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.
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.