# Is there an easier formulation for the Hadamard product of certain pair of series?

I'm interested in finding a recursion or simple representation for a "Hadamard product" of two power series.

The Hadamard product is defined on generating functions $f(x)$ and $f(y)$ as $f(z)$ below:

$f(x) = \sum_{i=0}^\infty{c_i x^i}$

$f(y) = \sum_{i=0}^\infty{d_i y^i}$

$f(z) = \sum_{i=0}^\infty{c_i \cdot d_i z^i}$

My Particular Interest

Now suppose that we work on the series above with only the first N terms as possibly nonzero, and the rest will be considered to be zero.

Take the coefficients of $f(x)$ to be the reciprocals of the first $N$ naturals, e.g.

$f(x) = \sum_{i=1}^N{\frac{1}{i} x^{i+1}}$

Now suppose that $f(y)$ is in the form of an expression, i.e.

$f(y) = \frac{1-(2y)^{N-1}}{1 - 2y}$ represents the first N terms of the form $(2y)^N$

or

$f(y) = \frac{1-(n+1)x^N+Nx^{N+1}}{(x-1)^2}$ represents the first N naturals in series

or some similar function...

My Question

Can we find a simple expression, such as a recursion or formula, to represent the "Hadamard product" of these new series?

We could simply write out the $N$ terms in the resulting function $f(z)$. However, I'm wondering if there is a more compact way of representing this. I'm particularly interested when $f(x)$ is the series of reciprocals mentioned above. However, I am interested in finding this result for almost any $f(y)$. Is there a way to do this?

I'd like to consider both the cases when coefficients of $f(y)$ take on real values and when it is complex-valued.

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It's $z \int_0^z \frac{f(y)}{y^2} dy$ after you remove the constant and linear terms.