I was wondering if I can calculate fractional sums (non-integer sums) with functional equations in the following manner
$$f(x):=\sum_{n=1}^xg(n)$$
$$f(x)=g(x)+f(x-1)\tag1$$
$(1)$ most certainly should be true, and in using such an equation, I've seen fractional summations calculated. (being wary that an extra constant will sneak in)
For instance,
$$f(x):=\sum_{n=1}^x\frac1n$$
$$\implies f(x)=\frac1x+f(x-1)$$
$$\implies \int_1^xf(t)dt=\int_1^x\frac1t+f(t-1)dt$$
$$\int_1^xf(t)dt=\ln(x)+\int_1^xf(t-1)dt$$
$$\implies\int_1^xf(t)-Cdt=\sum_{n=1}^x\ln(x)\tag{backwards $(1)$}$$
$$\implies\int_1^xf(t)-Cdt=\ln(\Gamma(x+1))$$
$$\implies f(x)=C+\frac{d}{dx}\ln(\Gamma(x+1))$$
$$\sum_{n=1}^x\frac1n=\gamma+\frac{d}{dx}\ln(\Gamma(x+1))$$
where we find $C$ by putting in a trivial value, like $x=1$.
So... I was wondering if this is a possibly good way to going about fractional sums?
EDIT
I also find that defining a summation this way allows me to differentiate in a neat way:
$$\frac{d}{dx}f(x)=\frac{d}{dx}\sum_{n=1}^xg(n)=C+\sum_{n=1}^xg'(n)?$$
Is that correct? Despite being a discrete type of thing, we simply assume it can be continuous and make sense of differentiating it?
It appears to pass well for easy/known sums that have closed forms and closed form derivatives.
Then my second thought.
Suppose we have a different approach utilizing infinite series:
$$p(x,k):=\sum_{n=k}^xq(n)$$
$$P:=\lim_{(x,k)\to(\infty,-\infty)}p(x,k)=\sum_{n=-\infty}^\infty q(n)$$
Let $P$ be some constant.
Since we can calculate $P$ through various methods, and numerically by adding terms, we could then manipulate things as follows,
$$p(x,k)=P-p(\infty,k)-p(x,-\infty)$$
And we can surely calculate $p(\infty,k)$ or $p(x,-\infty)$ by starting at $k$ or $x$, respectively, and approach $\infty$ term by term?
$$p(\infty,k)=q(k)+q(k+1)+q(k+2)+\dots$$
$$p(x,-\infty)=q(x)+q(x-1)+q(x-2)+\dots$$
Right? So long as $P$ is actually a finite constant and the summation doesn't diverge, we can numerically approach the problem this way?