Interpreting summation with fractional part? This summation is obvious if k is even.
$$\sum_{n=1}^{k/2} f(n)$$
How should this summation be interpreted if k is odd?
(update)
The context is deciding and writing a proof for the assertion that
$$\sum_{n=1}^{k} f(n)$$
and
$$\sum_{n=1}^{k/2} f(n) + \sum_{n=k/2}^{k} f(n)$$
are equivalent statements.  Of course for even k, the meaning is obvious. I suspect the proof will be "proof by cases" showing that however k/2 is interpreted, the fact that it's the upper bound in the first term and the lower bound in the second term will make it equivalent to the first equation.
But, I want to make sure that I understand what the normal (if any) interpretation of k/2 in the bounds is for non-even values of k.
 A: I've changed my mind
$\sum_{i=0}^k \ne \sum_{i=0}^{k/2} + \sum_{i=k/2}^k$.
If $k$ is even the $k/2$ term is added twice in the sum on the RHS.
If $k$ is odd then the RHS second sum starts on a fraction and every term is a fraction.
To correct this the following is always true:
$\sum_{i=0}^k \ne \sum_{i=0}^{\lfloor k/2 \rfloor} + \sum_{\lfloor k/2 \rfloor + 1}^k$.
That should be clear.
====somewhat wrong answer below ====
it has to be agreed upon by convention that if $m \not \in \mathbb N$ what $\sum_{i=0}^m$ means. As $i$ will never equal $m$. 
I think it is standard convention that $\sum_{i=k}^m$ actually means $\sum_{i \ge k;i++; i \le m}$.  Or in other words $\sum_{i=k}^m$ = $\sum_{i=\lceil k \rceil}^{\lfloor m \rfloor}$
Actually if $k$ is even that is NOT true as $f(k/2)$ is added twice.  If $k$ is odd it IS  true as $\sum_{i=0}^{k/2} = \sum_{i=0}^{\lfloor k/2 \rfloor}$ and $\sum_{i=k/2}^{k} = \sum_{i = \lceil k/2 \rceil}^{k}$.  If $k$ is even $\lfloor k/2 \rfloor =  \lceil k/2 \rceil$ and the term is counted twice.  If $k$ is odd it is not.
So the statement is true if $k$ is ODD and it is FALSE if $k$ is even.
A: The question you are asking is far from trivial and goes to the core of discrete mathematics,
thus deserving some dedicated space here: I'll try and summarize some major points.
The usual interpretation of such a sum  is through  indefinte sum.
Consider that, given $f(n)$, you know a $F(n)$ such that
$$
f(n) = F(n + 1) - F(n) = \Delta _{\,n} F(n) = \Delta _{\,n} \left( {F(n) + c} \right)
$$
Then $F(n)$ is the "anti-delta" or "Discrete Primitive" of $f(n)$
$$
F(n) + c = \Delta _{\,n} ^{ - 1} f(n) = \sum\nolimits_{\,n} {f(n)} 
$$
where $c$ is a constant.
Let's introduce this specific definition of summation
$$
\eqalign{
  & \sum\nolimits_{\,k\, = \,m}^{\;n} {f(k)} \quad \left| {\;{\rm integer}\,n,m} \right.\quad  = F\left( n \right) - F\left( m \right) =   \cr 
  &  = \left\{ {\matrix{
   {f(m) + f(m + 1) +  \cdots  + f(n - 1)} & {\left| {\;m < n} \right.}  \cr 
   0 & {\left| {\;m = n} \right.}  \cr 
   { - \left( {f(n) + f(n + 1) +  \cdots  + f(m - 1)} \right)} & {\left| {\;n < m} \right.}  \cr 
 } } \right.\;\; =   \cr 
  &  = \left\{ {\matrix{
   {\sum\limits_{m\, \le \,k\, < \,n} {f(k)}  = \sum\limits_{m\, \le \,k\, \le \,n - 1} {f(k)} } & {\left| {\;m < n} \right.}  \cr 
   0 & {\left| {\;m = n} \right.}  \cr 
   { - \sum\limits_{n\, \le \,k\, < \,m} {f(k)}  =  - \sum\limits_{n\, \le \,k\, \le \,m - 1} {f(k)} } & {\left| {\;n < m} \right.}  \cr 
 } } \right.\;\; \cr} 
$$
and we can see that
$$
\eqalign{
  & \sum\nolimits_{\,k\, = \,m}^{\;n} {f(k)}  =  - \sum\nolimits_{\,k\, = \,n}^{\;m} {f(k)}  =   \cr 
  &  = \sum\nolimits_{\,k\, = \,m}^{\;q} {f(k)}  + \sum\nolimits_{\,k\, = \,q}^{\;n} {f(k)}  = \sum\nolimits_{\,k\, = \,q}^{\;n} {f(k)}  - \sum\nolimits_{\,k\, = \,q}^{\;m} {f(k)} \quad \left| {\;{\rm integer}\,n,m,q} \right. \cr} 
$$
So, $F(n)$ exists and is defined, apart from a constant, over the definition domain of $f(n)$.
