Sum with a fractional number of terms??? I was playing around on desmos with the Bernoulli polynomials which are defined as
$$B(x,m)=\sum_{n=0}^{m}\frac1{n+1}\sum_{k=0}^{n}(-1)^k{n\choose k}(x+k)^m$$
And I noticed that graphs were being displayed for fractional $m$. I guess that is to be expected, as non-integer $m$ is usually interpreted as $\lfloor m\rfloor$. But the confusing part, as you will see in the link I provided, is that desmos thinks that 
$$B(x,5/3)\neq B(x,7/4)$$
even though $\lfloor5/3\rfloor=\lfloor7/4\rfloor=1$. I am confused. Could someone explain how summation can be extended to non-integer indices? 
 A: The interpretation that you give is in fact the correct one for a sum indicated as, e.g.
$$
\sum\limits_{0\, \le \,n\, \le \,3/2} {f(n)}  = \sum\limits_{0\, \le \,n\, \le \,\left\lfloor {3/2} \right\rfloor } {f(n)}  = f(0) + f(1)
$$
But there is a definition of Sum that applies to  fractional, real, or even complex indices.
That is based on the concept of Antidelta or Indefinite Summation.
If we can establish that
$$
f(n) = F(n + 1) - F(n) = \Delta F(n)
$$
then we have that
$$
\eqalign{
  & F(n) = \Delta ^{\left( { - 1} \right)} f(n) = \sum\nolimits_{k = 0}^n {f(k)}  + c = \sum\limits_{0\, \le \,k\, \le \,n - 1} {f(k)}  + c\quad  \Rightarrow   \cr 
  &  \Rightarrow \sum\limits_{m\, \le \,k\, \le \,n - 1} {f(k)}  = \sum\nolimits_{k = m}^n {f(k)}  = F(n) - F(m) \cr} 
$$
So, if $f(z), F(z)$ exist also for real or complex $z$, then it is natural to define
$$
\sum\nolimits_{\,k = a\;}^{\,b} {f(k)}  = F(b) - F(a)
$$
For example
$$
\eqalign{
  & F(x) = {{x\left( {x - 1} \right)} \over 2}\quad  \Rightarrow \quad \Delta F(x) = {{\left( {x + 1} \right)x} \over 2} - {{x\left( {x - 1} \right)} \over 2} = x  \cr 
  & \sum\nolimits_{k = 0}^x k  = {{x\left( {x - 1} \right)} \over 2} = \left( \matrix{
  x \cr 
  2 \cr}  \right) = {{\Gamma (x + 1)} \over {\Gamma (3)\Gamma (x - 1)}} \cr} 
$$
Regarding the Bernoulli polynomials the indicated link provides 
various relations that can be used to extend their definition to non-integral indices, including
$$
\sum\nolimits_{k = 0}^x {k^s }  = {{B_{\,s + 1} (x) - B_{\,s + 1} (0)} \over {s + 1}}
$$
