Leibniz integral rule for summation

Question

Context

Fundamental points of Feymann trick:

1. You have an integral $I_0=\int_a^b f(t)\mathrm{d}t$
2. Now consider a general integral $I(\alpha)=\int_a^b g(\alpha,t)\mathrm{d}t$ so that $I'(\alpha)=I_0$ for some $\alpha_0$, where $g(\alpha,t)$ is an easier integrative to compute than $f(t)$
3. You solve $I(\alpha)$
4. You calculate $I_0=I'(\alpha_0)$

My problem

If in the 3rd point I calculate the integral hase a formula like this: $$I(\alpha)=\sum_{k=0}^{\alpha}h(k,\alpha)$$

How do I calculate $I'(\alpha_0)$?

Question

Is there a formula similar to that of Leibniz integral rule in which instead of the integral there is a summation?

My idea

I'm not sure if this formula is correct, but I suspect the solution might be something like this

$$\frac{\mathrm{d}}{\mathrm{d}n}\sum_{k=0}^{n}f(n,k)=\sum_{k=0}^{\infty}\frac{\partial}{\partial n}f(n,k)+\text{something}$$

Example with disclaimer

I'm not sure there should be infinite in the series, but I think there should still be a "high" number for the specific problem

I guess the sum must not depend on n otherwise I would have the following problem: $$\sum_{k=0}^{n}f(n,k)=\sum_{k=0}^{n}f(n,n-k)$$ $$\text{But}$$ $$\sum_{k=0}^{n}\frac{\partial}{\partial n}f(k,n)\neq\sum_{k=0}^{n}\frac{\partial}{\partial n}f(n-k,n)$$

$$\sum_{k=0}^{n}\binom{n}{k}=2^n\qquad\overset{\partial_n}{\Rightarrow}\qquad\sum_{k=0}^{n}\binom{n}{k}(H_n-H_{n-k})\propto \ln(2)\cdot 2^n$$

$$\sum_{k=0}^{n}(-1)^k\binom{n}{k}=0\qquad\overset{\partial_n}{\Rightarrow}\qquad\sum_{k=0}^{n}(-1)^k\binom{n}{k}(H_n-H_{n-k})= \frac{(-1)^n}{n}\to 0 \text{ for }n\to \infty$$

In this case the problem is that the binomial coefficient can only be calculated between $0$ and $n$

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I was also trying to get it by transforming the summation into a series using the Dirac comb and appling Leibnitz:

$$\sum_{k=0}^{n}f(n,k)=\int_0^n f(n,k)\text{Ш}_1(k)\mathrm{d}k$$

$$\frac{\partial}{\partial n}\sum_{k=0}^{n}f(n,k)=\frac{\partial}{\partial n}\int_0^n f(n,k)\text{Ш}_1(k)\mathrm{d}k$$

$$f(n,n)\text{Ш}_1(n) \frac{\partial}{\partial n}n-f(n,n)\text{Ш}_1(0) \frac{\partial}{\partial n}0+\int_{0}^{n}\frac{\partial}{\partial n}f(n,k)\text{Ш}_1(k)dk$$

$$f(n,n)\text{Ш}_1(n)+\sum_{k=0}^{n}\frac{\partial}{\partial n}f(n,k)$$

But that's clearly wrong.