===== Niemand seqeunce =====
==== Note ====
Let's define a Niemand sequence $(a_{n,N})$ as a sequence (in $N$) of sequences (in $n$).

The map

$(a_{n,N}) \mapsto (S_N):=\sum_{n=0}^N a_{n,N}$

removes the $n$-index. And then

$(S_N) \mapsto \sum_{n=0}^N S_N$

removes the second. 

=== Examples ===
== a{n,N} constant in N ==
For $a_{n,N}$ constant in $N$, the series $\sum_{n=0}^N a_{n,N}$ is just the sequence of partial sums.

== Riemann integral ==
$f$ a function and $x_0,x_1$ numbers.

With $h(N):=\dfrac{x_1-x_0}{N}$

$a_{n,N}:=h(N)\,f\left(x_0+n\,h(N)\right)$

we have that 

$\sum_{n=0}^N a_{n,N}$

is the Riemann sum and

$\lim_{N\to\infty}\sum_{n=0}^N a_{n,N}=\int_{x_0}^{x_1}f(x)\,{\mathrm d}x$

(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

$\lim_{N\to\infty}\sum_{n=0}^N a_{n,N}={\mathrm e}^{k\,z}$

<code>
Sum[Product[1 - (k - 1)/N, {k, 1, n}] ((C[N] k z)/(1 - (k \[Zeta])/N))^n/n!, {n, 0, N}] // simple
</code>

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 $\dfrac{1-{\mathrm e}^{-c\,z}}{z}$. 

<code>
Sum[-Product[1 - k/N, {k, 1, n}] ((-c[N])^(n+1)/(n + 1)) z^n/n!, {n, 0, N - 1}] // Simplify
</code>

(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 $z=0$.

=== References ===
-----
=== Context ===
[[Infinite series]]
=== Related ===
[[Infinite sum of complex numbers]], 
[[Finite geometric series]]