this post was submitted on 26 Mar 2024
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[โ€“] Jayjader@jlai.lu 3 points 8 months ago (1 children)

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.

[โ€“] mitrosus@discuss.tchncs.de 2 points 7 months ago (1 children)

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.

[โ€“] Jayjader@jlai.lu 1 points 7 months ago* (last edited 7 months ago) (1 children)

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/โˆšn, 1/โˆš(โˆšn), etc.

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

[โ€“] mitrosus@discuss.tchncs.de 1 points 7 months ago (1 children)

Hang on, that's not a decreasing trend. 1/โˆš4 is not smaller, but larger than 1/4...?

[โ€“] Jayjader@jlai.lu 1 points 7 months ago

From 1/โˆš3 to 1/โˆš4 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/(3ยฒ) to 1/(4ยฒ).

The curve here is not mapping 1/4 -> 1/โˆš4, but rather 4 -> 1/โˆš4 (and 3 -> 1/โˆš3, and so on).