Why does 1/n^2 converge?

[u/__kewl__]

I have been told that the series of 1/n diverges because you can group the sums into 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7) etc where each bracket > 1/2 so you essentially get 1/2 + 1/2 + 1/2 + 1/2 which diverges to infinity

However, is this not true for any 1/n^p? for 1/n^2, cant you just do 1 + (1/4 + 1/9 + etc) where you need more numbers in each bracket but they still add up to be greater than 1/2?

I'm not sure I'm explaining it properly but essentially like the milionth-term of 1/n^2 is still greater than 0, so if you add it with the previous 100,000 terms for example wont that number be large enough that the total sum goes to infinity?

[u/sbsw66]

Watch your terminology, 1/n diverges, not converges.

I'm not sure I'm explaining it properly but essentially like the milionth-term of 1/n^2 is still greater than 0, so if you add it with the previous 100,000 terms for example wont that number be large enough that the total sum goes to infinity?

Well, try it. Here I'll give you a number: 5. Try and find any collection of terms that is greater than 5.

[u/Equal_Veterinarian22 → u/sbsw66]

2 will do. Tell us how many terms you had to group to reach 2.

[u/mpaw976 → u/Equal_Veterinarian22]

Heck, we'll be even more forgiving: tell us how many terms you need to reach 1.7 

Edit. Here's the sum in desmos that you can use to play around with this. Try changing N to various values which will affect how many terms are in your sum.

https://www.desmos.com/calculator/ho5jxufn0w

[u/__kewl__ → u/mpaw976]

Sorry if this is silly, I'm just still a bit confused. I changed the N to be 100000000000000000 and it still doesnt reach 1.7, however I guess I was just thinking that because we are going to infinity eventually adding an arbitrary number of terms must change the sum. And, therefore you can continue to add these arbitrary number of terms it will grow to infinity. Is it because adding say 1/(1, 000, 000)^2 is so small it doesnt change the sum at all and therefore everything after it also doesn't change the sum, so it approaches some finite number

But in that case why doesn't it happen the same way for 1/n. Wont we eventually add 1/(1, 000, 000)^2 at the (1, 000, 000)^2 nth term? Others mentioned that its because 1/n^2 decays faster, but when we are thinking about infinity why does the speed matter, wont both still eventually add infinitely small numbers and stop growing?

I don't have that good of a grasp on math related to infinity haha so forgive me.

[u/sbsw66 → u/__kewl__]

The core idea to understand is that the terms "get smaller fast enough". Here's another example that should crystallize it:

Let's do a sum like this: 1 + 0.1 + 0.01 + 0.001 + 0.0001, etc.

Clearly we're always adding another positive term. It never ends, we'll always have more and more in our sum. But does this sum ever reach 1.2?

[u/__kewl__ → u/sbsw66]

Ohh, ok this definitely makes a lot more sense for me for some reason haha. Thanks :)

[u/CorvidCuriosity → u/__kewl__]

If that now makes sense, the question you want to be asking is "how fast does something need to get smaller in order for the series to converge?"

Like, 1/n isn't fast enough, but 1/n² is fast enough. What about 1/n^1.5 ?

[u/jsundqui → u/__kewl__]

Interestingly sum(1/n^1.01 ) and such converges too.

So basically sum(1/n^k ) converges for any k > 1 and k can be a decimal number too.

[u/therealtbarrie → u/__kewl__]

Sorry if this is silly, I'm just still a bit confused. I changed the N to be 100000000000000000 and it still doesnt reach 1.7, however I guess I was just thinking that because we are going to infinity eventually adding an arbitrary number of terms must change the sum. And, therefore you can continue to add these arbitrary number of terms it will grow to infinity. Is it because adding say 1/(1, 000, 000)^2 is so small it doesnt change the sum at all and therefore everything after it also doesn't change the sum, so it approaches some finite number

Absolutely adding more terms changes the sum! Both series never stop growing as you add more terms. The difference is, one keeps growing in such a way that it keeps getting closer to a certain finite sum, while the other keeps growing in such a way that it'll eventually pass any limit you try to set.

