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Not really a simple regularization analogy

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Last year I posted about an infinite sum involving the mean of the harmonic numbers,

\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{k} = \lim_{n\to\infty}\frac{\ln n}{n}

The method I used to evaluate this was to approximate the sum by an integral. This is a technique that is used in various places in physics, such as in the computation of the Fermi surface in metals.

In this particular case, we only cared about the large-n limiting behavior of the sum, namely that it grows sublinearly for large n. But suppose you wanted to know whether there was, say, a constant term as well. Here's one way to figure that out. Instead of converting the entire sum to an integral, you choose some cutoff value a, and keep the first a terms in the sum explicitly.

\sum_{k=1}^{n}\frac{1}{k} \approx \sum_{k=1}^{a}\frac{1}{k} + \sum_{k=a+1}^{n}\frac{1}{k} = \sum_{k=1}^{a}\frac{1}{k} + \ln\frac{n}{a+1}

Now you have a value, a, that represents a "break" in your sequence, but it's an arbitrary value, so you can choose it to be anything you want. In fact, you can even think of a as a "sliding" cutoff, whose value can go up and down arbitrarily. This suggests that you might want to figure out the effect on your sum each time you bump up a by one.

  • The sum goes from \sum_{k=1}^{a}\frac{1}{k} to \sum_{k=1}^{a+1}\frac{1}{k}, for a net increase of \frac{1}{a+1}
  • The logarithm goes from \ln\frac{n}{a+1} to \ln\frac{n}{a+2}, for a net decrease of \ln\frac{a+2}{a+1}

This whole thing gets plugged into the limit expression from above, giving

\lim_{n\to\infty}\frac{1}{n}\biggl(\sum_{k=1}{a}\frac{1}{k} + \ln\frac{n}{a+1}\biggr)

Notice that you have a finite value (the sum) divided by n, added to \frac{1}{n}\ln\frac{n}{a+1}. Both of these terms obviously go to zero in the limit as n\to\infty — and specifically, this is true regardless of what value you choose for the cutoff a. So now you know you have the freedom to make the cutoff as large as you want, including as many terms as you feel necessary in the actual sum (instead of the integral) to get the error as low as you like.

This behavior where you have a cutoff that you can slide up and down without affecting your result (much) is kind of similar to regularization, but not exactly. Later on this month I'll post a much better example that shows how regularization can actually convert an undefined difference of two infinite expressions into a finite value.

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