Now, if the domain of definition of $F$ extends to the reals, it comes natural to define
$$
\sum\nolimits_{k\, = \,m}^{\;x} {f(k)}  = F(x) - F(m) = \left( {F(x) + c(x)} \right) - \left( {F(m) + c(m)} \right)
$$
where, this time, we put $c(x)$ as it might actually be any periodic function of $x$ with period $1$.
If $f$ can also be extended to reals, it will be
$$
\sum\nolimits_{k\, = \,a}^{\;x} {f(k)} \quad \left| {\;\;a,x \in \;\;\mathbb{R}} \right.\quad  = F(x) - F(a) 
$$
and
$$
\sum\nolimits_{k\, = \,x}^{\;x + 1} {f(k)}  = F(x + 1) - F(x) = \Delta _{\,x} F(x) = f(x)
$$
Let's take an example to fix the concept.
$$
\begin{gathered}
  \sum\nolimits_{\,x} {\frac{1}
{x}} \, = \psi \left( x \right) + c = \frac{{\Gamma '(x)}}
{{\Gamma (x)}} + c\quad  \Rightarrow  \hfill \\
   \Rightarrow \quad \sum\limits_{1\, \leqslant \,k\, \leqslant \,\,3/2} {\frac{1}
{k}}  = \sum\nolimits_{\,k\, = \,1}^{\;5/2} {\frac{1}
{k}}  = \psi \left( {5/2} \right) - \psi \left( 1 \right) = \frac{8}
{3} - 2\ln (2) \hfill \\ 
\end{gathered} 
$$
but we can also legitimally write
$$
\sum\nolimits_{\,x} {\frac{1}
{x}} \, = \psi \left( x \right) + \sin (2\pi x)
$$
or
$$
\sum\nolimits_{\,x} {\frac{1}
{x}} \, = \psi \left( x \right) + \frac{1}
{{1 + x - \left\lfloor x \right\rfloor }}
$$
which implies that the value we gave to the summation above could actually be whatever else,
unless we agree to exclude the periodic component, and keep only the "smoother" component as it is normally done.
Many functions - but not all - admit a Discrete Primitive, i.e. a function $F(x)$ such that $f(x)=F(x+1)-F(x)$
for all the $x$ in the definition domain of $f$.
In conclusion, the standard answer to your question is
$$
\sum\limits_{1\, \leqslant \,k\, \leqslant \,n/2} {f(k)}  = F(1 + n/2) - F(1)
$$
Concerning an intuitive blick at "what is making up the total" we have that
$$
\sum\nolimits_{k\, = \,m}^{\;x} {f(k)}  = \sum\nolimits_{k\, = \,m}^{\;\left\lfloor x \right\rfloor } {f(k)}  + \sum\nolimits_{k\, = \,\left\lfloor x \right\rfloor }^{\;x} {f(k)}  = \sum\nolimits_{k\, = \,m}^{\;\left\lfloor x \right\rfloor } {f(k)}  + \sum\nolimits_{k\, = \,0}^{\;\left\{ x \right\}} {f(x + k)} 
$$
however, when the limits are $x$ and $x+m$ we obtain
$$
\sum\nolimits_{k\, = \,x}^{\;x + m} {f(k)}  = \sum\nolimits_{k\, = \,x}^{\;x + 1} {f(k)}  +  \cdots  + \sum\nolimits_{k\, = \,x + m - 1}^{\;x + m} {f(k)}  = \sum\nolimits_{j\, = \,0}^{\;m} {f(x + j)}  = \left\{ {\begin{array}{*{20}c}
   { - \sum\limits_{m\, \leqslant \,j\, \leqslant \, - 1} {f(x + j)}  =  - \sum\limits_{0\, \leqslant \,l\, \leqslant \, - m - 1} {f(x - l + 1)} } & {m < 0}  \\
   0 & {0 = m}  \\
   {\sum\limits_{0\, \leqslant \,j\, \leqslant \,m - 1} {f(x + j)} } & {0 < m}  \\
 \end{array} } \right.