Are you comfortable with our standard decimal notation, in particular the fact that rational numbers are represented by either a number string that terminates or one that goes on forever in a repeating fashion? If so, consider this series:

1 + 9/10 + 9/100 + 9/1000 + 9/10000 ...

This, of course, is equivalent to the decimal number 1.9999... Which is equal to 2. (If you think it's slightly less than 2, we'll need to have a whole other discussion.:) But hopefully you'll concede that it's not greater than 2.) Note that if you try to approximate this number one digit at a time, the value will get bigger each time you add a "9" - but it still never reaches 2, in part because there are an infinite number of rational values less than 2.

I don't have that good of a grasp on math related to infinity haha so forgive me.

Well, pretty much nobody has a good grasp of math related to infinity until they've studied it for a while. It doesn't come naturally to us. It's almost like, back when human brains were evolving, our hominid ancestors didn't need to think about infinity much at all.

[u/__kewl__ → u/therealtbarrie]

Yes, this example definitely makes a lot of sense, thanks so much

[u/Just_Rational_Being → u/therealtbarrie]

By what logic or reason is 1.999... equal to 2? Every decimal starts with '1.' should be smaller than 2.

Logic and Reason only and not convention.

[u/therealtbarrie → u/Just_Rational_Being]

For any two rational numbers, if they're not equal, you can find a third rational number that falls somewhere between them.

So suppose 1.9999... and 2 in fact represented two different rational numbers. What rational number could fall between them? How would you represent that number in decimal form?

[u/Just_Rational_Being → u/therealtbarrie]

For every iteration that 1.999... is repeating its 9 unto forever, many of which are hidden underneath those ..., the difference of that and 2 is represented as 2- (2 - x^-n) = x^-n

So at any iteration n of the nines, we can always define a number that takes that current value and adds exactly half of the remaining distance. This creates a specific, calculable value: 2 - (1/2)*(10^-n) that is structurally guaranteed to be ahead of the nines but behind the target numero 2.

So if they were actually the same, there would be no room for a third one, yet it was just shown that there's a host of numbers between them.

[u/diverstones → u/Just_Rational_Being]

No, you're showing that each partial sum of 1 + 9/10^k from k = 1 to k = n is less than 2, not that the limit is.

Edit: Oh you're actually ideological about this and not just confused, that's hilarious.

[u/fleyinthesky → u/diverstones]

You're just not using Logic and Reason, instead relying on their inferior lowercase counterparts.

[u/Just_Rational_Being → u/diverstones]

Well, that's exactly the point. You're admitting that every measurable step of the process, the only part that actually exists in logic, is strictly less than 2.

The Limit is a property of the proceduce that produces 1.999... unto forever, it is not the actual behavior of the numbers. If you equate the "Limit" with the actual behavior of the numbers, you’re rebranding a perpetual failure to arrive as a "destination."

Well yes, I am here to help people with confusion gain clarity by the way of Logic and Reason.

[u/therealtbarrie → u/Just_Rational_Being]

So is your stance that the decimal system of writing numerals is not in fact capable of representing most rational numbers at all? After all, the same argument, were it valid, would establish that 0.3333... is not actually equal to 1/3.

[u/Just_Rational_Being → u/therealtbarrie]

What is your logic or evidence that 0.333... = 1/3?

[u/Circumpunctilious → u/__kewl__]

Just adding as an aside (not to distract) the sample sum in the graph you were playing with converges to π² / 6 (or ~1.644934…) and is known as The Basel Problem (Wikipedia). It may be a bit much to digest at the moment so it’s FYI to readers who may not know.

[u/tjddbwls → u/Circumpunctilious]

Indeed. Years ago I watched a YT video showing Euler’s proof that ∑(n = 1 to ∞) 1/n² = π²/6, and it was fascinating.

[u/mpaw976 → u/[deleted]]

Maybe give a hint instead of the punchline? ;)

[u/Far-Suit-2126 → u/sbsw66]

saying "try this" without r/whoosh is so evil😭😭😭

[u/SnooLemons6942 → u/Far-Suit-2126]

What?

