Relation between roots of $f$ and roots of the partial sums of its power-series It is well known we can write any holomorphic function as a local power series, for example
$$\exp(z) = \sum_{n=0}^\infty \frac{z^n}{n!}$$
Obviously each partial sums are polynomials of degree $n$ which have always $n$ roots. I now wonder what is the relation or if there is any relation between the roots of a holomorphic function and the roots of its partial sums. For exmample it is well known that the exponential function does not have any roots, but for the partial sums
$$(p_1(z)=1,) p_2(z)=1+z, p_3(z)=1+z+\frac{z^2}{2}, p_4(z)=1+z+\frac{z^2}{2}+ \frac{z^3}{6},..$$
we have the zero sets $$\{-1\}, \{-1-i,-1+i\},\{\approx-1,5961,\approx -0.702\pm1.807i\},..$$
(exact forms of the last roots are very poor)
For other holomorphic function that have roots, for example the Sine, we know
$$\sin(z)=\sum_{n=0}^\infty(-1)^n \frac{z^{2n+1}}{(2n+1)!}$$
It has roots in $k\pi$, $k\in\mathbb{Z}$, and its partial sums
$$ p_1(z)=z, p_2(z)=z-\frac{z^3}{6}, p_3(z)=z-\frac{z^3}{6}+ \frac{z^5}{120},..$$
have the zero sets
$$\{0\},\{0,\pm\sqrt{6}\},\{0,\pm3.2369 \pm 0.6908 i\},..$$
Obviously $0$ is always gonna be a root, but whats is about the other roots?
Do the roots of the partial sums converge towards the roots of the limit function (what sounds reasonable)? And if so, what happens in the case where the limit function itself has no roots (e.g. exponential function?) May the zero set be diverging?
Is there any relation between the zeros of the partial sums (that always exists) and the zeros of the limit function?
Edit:
I made some mathematica code to plot the zeros for increasing partial sums and indeed it looks like in the case of the exponential function its zeros are diverging. Since I have no idea how (or if its even possible) to upload mathematica code here, here some images
For the sine on the other hand it seems like that all the sine zeros are generated, but also there are many diverging roots.
 
However I could only calculate up to $n=50$ because for $n$ too big mathematica really needs some time to actually calculate the roots.

You may look for Hurwitz's Theorem. Since power series of $\exp$ and $\sin$ converges uniformly on every compact subset of $\mathbb{C}$, we can take small neighborhoods around the zeros of original function, to see that zeros converge to original function's zeros. For the exponential function, I think all the zeros will diverge, since $\exp$ has no zero in $\mathbb{C}$(and therefore $p_n(z)$ has no zero in each compact disk $D_r = \{z:|z|<r\}$ for large enough $n$). If you apply the fundamental theorem of algebra, the zeros would accumulate to the original zero, and many other zeros(there are many since $n$-degree polynomials have $n$ roots) will diverge to $\infty$.
