"Natural" interpolation between partial sums of a power series Suppose $f(z)=\sum_{n=0}^\infty a_n z^n$ has a radius of convergence of $R$. Let the $N$-th partial sum be $f_N (z)=\sum_{n=0}^N a_n z^n$. What smooth (analytic) function interpolates between $f_N(z)$. In other words what "natural" analytic function $F(\alpha,z)$ has the property that $F(N,z)=f_N(z)$ for $N=0,1,\cdots$.
Related Question for $f=e^z$
 A: If $a_k$ happens to be a continuous function in $k$, say $a(k)$, one could introduce a parameter $\omega$ and define
$$
f(z,\omega)=\sum_{k=0}^\infty a(k+\omega)\, z^k,
$$
where $f(z):=f(z,0)$. Introducing the parameter $n\in\Bbb N_0$ we could then express the the truncated series in the question as
$$
f_n(z)=\sum_{k=0}^na(k)\,z^k.
$$
Now writing the sum as the difference of series (assuming convergence of series) yields
$$
f_n(z)=\left(\sum_{k=0}^\infty-\sum_{k=n+1}^\infty\right)a(k)\,z^k=f(z)-z^{n+1}\sum_{n=0}^\infty a(k+n+1)\,z^k=f(z)-z^{n+1}f(z,n+1).
$$
We can now relax the assumption $n\in\Bbb N_0$ to find an interpolation of $f_n$.
As a simple example consider the truncated geometric series
$$
f_n(z)=\sum_{k=0}^n z^k.
$$
Clearly, $a(k)=1$ is continuous everywhere so we write the interpolation function as
$$
f_\nu(z)=\sum_{k=0}^\infty z^k-z^{\nu+1}\sum_{k=0}^\infty z^k=\frac{1}{1-z}-z^{\nu+1}\frac{1}{1-z}=\frac{1-z^{\nu+1}}{1-z},
$$
which provides an interpolation for $|z|<1$ and $\nu\in\Bbb C$.