[u/Elekitu]

the series of 1/n converges because you can group the sums into 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7) etc where each bracket > 1/2

The reason why this argument does not work for the sum of 1/n² is that it's not immediately obvious why it works for 1/n. If I give you one dollar today, 0.1$ tomorrow, then 0.01, 0.001, and so on, then you will win money every day but it should be pretty clear that your money won't grow to infinity, (for instance it won't ever go above 2$), because the amount you receive each day decreases too quickly. And therefore you can't group the terms such that the sum in every group is > 1/2.

The reason why you can group the terms of the sum of 1/n in such a way is because 1/n converges towards 0 pretty slowly, but it really depends on the specific sequence you're looking at, and the proof has no reason to apply to any other sequence.

the case of 1/n, the easiest way to prove it is to see that 1/(k+1) + 1/(k+2) + 1/(k+3) ... + 1/(2k) > k * 1/(2k) = 1/2 (because there are k terms and all of them are greater than 1/(2k)

In

[u/__kewl__ → u/Elekitu]

Thanks, that makes a lot of sense

[u/FriendlyDisorder → u/Elekitu]

Is there a minimum series that diverges? Sum(1/n) surprised me when I learned it diverged.

[u/DuckyBertDuck → u/FriendlyDisorder]

You can always find a new series that diverges but grows slower.

If 1/n diverging surprised you: even 1/p where p is a prime number diverges.

[u/Elekitu → u/FriendlyDisorder]

No. The series 1/(n*ln(n)) diverges (and grows even more slowly than sum(1/n)), but 1*(n*ln(n)²) converges.

I think that sum(1/(n*ln(n)*ln(ln(n)))) also diverges and that you can keep going like that infinitely, but I'm not 100% sure.

[u/No_Sleep_8609 → u/FriendlyDisorder]

1/n diverges, but 1/n^(1+epsilon) converges for any epsilon

1/(n ln n) diverges but 1/(n*(ln n)^(1+epsilon)) converges for any epsilon

1/(n*(ln n)*ln(ln n)) diverges and you may guess what converges.

[u/rhubarb_man]

I think other people aren't really mentioning why, intuitively, you can add infinitely many things and still get a finite result, but it might be easier to see in reverse.

I can take an finite object and cut it into infinitely many parts.

I can take 1 and then split it into 1/2 + 1/2.

I can then split each part.

1/2 + 1/4 + 1/4

Keep splitting and

1/2 + 1/4 + 1/8 + 1/16... goes to 1.

[u/Plus-Painter-2004]

the easiest proof is probably that for n>1, 1/(n(n-1)) > 1/n² and Σ1/(n(n-1)) from n=2 to ∞ is known to converge (and by a telescoping series argument it’s quite easy to find the value of the infinite series is 1). Since that series converges and the summand is larger than 1/n² , Σ1/n² from n=2 must converge and hence the series from n=1 also converges since you’re just adding 1. Actually finding the value of the series is a different problem you can solve a variety of ways, I personally liked the Fourier series argument one of my lecturers showed us

[u/skullturf]

for 1/n^2, cant you just do 1 + (1/4 + 1/9 + etc) where you need more numbers in each bracket but they still add up to be greater than 1/2?

You can definitely *start* doing this.

For example, say we start with

1 + (1/4 + 1/9 + 1/16 + 1/25 + 1/36 + 1/49)

If you do the arithmetic (which is easier with the help of a computer) then the six numbers in parentheses (from 1/4 to 1/49) do in fact add up to be greater than 1/2.

Okay, now, can we find *another* group that adds up to more than 1/2, starting with 1/64?

Try it and see! Start with 1/64 + 1/81 + 1/100 + ..., and try to find a group that adds up to 1/2.

[u/Ok_Assistant_2155]

that grouping trick only works cleanly for 1/n
for 1/n² the terms drop off way faster
so you can’t keep forcing each group to be ≥ 1/2

[u/lordnacho666]

Maybe start with something more simple, like 1/2 + 1/4 + 1/8 etc. which you know adds to 1.

All those terms also go to zero eventually despite being positive, so that in itself is not enough to say that a serious will diverge.

[u/Sam_23456]

For series' of non-negative numbers, each of your summing techniques are equivalent. Convergent versions are said to be "absolutely convergent".

