Niemand seqeunce
Note
Let's define a Niemand sequence (an,N) as a sequence (in N) of sequences (in n).
The map
(an,N)↦(SN):=∑Nn=0an,N
removes the n-index. And then
(SN)↦∑Nn=0SN
removes the second.
Examples
a{n,N} constant in N
For an,N constant in N, the series ∑Nn=0an,N is just the sequence of partial sums.
Riemann integral
f a function and x0,x1 numbers.
With h(N):=x1−x0N
an,N:=h(N)f(x0+nh(N))
we have that
∑Nn=0an,N
is the Riemann sum and
lim
(see Limit in a metric space)
Exponential function
todo: compare the following construction with the two q-exponentials
a_{n,N} = \left(\prod_{k=1}^n\left(1-\dfrac{k-1}{N}\right)\right) \dfrac{1}{n!}\left(\dfrac{C(N)\,k\,z}{1-\frac{k\zeta}{N}}\right)^n
{\mathrm e}_N(z):=\sum_{n=0}^N a_{n,N} = \left(1+\dfrac{1}{1-\frac{k\zeta}{N}}c(N)\dfrac{z}{N}\right)^N
Fulfills
{\mathrm e}_N'(\zeta)=k\,{\mathrm e}_N(\zeta)
(only for z=\zeta, not for all of z like the exponential function)
and with c(N)=1, we have
\lim_{N\to\infty}\sum_{n=0}^N a_{n,N}={\mathrm e}^{k\,z}
Sum[Product[1 - (k - 1)/N, {k, 1, n}] ((C[N] k z)/(1 - (k \[Zeta])/N))^n/n!, {n, 0, N}] // simple
What's nice here is that also
\sum_{n=0}^\infty \lim_{N\to\infty} a_{n,N} = {\mathrm e}^{k\,z}
(see Exponential function)
Inverting z
a_{n,N} = -\left(\prod_{k=1}^n\left(1-\frac{k}{N}\right)\right)\dfrac{(-c(N))^{n+1}}{n+1}\dfrac{z^n}{n!}
\sum_{n=0}^{N-1} a_{n,N} = \dfrac{1}{z} \left(1 - (1-\frac{c(N)\,z}{N})^N\right)
For c(N):=c a constant, this is a sort of regularization of \dfrac{1}{z} and has the limit N\to\infty of \dfrac{1-{\mathrm e}^{-c\,z}}{z}.
Sum[-Product[1 - k/N, {k, 1, n}] ((-c[N])^(n+1)/(n + 1)) z^n/n!, {n, 0, N - 1}] // Simplify
(see Infinite geometric series)
In fact, let
\phi(z):=-\left(1-\dfrac{c(N)\,z}{N}\right)^N
Then
\dfrac{1}{z} \left(1 - (1-\frac{c(N)\,z}{N})^N\right) = \dfrac{\phi(0+z)-\phi(0)}{z}
is just the finite difference quotient of this at z=0.