[ForOldHack]

This needs to be repeated time and time again for people who deny the basic tools of Calculus, and will suffer the misuse of them. ( specifically the sum of the sequence of the reciprocals of the natural numbers is equal to 1/12. I get that it is a useful tool for quantum chromodynamics, but it makes my skin crawl. )

[eszed]

It's one of the regrets of my life that I didn't take start calculus my first term in college. I'd done pre-calc in high school, but I'm a Humanities guy, and maths - even when I "get" it (and I'd done fine in high school) - drops out of my head pretty quickly. By the time I had an opportunity to continue, and signed up for a "refresher" pre-calc course, I realized I'd have had to go right the way back and re-learn enough algebra and geometry stuff that it was too heavy a lift. Maybe when I'm retired and have time on my hands I'll sign up for some courses at a community college. I still remember how satisfying it was to grok a problem or a concept that had seemed impossible.

So, ahem: is there any possibility that I'll be able to understand "the sum of the sequence of the reciprocals of the natural numbers is equal to 1/12"? I understand (I think!) each of the words in that sentence, but I can't make them make sense! I've got as far as

1 + 1/2 + 1/3, etc.

So how does something that starts off with 1 + [something] end up < 1?

[rachofsunshine]

It doesn't, and in fact I believe the person you're replying to has it confused with another result where the answer would also be "it doesn't".

The sum 1 + 1/2 + 1/3 + 1/4 ... - what we call the harmonic series - runs off to infinity, although it does so very slowly (the nth partial sum is approximately equal to ln(n) plus a constant whose value is about 0.58). Showing that this series diverges is standard calc-textbook stuff (it's the textbook example of the integral test for convergence, although there are plenty of other ways to show it).

However, in math well beyond basic calculus, there are methods for assigning meaningful values to series that don't converge in the conventional sense. Those methods assign the same value to convergent series as the regular old calculus arguments would, but they can also assign values to some divergent series in a way that is consistent and useful in some contexts.

For example, the series 1 + 1/2 + 1/3 + 1/4 + ... is a specific example of a more general series 1/1^z + 1/2^z + 1/3^z + 1/4^z + ..., for some arbitrary number z. This series only converges in the standard calculus sense when the real part of z is > 1, but it turns out that that's enough to define a function called the Riemann zeta function, whose input is the value z in the series and whose output is the sum of the series.

The zeta function, as it turns out, can be extended to values where the original series didn't converge. And doing so gives you a method for assigning values to "sums" that aren't really sums at all.

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It turns out that even THAT won't get you a completely nice value for 1 + 1/2 + 1/3 + 1/4 + ..., because that sum corresponds to the zeta function's value at z = 1. But the zeta function isn't well-behaved at z = 1. If you do even more massaging to beat a number out of it, you don't get 1/12, you get the about-0.58 constant mentioned in the previous section.

Which brings me to the way the poster you were replying to is probably confused. I think the sum they meant to refer to was the even-more-obviously-divergent sum 1 + 2 + 3 + 4 + ... This sum happens to correspond loosely to the z = -1 value of the zeta function, since 1/2^-1 is just 2, 1/3^-1 is just 3, and so on.

Again, the sum 1 + 2 + 3 + 4 + ... does not converge, but if we choose to identify it with a zeta function value, we'd identify the sum 1 + 2 + 3 + 4 + ... with zeta(-1). And the value of zeta(-1) happens to be -1/12 (yes, minus 1/12).

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So the answer here is just that no, 1 + a bunch of positive numbers is never < 1. It's actually a decent basic calculus exercise to prove that any series a_0 + a_1 + ... a_n, where all the terms are non-negative, either fails to converge or converges to a value >= a_0. But because we're taking some extra steps here to leave the domain of convergent series in the first place, it turns out we can get results that (for conventionally convergent series) would be impossible.