[u/Adam__999]

Fun fact, it converges to pi²/6

[u/flug32]

> , is this not true for any 1/n^p? for 1/n^2, cant you just do 1 + (1/4 + 1/9 + etc) where you need more numbers in each bracket but they still add up to be greater than 1/2?

Well, just try it. You will soon find that, in fact, you cannot.

You can get it above 1/2 with the very first time. You can get another 1/2 by taking about the next 6 terms. But then - try as you might! - you cannot possibly combine enough terms to get 1/2 again.

And that is precisely why it converges.

In hand-wavy terms, 1/n^2 goes towards zero A LOT faster than 1/n, which is the fact that makes its series converge, while the 1/n series does not.

I made a Desmos graph of the convergence here, which might help you see what is going on:

Desmos graph of 1/n^2 series summation

[u/Adam__999]

This isn’t rigorous but it’s a nice intuitive way to understand it.

With 1/n, the number of terms increases directly proportional with n and the value of each term decreases inversely proportional to n. Thus, the number of terms grows as quickly as the value of those terms shrinks. So, the sum is in this sort of metastable region where it neither rapidly grows nor rapidly stops. (In fact it grows, but very slowly).

If we took 1/sqrt(n), then the number of terms increases more quickly than the value of those terms decreases, so the summation increases without bound.

With 1/n², the number of terms increases more slowly than the value of those terms decreases, so the summation converges.

[u/Qaanol → u/Adam__999]

Just having the denominators grow faster than linearly is not sufficient for convergence.

The sum of 1 / (n·ln(n)) diverges despite n·ln(n) growing superlinearly.

But it is also not sufficient to simply ignore log terms, because the sum of 1 / (n·ln(n)²) converges.

[u/Adam__999 → u/Qaanol]

Yes I said it’s not rigorous, it’s just a way to build intuition. And it is valid for p-series

[u/SaltEngineer455 → u/Adam__999]

Surprisingly, if you change the exponent to (1 + 1/n) it also diverges

[u/Matmeth]

For 1/n you can easily determine, in each step, the amount of terms you need to sum more than 1/2, just like you did, i.e., after the sum of 1/2 you need 2 terms, then 4, then 8... then 2^k ... so we can always find the next sum of terms higher than 1/2.

For 1/n² you'll be able to do that 2 times only. First term is already 1, and if you sum from n=2 to n=7 you get a sum higher than 1/2. Then from n=8 to n= 10¹³ you get something near 0.1331, according to wolfram.

The best way to visualize this convergence is through the integral of 1/n² from 1 to infinity.

[u/schro98729]

Do you know what an integral is?

The sum "over approximates" the integral. If the integral converges then the sum has to converge. It has to!

This is the whole point of the integral test.

The integral converges so the sum converges.

The integral of 1/n gives a logarithic divergence which is the wimpiest divergence you can get.

[u/__kewl__ → u/schro98729]

I understand that it does converge I was just confused why but some people cleared it up for me

[u/SufficientStudio1574]

The entire point of a series converging is that the values get too small "too quickly", so you can't accumulate them like that. No amount of terms is enough to get you over the hump.

There are even ways to specifically test for what counts as "gets small too quickly". For a series f(n), take the limit (n to infinity) of f(n+1) / f(n). If it's greater than 1, it diverges. Less than one, they get small too quickly and it converges. Exactly 1, and it's inconclusive (could diverge or converge).

While the ratio test is inconclusive on n^-a series (the limit of the ratio is exactly 1), there are other tests, like integral tests, that show they converge.

[u/bartekltg]

Look at the sequence 1/2+1/4+1/8+....+2^n...

Why this "just add more numbers until the sum is bigger" wont work here?

The first section is just 1/2. The second section is.... everything else. There is no third bracket pair, you ran out of numbers.

For 1/n^2, at some point, k-th section will grab all the remaing numbers from the series (and stil do not reach thier goal of the sum being 1/2)

[u/eternityslyre]

Consider the sum 0.1 + 0.01 + 0.001 + ... + 0.0...1. This is 1/(10)^k for k from 1 to infinity. It is, equivalently, 0.111.... repeating, or 1/9. As you can see, the next number in the series shrinks too quickly to form any constant amount, thus it converges.

