 0 to 1. Then the next term gets us half way from 1/2 to 1. Then the next term gets us half way from 3/4 to 1…
Each time, we’re adding half of what we need to add to get to 1…so it’ll never quite get to 1, it’ll just get closer and closer, so we say that converges to 1. Now we said we’d add that 1 we ignored so what you described converge to 2. Basically, all sums of infinite sequences do this, where each number added will always get you only part of the way you need to go to get to a certain number
If you’re familiar with limits, there’s a more general way to explain it. Some of how you phrase this leads me to think you know a bit about limits maybe. Even if not, this explanation may still work out:
Let’s call S(n) the sum of the first n terms in this sequence. Note this is a finite sequence, since it’s just n terms long. Now, let’s find this formula. For your above example,
S(1) = 1 = 1/1
S(2) = 1 + 1/2 = 3/2
S(3) = 1 + 1/2 + 1/4 = 15/8
S(4) = 1 + 1/2 + 1/4 + 1/8 = 31/16
…
So, now we look for a pattern here, we are looking for a formula for calculating the finite sum S(n) for the first n terms, so if you want me to find the sum of the first 127 terms, I can just plug in 127 and find the answer
Well the formula we get is:
S(n) = (2^(n+1) – 1) / 2^n
Or more simply:
S(n) = 2 – 1/2^n
Now, if you’re familiar with limits, it’s quite clear that this one approaches 2 as n gets infinitely large.
This is actually the quite rigorous definition of a sum of an infinite sequence converging (meaning it yields a finite number like you asked about), where you create a formula for the sun of the first n terms (again, it’ll always be a finite number because it’s just adding up n numbers), then look at the limit as n approaches infinity of this formula, if that limit exists, the infinite sum converges and will equal a finite number, if that limit isn’t a finite number, it’s said to diverge, which means it doesn’t equal a finite number.
Shortcuts exist to make determining this easier in many cases, but this definition of finding the formula for the sum of the first n terms then seeing what that formula approaches as n approaches infinity REALLY what it means for this infinite sum to “equal” a finite number (although we don’t say it equals it, we typically say it converges to that number)
In a general sense, there is nothing special about an infinite series, some converge on a number, some climb to infinity. This idea is known as a limit, and really only applies in the abstract, because you can’t literally do something an infinite number of times, but you can imagine that you can. If at any point you decide to stop adding stuff, it’s no longer an infinite series and instead actually does have a discrete answer.
As for your second question, no it goes not only apply to numbers less than zero, limits of complex formulas can converge on anything, including infinity itself.
To give real life context to your own example, say you start with a glass of water. Every time you take a sip, you drink half of what is left. Immediately you can see that this will become a problem in a practical sense, because eventually you’ll have a tiny drop of water, and your next sip would only drink half of that drop. No matter how many times you go back in for a sip, you don’t actually drink the rest of the water, there is always a small amount left. We both know that in real life you’ll eventually get to 0 water left, but in the strictest since of the example, that’s not what theoretically happens. Luckily though, we can describe this problem in terms of a limit, by asking ‘what is the limit of water left if you were to take an infinite number of sips where every sip you drink half of what was remaining.’ Mathematically every sip gets you a little closer and a little closer and a little closer to zero, so we say the limit is zero. If at any point you stop drinking, let’s say after 100 sips, there is a tiny microscopic spec of water still in the glass, and we could quantify that as a traditional answer instead of calling it a limit.
There is a bar joke that comes in many different forms, but uses this same principle that goes; A man goes to a bar but is afraid of getting too drunk to make it home, so he comes up with a plan to pace himself. When he first arrives, he orders a beer. After he finished the beer, he asks the bartender to only give him half of a beer. After he finishes that, he asks the bartender to give him only a quarter of a beer, then an eighth, then a sixteenth. The bartender starts getting fed up and asks the man how long he plans to keep it up, to which the man says ‘oh I can keep this up forever’. The next day the same guy comes back to the bar and orders a beer with a grin on his face. The bartender pours the man two glasses of beer and tells him, ‘you really gotta know your limits’.