$$
If $f(x)$ is "summable on the right", that is if it exists and is finite for all the $x$ in the domain of $f$ the following limit
$$
\mathop {\lim }\limits_{m\; \to \;\infty } \sum\nolimits_{k\, = \,x}^{\;m} {f(k)}  = \sum\nolimits_{k\, = \,x}^{\;\infty } {f(k)}  = \sum\limits_{0\, \leqslant \,j\, < \infty } {f(x + j)} 
$$
then we can write
$$
\begin{gathered}
  \sum\nolimits_{k\, = \,m}^{\;x} {f(k)}  = \sum\nolimits_{k\, = \,m}^{\;\infty } {f(k)}  - \sum\nolimits_{k\, = \,x}^{\;\infty } {f(k)}  = \sum\nolimits_{j\, = \,0}^{\;\infty } {f(m + j)}  - \sum\nolimits_{j\, = \,0}^{\;\infty } {f(x + j)}  =  \hfill \\
   = \sum\nolimits_{j\, = \,0}^{\;\infty } {\left( {f(m + j) - f(x + j)} \right)}  \hfill \\ 
\end{gathered} 
$$
which is the Mueller's Formula.
It goes in an analogue way if $f(x)$ is "summable on the left", and we can summarize the conclusions therefrom as
$$
\begin{gathered}
  F(x) + c = \sum\nolimits_{\,x} {f(x)}  = \Delta _{\,x} ^{ - 1} f(x) =  \hfill \\
   = \sum\limits_{0\, \leqslant \,k} {f(x - (k + 1))} \quad \left| {\;f\;\text{summable}\,\text{left}} \right.\quad  =  \hfill \\
   =  - \sum\limits_{0\, \leqslant \,k} {f(x + k)} \quad \left| {\;f\;\text{summable}\,\text{right}} \right.\quad  =  \hfill \\
   = \sum\nolimits_{\,k\, = \,0\;}^{\,\;\left\lfloor x \right\rfloor } {f(x - (k + 1))} \quad \; + \Delta _{\,x} ^{ - 1} f(\left\{ x \right\})\quad \left| {\;f\;\text{summable}\,\text{left}\,\text{and/or}\,\text{right}} \right.\quad  =  \hfill \\
   = \sum\nolimits_{\,j\, = \,0\;}^{\,\;\left\lfloor x \right\rfloor } {f(\left\{ x \right\} + j)} \quad \;\;\quad  + \Delta _{\,x} ^{ - 1} f(\left\{ x \right\})\quad \left| {\;f\;\text{summable}\,\text{left}\,\text{and/or}\,\text{right}} \right. \hfill \
\end{gathered} 
$$
where the 3rd = 4th line can be easily derived from the precedent.
The term $\Delta _{\,x} ^{ - 1} f(\left\{ x \right\})$ is a period-1 function , which normally does not have a simple closed form
and which is cancelling an equivalent component implicit in the first term. So if taken out, leaves a non-standard $F(x)$.
A: In my opinion, this summation should be interpreted as invalid if $k$ is odd. But as the various commenters have already noted, the problem is bypassed easily enough by the use of the floor function.
The notational convention for iterations like $$\sum_{n = \textrm{start}}^{\textrm{end}} f(n)$$ is that both $\textrm{start}$ and $\textrm{end}$ are integers, that the step is $1$ and that $\textrm{start} \leq \textrm{end}$.
What happens when you need a step other than $1$? You can do something like $$\sum_{n = \textrm{start}}^{\textrm{end}} f(10n)$$ or $$\sum_{n = \textrm{start}}^{\textrm{end}} f\left(\frac{n}{10}\right).$$
The reason I bring both of these up is that technically the step is $1$ even though in a computer program you might specifically write something like step = 10 or step = 0.1.
None of these options in mathematical notation explicitly allow for $\textrm{start}$ orr $\textrm{end}$ to be anything other than integers. To assume that non-integer values are implicitly allowed, or to assume that they will be interpreted in the same particular way that you intend them to be, is a needless sacrifice of clarity, in my opinion.
Just because a couple of people tell you that it's interpreted $\textrm{end} = \lfloor \frac{k}{2} \rfloor$ for one and $\textrm{start} = \lfloor \frac{k}{2} \rfloor + 1$ for the other does not guarantee that everyone will see it that way, at first anyway. In fact, one of the answerers seems to think that if $n = \frac{k}{2}$ is not an integer, then that is the starting value but the step is still $1$, e.g., $n$ takes on these values:  $$\frac{k}{2}, \frac{k}{2} + 1, \frac{k}{2} + 2, \ldots, k - \frac{1}{2} \textrm{ or } k + \frac{1}{2}.$$ For example, if $k = 7$, would we have $\textrm{end} = 6.5$ or $\textrm{end} = 7.5$?
We can certainly argue and argue that one choice is more valid than the other, but do you care more about being right on this issue or do you care more about getting your point across? There's just too much potential for ambiguity, and it can all be avoided by the simple use of the floor or ceiling function as needed.
