How to extend summations to fractional orders? I was going through some old stuff when I happened upon fractional summations.  They come of the form $\sum_{k=a}^bf(k)$ for $b-a\notin\mathbb Z$.  One such example is the extended harmonic numbers, which has:
$$\sum_{k=1}^n\frac1k=\sum_{k=1}^\infty\frac1k-\frac1{k+n}$$
Which got me thinking.  If we had $f(x)$ be continuous and $\lim\limits_{x\to\infty}f(x)=0$, then one might perchance define fractional summations as follows:
$$\sum_{k=1}^xf(k)\equiv\sum_{k=1}^\infty f(k)-f(k+x)\forall x\notin\mathbb Z$$
For example,
$$\sum_{k=1}^x\frac1{k^s}=\sum_{k=1}^\infty\frac1{k^s}-\frac1{(k+x)^s}=\zeta(s)-\zeta(s,x+1),s\ne1,s>0$$

This got me thinking:

What other ways can we extend summations to fractional orders?
Anything to worry about the above definition?
Where would these fractional summations appear?


Related: Indefinite summation
 A: I loved your proposal, because I had the same observation that you made when I studied the Generalized Zeta function. Let's go to the point:

What other ways can we extend summations to fractional orders?

There is another way to extend it using the Fundamental theorem of discrete calculus. But using this formula does not make much sense since it is very difficult to find the telescopic sum (I've been trying for more than 10 years to find one telescopic sum for the Riemann Zeta Function, for example...).
Instead, I offer you a demonstration of the formula you pose, using it for any complex number. 
Proof. Taking into account the following property:
$$ \quad\quad \sum\limits_{k=1}^{n} f(k) = \sum\limits_{k=1}^{z} f(k) + \sum\limits_{k=z+1}^{n} f(k)$$
which should be valid for integers, it is also valid for $z\in\mathbb{C}$ (using the Fundamental theorem of discrete calculus). Changing the variable $k$, without loss of generality:
$$ u=k-z \iff k=u+z$$
$$\quad\therefore\quad
\left\{\begin{aligned}
k_1 &=n & \iff& \quad u_1=n-z \\
k_0 &=z+1 & \iff& \quad u_0=1
\end{aligned}\right. $$
you can arrive to (using the Fundamental theorem of discrete calculus again):
$$ \sum\limits_{k=z+1}^{n} f(k) = \sum\limits_{u=1}^{n-z} f(u+z) $$
and finally:
$$\begin{aligned}
\sum\limits_{k=1}^{n} f(k) &= \sum\limits_{k=1}^{z} f(k) + \sum\limits_{u=1}^{n-z} f(u+z) \\
\sum\limits_{k=1}^{n} f(k) - \sum\limits_{u=1}^{n-z} f(u+z) &= \sum\limits_{k=1}^{z} f(k)
\end{aligned}$$
Then the $n\to\infty$ limit can be taken:
$$\begin{aligned}
\sum\limits_{k=1}^{z} f(k) &= \lim_{n\to\infty} \sum\limits_{k=1}^{n} f(k) - \sum\limits_{u=1}^{n-z} f(u+z) \\
 &= \sum\limits_{k=1}^{\infty} f(k) - \sum\limits_{u=1}^{\infty} f(u+z)
\end{aligned}$$
But both sums starts at $1$. Therefore, you can abuse the notation of index $u$, changing it to $k$:
$$ \forall z\in\mathbb{C},\quad \sum\limits_{k=1}^{z} f(k) = \sum\limits_{k=1}^{\infty} f(k) - f(k+z)$$

Where would these fractional summations appear?

In effect, there are several papers with this formula and interesting uses of it. One of them is How to Add a Non-Integer Number of Terms, and How to Produce Unusual Infinite Summations. There appear ingenious ways to evaluate infinite series with sums of fractional indexes and vice versa.

Anything to worry about the above definition?



*

*$\lim_{x\to\infty} f(x) = 0$ is a necessary but not sufficient condition to ensure the convergence of the series. Be very careful with that.

*In the case that the series diverges, the formula defines an Analytic continuation of it.

