series: can the result be zero for a continuous interval of its argument?

I'm considering the series $$f_c(x) = \sum_{k=c}^\infty \left( c^{k-1} \binom{k}{c} \cdot \prod_{j=1}^{k-1} (x-1/j) \right)$$ where the parameter $c \in \mathbb N ,c \gt 0$ and fixed for a certain discussion.
Let's denote the product-term $\prod_{j=1}^{k-1} (x-1/j)$ as $y_k(x)$

First, I see, that for $x =1/m, m\in \mathbb N$ the series is finite (the higher terms $y_{k \ge m}(1/m) = 0$) and gives an exact value. While this is obvious, it is not obvious to me, that, for all such x , and even for all c (!), it seems that it is also true that $f_{c \gt 0} (1/m) = 0$ which I can approximate faily well using Pari/GP.

Q1: This looks somehow like a telescoping effect. Can it be shown, that for the so-constructed finite sums the result is always zero?

Next, if $|x| \lt 1 \ne 1/m$ is a rational value with a non-simple fraction or is even irrational (the single terms do not vanish) then $f_c(x)$ becomes a series with infinitely many terms. Still it seems, that for any such x the series converges for c=1 and again it converges to zero. For example, using 400 terms, I got $f_1(0.8+\pi /1000,400) \lt 1e-38$ . Increasing the number of terms to 800 I got $f_1(0.8+\pi /1000,800) \lt 1e-76$ . Letting x approach 1 from below, I just needed more terms to always approximate zero (for negative x the terms have alternating signs, and their absolute values increase first before they begin to decrease).
For c=2 it is similar, only that it must be $|x| \lt 1/2$ and in general it seems $f_c(x)$ approaches zero whenever we have $|x| \lt 1/c$

Q2: True? Is there an algebraic proof?

If $x \lt -1/c$ the absolute value of the terms increase, but it seems, the series can always summed by Euler-summation, because the terms alternate in sign. Using that summation: again it seems, that the series always approaches zero.

Q3: From the properties of power series I know, that when a continuous interval of the argument leads to the same function-value, then the function must be the constant function. But this seems to be different here. Is there possibly a trivial answer for this?

Remark: This is an extension of an earlier discussion of mine where I considered the construction of the formal series $f_c(x)$ based on a matrix-problem. That older question was not yet answered, so I thought, to develop it into a question on series (here with a Newton-basis) might help to get more insight.
Here are some references to older questions:
The question as problem with an infinite matrix
Learning the expression "Newton-basis"

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