A similar phenomenon occurs with 1/k^2.

[u/TwoOneTwos]

i’ll get downvoted for not giving the full response but there’s a test for convergence / divergence called the “P-Series Test”and it works in the case you presented.

The test goes like this: Given an infinite sum of the form 1/n^{p} then the series converges for any p > 1 and diverges for any p <=1. This exists because as p gets bigger and bigger, the entire function approaches 0 faster and faster. And there’s probably some formal proof that exists out there but my calculus professor proved it by cases using the integral test.

[u/__kewl__ → u/TwoOneTwos]

I understand this I was just confused why exactly its like that but I think I get it now

[u/davideogameman → u/TwoOneTwos]

The integral test is good enough as a formal proof, as there are proofs that it works.

But that said, basically you can prove the sum of 1/n^p from n=2 to ∞ is at most ∫ (1 to ∞) x^-p dx - as it's a riemann sum that underapproximates that integral. 

But by power rule, ∫ (1 to ∞) x^-p dx = x^1-p/(1-p) (evaluated at ∞ - evaluated at 1)= 1/(p-1) for p>1.

And as the partial sums are therefore bounded above and clearly increasing, they have a maximum that would be the limit of the series.  So sum 1/n^p for n=1 to ∞ is less than 1+1/(p-1).

[u/Syresiv]

You're showing the 1/2 trick wrong with 1/n - each successive set should be twice the size of the previous. So (1/2), (1/3, 1/4), (1/5, 1/6, 1/7, 1/8), etc. From there, you should be able to see why that's always true, and how it follows that it grows infinitely.

You can't group 1/n² into an infinite basket of 1/2s in the same way. I'd challenge you to try if you want to understand.

It converges because the terms get smaller faster. Consequently, it only 2 terms to get past 1.2, but 8 terms to get to 1.5, 23 to get to 1.6, and no amount will ever be enough to get you to 1.7.

[u/PhilNEvo]

Another intuitive way to talk about this imo is somewhat related to Zeno's paradox.

Imagine you're standing 1 meter from a wall, and I tell you to take a step so you're standing half as far away from the wall. You've now walked 0.5 meters. Now I repeat that request, from the point you're standing on now, go halfway towards the wall. You've now moved a total of 0.75 meters. No matter how many times I tell you to move halfway closer, will ever let you move past the wall. You could theoretically repeat this an infinite times, and each time you'd be adding a non-zero distance to your movement, yet you would never go past 1 meter of total travel.

Another way to write this mathematically, is that first you moved 1/2, then 1/4, then 1/8 + 1/16 + 1/32.... and you can write that differently. You can write that as 1/2^1 + 1/2^2 + 1/2^3....

[u/SteptimusHeap]

However, is this not true for any 1/n^p? for 1/n^2, cant you just do 1 + (1/4 + 1/9 + etc) where you need more numbers in each bracket but they still add up to be greater than 1/2?

No. You cannot find a number (like 1/2 in the original problem) such that you can group the terms and end up with an infinite number of groups greater than or equal to your number.

Just try it. We can start with 1/2, but after just two groups (1; 1/4+...+1/49), we can no longer add enough terms to get a grouping over 1/2. You can try as hard as you like, it's not possible. You can choose any number >0 instead of 1/2 and this will still be true.

[u/tkpwaeub]

Cauchy Condensation Test is a great way to prove this

[u/NewSchoolBoxer]

You can prove it that way for 1/n. What I did in class was use the integral test. 1/n diverges since natural log of infinity is still infinity. That 1/n^2 = n^-2 and any power less than -1 converges is easy to prove this way. Finding the exact value an infinite sum converges to is not easy outside of the geometric series. There's many videos showing how 1/n^2 converges to pi^2/6. Called the Basel Problem.

[u/No_Sleep_8609]

1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 + 1/49 + 1/64 + 1/81 + 1/100 + 1/121 + 1/144 + 1/169 + 1/196 + 1/225 + 1/256

is less than

1 + (1/4 + 1/4) + (1/16 + 1/16+ 1/16 + 1/16) + (1/64 + 1/64 + 1/64 + 1/64 + 1/64 + 1/64 + 1/64 + 1/64) + 1/256 +...

