tommy's "linear" summability method

[Caleb]

tommy's summability method
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I might not be the first to consider it so the name is perhaps disputable or temporarily anyway ,


This method is for divergeant strictly increasing positive integer sums

consider the geometric sum : 1 + 2 + 4 + 8 + ... + 2^n + ... = x

now we see 2 x = 2 + 4 + 8 + ...

notice x - 1 = 2 + 4 + 8 + ...

so

2 x = x - 1.

thus

x = -1.


lets generalize for real a > 1 :

1 + a + a^2 + ... = x

then

a x = x - 1

so

x = -1/(a-1) = 1/(1-a).

This is all old stuff.

So whats next ?

consider an entire function f(x) given by a taylor series : f_1 x + f_2 x^2 + f_3 x^3 + ... + f_i x^i + ...

such that f(i+1) > f(i) for every integer i > 0.

Notice f(0) = 0.


Now we can sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x

Here is how

sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x

sum_n sum_i  f_i (a^n)^i = x

so

x = sum_n sum_i  f_i (a^n)^i = sum_i sum_n  f_i (a^i)^n

= sum_i f_i * - 1/(a^i - 1) = sum_i f_i / (1 - a^i)



As example consider

dexp(x) := exp(x) - 1.

dexp(1) + dexp(2) + dexp(2^2) + dexp(2^3) + dexp(2^4) + ... = x

Then

x  = sum_i f_i /( 1 - 2^i ) = sum_i  1/( i! * (1 - 2^i) )


this clearly converges.

(I got an estimate of x = -1.19355 btw)

I call this a linear summability method because it has some linear properties.



regards


tommy1729

This derivation is almost correct, though it possesses a problem that only emerges when considering the right functions. You noted that
\[\sum_{n=0}^\infty  f(a^n) = \sum_k f_k \sum_n a^{nk} = \sum_k f_k \frac{1}{1-a^k}\]
However, note that the last step involves evaluating an analytical continuation outside the domain of the convergence of the series. When this happens, things can do wrong. Consider \(f(z,x) = z^x\) and \[\sum (-1)^n f(z,a^n) = \sum (-1)^n z^{a^n} \text{ } {}^{"}=^{"} \text{ } \sum \frac{\ln(z)^k}{k!(1+a^k)}\] Actually, for \(a>1\) these things are not equal, Gottfried talks about the difference for \(a=2\) here: https://mathoverflow.net/questions/19866...871#198871. However, for \(a<1\), the inner series converges and so in many cases the reasoning is valid and the series converges properly. 

Now, one could stop there, and say perhaps the summation method simply does not work for \(a>1\) since we are attempting to sum divergent series. However, we don't need to be so closed minded-- maybe there is a way to recover the correct function from these evaluations. I consider this view, which you may find interesting, here: https://mathoverflow.net/questions/43869...tical-cont

If you believe the arguemnt I give in the above link is valid, then you evaulation of the dexp is not quite correct, and part of the correction factor can be obtained by evaulating the residues that emerge when \(1-2^z=0\). However,