Agreed, and I hope I wasn’t guilty of this. The OP asked “Did I do this right?”; and neither an unqualified Yes nor an unqualified No would have been an appropriate answer.
Thudlow, no, I didn’t perceive you shooting me down at all! Over the years, you’ve consistently answered many of my math questions in a way that I can understand, and so I knew where you were coming from. I know from seeing answers in areas where I have a certain level of expertise (art, music and marketing), that there often are nuances at play where as you put it “neither an unqualified yes nor an unqualified no would have been the appropriate answer”
Indistinguishable, thank you for comments and perspective. I remember reading your post on the various means of summation in the “sum of natural numbers is -1/12” thread and found that very helpful.
So let’s go down the rabbit hole further, shall we? I thought of this as I was falling asleep last night. Apologies, because I don’t know how to express this robustly, but:
Given a Ramanujan summation of the series of the powers of n, n>1 (although I noted Indistinguishable’s point about it works for n>0 and < 1 but don’t fully understand that yet), we get the following:
n=2, S = -1
n=3, S = -1/2
n=4, S = -1/3
n=5, S = -1/4
n=6, S = -1/5
So, ladies and gentlemen, that’s the harmonic series in a negative direction. As even I know, the harmonic series is divergent, and so adds to negative infinity (by an intuitive method of summation). So what that means is by this summation method, when you add a shitload of non-prime, positive integers together, you get negative infinity.
Jesus Christ, that’s amazing! Assuming one of youse don’t come in here and tell me I made (yet another) mistake.
You’re alright; my ire was focused elsewhere. Your post #12 put it fine:
Yes, that’s all correct. In the same sense that 1 + 2 + 4 + 8 + … = -1, that 1 + 10 + 100 + 1000 + … = -1/9, and so on, we may say that the sum of n^k over all n >= 2 and k >= 0 is -∞.
Of course, there is another, more straightforward sense in which 1 + 2 + 4 + 8 + … = +∞, and in that sense, the sum of n^k over all n >= 2 and k >= 0 is +∞ as well. In this more straightforward sense, adding a lot of increasingly large positive values gives a positively infinite result, not a negatively infinite result. Math doesn’t deny this and no one is wrong to say this; there are just multiple different perspectives available to us.
You may find it interesting to consider now the sum n + 2n^2 + 3n^3 + 4n^4 + …. Can you figure out what this should come out to?
[And once you’ve done that, you may find it interesting to consider the sum of (k + 1) n^k over all n >= 2 and k >= 0. But that’s rather a challenge…]
Cool! I hit a roadblock on n + 2n^2 + 3n^3 + 4n^4 + …., but when I did n^0 + n + 2n^2 + 3n^3 + 4n^4 + …., I got it back to -1/(1-n), so the positive of n^0+n^1+n^2. Which is pretty cool, if correct. Is that what you had in mind?
You got that 1 + n + 2n^2 + 3n^3 + 4n^4 + … = -1/(1-n)? That doesn’t seem correct; for example, at n = 0, the left-hand side is 1 and the right-hand side is -1. What was your reasoning?
OK, you’re right. Made a mistake in the subtraction. (However, isn’t the implication n > 1, here?)
I just played around with it again, and think I have the answer, but want to sleep on it and check my math in the morning. This is fun!
Before I answer this, can you please confirm that I have this correct. Given:
S1 = n1 + n2 + n3 + …
S2 = a1 + a2 + a3 + …
That it is cool to subtract them in this way:
S1 - S2 = n1 + (n2 - a1) + (n3 - a2) + …
Tony in the infamous Numberphile video of the natural numbers = -1/12 did this in one step of his proof, but he was adding. So I just wanted to make sure that this is right and holy with subtracting.
For the notions of summation invoked in this thread, yeah, that would be fine. That’s certainly how I would happily expect you to be reasoning about these series.
Very good!
Cecil help me, here goes:
S = n + 2n^2 + 3n^3 + 4n^4 + …
nS = n^2 + 2n^3 + 3n^4 + 4n^5 + …
S – nS = n + (2n^2 – n^2) + (3n^3 – 2n^3) + (4n^4 – 3n^4) + …
(1-n)S = n + n^2 + n^3 + n^4 + …
(1-n)S = n (1 + n + n^2 + n^3 + …)
Using the sum demonstrated up-thread for 1 + n + n^2 + n^3 + …
(1-n)S = n (1/(1-n))
S = n
Perfect all the way up through this penultimate line!
Whoops, you made a slight slip in the algebra moving from the previous line to this one.
:smack:
S = n/(1-n)^2
It is awesome how these series can turn out! I keep thinking “Oh my God, it’s full of stars!”