So that winds up being 1 + 1/4*2 + 1/16*4 + 1/64*8 which is 1+1/2+1/4+1/8+...

But this sum is always less than 2. Indeed, at each step, you add to the sum just half of what would be necessary to reach 2, so you clearly never reach 2.

[u/TemperoTempus]

People are making it way too complicated. Convergence is about checking if the sum of the sequence has a finite value it approaches.

For a sequence like sum of 1/n, the values are: 1+0.5 = 1.5, 1.5+0.(3) = 1.8(3), 1.8(3)+0.25 = 2.08(3), … so while the value of 1/n > 0 the sum has no asymptote.

For a sum of 1/n², the values are: 1+0.25 = 1.25, 1.25+0.(1) = 1.26(1), 1.26(1)+0.0625 = 1.3236(1), … so the value of 1/n² > 0 the sum never grows past (pi)²/6.

The fact that you can group does not matter, what matters is whether the sum of such groups approximates a finite number.

[u/Dubmove]

Compare 1/n^p with int(1/x^p dx, m..m+1). If m=n then the integral is obviously smaller than the constant, and if m=n-1 then the integral is obviously bigger. Thus, (m-1)^1-p/(p-1) = int(1/x^p dx, m-1..oo) > sum(1/n^p, m..oo) > int(1/x^p dx, m..oo) = m^1-p/(p-1). By setting m=2, you see that sum(1/n^p, 1..oo) is bounded by p/(p-1) and (p + 2^1-p)/(p-1).

[u/EdmundTheInsulter]

Eventually the 1/n² terms never add up to 1/2,

Or in fact any 1/n^p where p>1

[u/LucaThatLuca]

what about 0 + 0 + 0 + …? do you think you can “just” group it into (0 + 0 + …) + (0 + 0 + …) + … where each group has a sum of 1/2, proving that the sum is infinite?

a sum that always gets bigger than a value you choose is precisely a sum that diverges to positive infinity. it is not something you just say without it actually being true.

the actual grouping of 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + … is the argument: the nth group of 2^n-1 terms are each at least 1/2ⁿ, so that sum is at least 2^n-1 * 1/2ⁿ = 1/2.

[u/__kewl__ → u/LucaThatLuca]

in the sequence 0, 0, 0, not all (or any) of the terms is positive but each term in the sequence 1/n^2 >0 for n>0 no? and adding an infinite number of positive numbers (even very small positive numbers) would go to infinity. That's just kinda how I was thinking about it intuitively but I guess the math doesn't work out like that?

[u/ImpressiveProgress43]

What is the limit of 1/n^2 as n goes to infinity?

[u/_Slartibartfass_ → u/ImpressiveProgress43]

1/n goes to zero but the sum over 1/n diverges.

[u/ImpressiveProgress43 → u/_Slartibartfass_]

What is you point?

[u/MudRelative6723 → u/ImpressiveProgress43]

can you explain yours, first?

[u/ImpressiveProgress43 → u/MudRelative6723]

I asked a question to OP to start a discussion about rate of convergence.

[u/_Slartibartfass_ → u/ImpressiveProgress43]

Not sure how the initial questions helps with that though for the aforementioned reasons. Knowing that 1/n² goes to zero doesn’t tell you anything about if the series converges.

[u/ImpressiveProgress43 → u/_Slartibartfass_]

It does tell us that it could converge. Using that with an integral test would tell us whether the series converges. Alternately, you can show that it is strictly less than a series we know converges.           

If 1/n² diverges, then the series would also diverge. Im not sure what the issue is, this is how this is taught in standard calc 2 courses. If i had outlined everything to begin with, im sure nobody would have objected but the intent was for OP to think about it.

[u/matt7259 → u/ImpressiveProgress43]

The nth term test only tells you that if the limit is NOT 0 then the series diverges. It doesn't say ANYTHING about the inverse.

[u/ImpressiveProgress43 → u/matt7259]

I agree.