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