The monotheistic all powerful one.

    • Jayjader@jlai.lu
      link
      fedilink
      arrow-up
      3
      ·
      9 months ago

      If I remember my series analysis math classes correctly: technically, summing a decreasing trend up to infinity will give you a finite value if and only if the trend decreases faster than the function/curve x -> 1/x.

      • mitrosus
        link
        fedilink
        arrow-up
        2
        ·
        9 months ago

        Great. Can you give me example of decreasing trend slower than that function curve?, where summation doesn’t give finite value? A simple example please, I am not math scholar.

        • Jayjader@jlai.lu
          link
          fedilink
          arrow-up
          1
          ·
          edit-2
          9 months ago

          So, for starters, any exponentiation “greater than 1” is a valid candidate, in the sense that 1/(n^2), 1/(n^3), etc will all give a finite sum over infinite values of n.

          From that, inverting the exponentiation “rule” gives us the “simple” examples you are looking for: 1/√n, 1/√(√n), etc.

          Knowing that √n = n^(1/2), and so that 1/√n can be written as 1/(n^(1/2)), might help make these examples more obvious.

          • mitrosus
            link
            fedilink
            arrow-up
            1
            ·
            9 months ago

            Hang on, that’s not a decreasing trend. 1/√4 is not smaller, but larger than 1/4…?

            • Jayjader@jlai.lu
              link
              fedilink
              arrow-up
              1
              ·
              9 months ago

              From 1/√3 to 1/√4 is less of a decrease than from 1/3 to 1/4, just as from 1/3 to 1/4 is less of a decrease than from 1/(3²) to 1/(4²).

              The curve here is not mapping 1/4 -> 1/√4, but rather 4 -> 1/√4 (and 3 -> 1/√3, and so on).