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
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.
[removed]
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)
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.