(If it’s not painfully obvious, the limit of this infinite sum is 2)
If this interests you further, I recommend checking out this old [Numberphile video](https://youtu.be/w-I6XTVZXww?si=Sy51w-sLcN–TLip) about why the infinite sum of (1+2+3+4+ …) equals -1/12
Since you mentioned studying this before, for geometric series (including the one you mentioned) it converges (or adds up to a finite number) if r is between -1 and 1. This is not true “only for terms that are less than 1”; the classic counterexample being the harmonic series (1 + 1/2 + 1/3 + 1/4 + 1/5 …..). There are also infinite series that don’t go to infinity, but also don’t converge, like 1 – 1 + 1 – 1 …..
If your confusion is how adding infinitely many numbers doesn’t go to infinity, ala Zeno’s paradox, it basically comes down to the extra terms are so small they don’t grow the sum to infinity. Let’s consider the infinite series .3 + .03 + .003 + .0003 ….. It’s pretty clear this number will not rise above even .4. In fact, if you remember long division, this sum is just 1/3.
Say you try to sweep a pile of dust into a dustpan, but there is still some on the ground. You sweep the smaller pile, but there is still some on the ground. Now let’s pretend the dust has no minimum size and you also aren’t allowed to just give up and say “good enough.” If there is always some dust left, you can continue the cycle infinitely despite starting with a finite pile of dust.
If you want a physical example with something you might have experience manipulating with your own hands, think about paper sizes. Specifically the A0, A1, A2, A3 etc. ones as you see in the second image here [https://en.wikipedia.org/wiki/Paper_size](https://en.wikipedia.org/wiki/Paper_size).
The idea is that A1 is half the size of A0, A2 is half the size of A1 etc.
So take a sheet of A0 paper.
Cut it in half to get 2 A1 sheets and put one on a table.
Take the remainder (now also of size A1), and cut it in half to get 2 A2 sheets and put one on the table.
Etc. Etc.
Two things should be clear if you keep doing this pattern (or series):
1. We will never run out of paper – we only ever put down half of what we have left. So this series is *infinite*. There is no “end”.
2. The amount of paper on the table must be less than the original amount we started with because we always have some left. And so the sum of those we have placed is *finite*.
We can get arbitrarily close to putting down all of the paper we started with by completing the pattern up to any paper size you want, e.g. A10000. And if that is not close enough we can just go further along the series, e.g. A10001, A100002 etc. By this method we can put down more than any amount of paper which is less than the original amount. But we can never actually put all the paper down.
This, basically, is what a limit is, and the value of that limit is finite – it’s the amount of paper we started with.
The easiest way you can visualize it is by imagining the distance between numbers as a rug. If someone asked you to stand on one side of the rug (1) and take a big step towards the middle of the rug (1 & 1/2) and then the next step would be half that distance (1 & 3/4) and then half that distance again, you would eventually reach the other end of the rug (2) if you had infinite steps but you would never go beyond the rug because even though there are infinite steps from one end of the rug to the other the total length of the rug is finite.
So this is a difficult question but here I go:
– Let us say we look at the infinite sum where each term being added is 1:
1+1+1+1…=x
We can easily see that no matter how many 1s you add, the number will just keep on growing and growing. We call this “divergent”
– Let us say we look at the infinite sum where each term being added is a tenth of the previous term:
1+0.1+0.01+0.001…=x
We can easily see that no matter how far you go down the line you will only ever get to the number 1.11111111… . We call the fact that some infinite sums settle at a particular value “convergent”
It is not always easy to find out if a sum is convergent or divergent, but we have tricks that help us figure it out.
It has nothing to do with all terms being less than 1. For example the sum of an infinite 1/2 is divergent. 1/2+1/2+1/2+1/2=x. This keeps growing and growing at half the rate of 1+1+1+1…, but the 1/2 sum will reach all numbers that the 1 sum will reach.
Similarly, the inverses of the integers: 1+1/2+1/3+1/4+1/5+1/6… Is well known to diverge.
How do we know this:
Look at this sum:
– 1+1/2+1/4+1/4+1/8+1/8+1/8+1/8+1/16….
At each point in the sum, that sum is smaller than this sum:
– 1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9…
However, notice that if we group the smaller sum like this and add the terms in the brackets you get: 1+(1/2)+(1/4+1/4)+(1/8+1/8+1/8+1/8)+(1/16….
1+1/2+1/2+1/2+… (we have already shown that An infinite sum of 1/2 diverges, so the other larger sum must also diverge)
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