Help identify this series I'm obtaining the following series as an analytical solution to a problem using a differential time $dt$
$$h(t)  = \lim_{n\rightarrow\infty}_{m\rightarrow-\infty}_{\Delta t\rightarrow 0}\sum_{i=m}^n f(t-i\ \Delta t)\ g(i\ \Delta t)\ \Delta t$$
$f(t)$ is a function defined completely in $\mathbb{R}$.
I'm deriving expressions from scratch, and got stuck at this step. Does this series seem familiar to you in any way? Is there a  way to lose the limit and summation expressions? I'd appreciate small hints.
Some background: water resources
It is based on the method of constructing arbitrary hydrographs from unit hyetographs in water resources. Basically, you have a function $i(t)$ that defines the amount of rainfall over a basin, and $u(t)$ that defines the water discharge through an exit point. Note that both are functions of time. $u(t)$ is also a functional of $i(t)$. A unit hyetograph is a function of $t$ defined through:
$$i_u(t) = \left\{
  \begin{array}{l l}
    0 & \quad t<0 \ \ {\rm or}\ \ t>\Delta t\\
    1 & \quad t>0\ \  {\rm and}\ \ t<\Delta t 
  \end{array} \right.
$$
The corresponding hydrograph is obtained empirically. Just suppose that there is a function $u_u(t)$ that corresponds to this unit hyetograph. Note that the only characteristic of $i_u(t)$ is $\Delta t$.
Now that we have defined a unit hyetograph $i_u(t)$, we want to construct $u(t)$ from arbitrary $i(t)$ by superposition, which explains why $u(t)$ is defined as a functional of $i(t)$ and $u_u(t)$. 
Below is a visual example (Implemented in MATLAB):

After Vincent Tjeng's answer, I've considered taking the convolution of both functions. I've used the conv function from MATLAB. The result can be seen in the following figure. I've taken $\Delta t=0.25$ to demonstrate how the series expression converges to the convolution:

EDIT: Non-calculus example on the subject:



*

*The unit hyetograph and the corresponding unit hydrograph are given.

*An arbitrary hyetograph is given. To determine the corresponding hydrograph, the unit hydrograph is multiplied by 2, offset by $t_r$. This is equal to $u(t)=u_u(t)+2\ u_u(t-t_r)$. The series above is obtained in a similar way. 

*The two steps are summed together.

*The resulting hydrograph.


Some remarks: This example (which follows the convention used by all engineers), is defined over a discrete timestep $\Delta t = t_r$. However, I now realize that while I am trying to obtain a general mathematical solution, I am ending up with a system where the $t_r$ is differential. Furthermore, I now realize that the expression I derived above assumues $i_u$, the unit hydrograph to be equal to the Dirac Delta function, which is not the case in practice. Nevertheless, I find the following paragraph in Wikipedia:

An instantaneous unit hydrograph is a further refinement of the
  concept; for an IUH, the input rainfall is assumed to all take place
  at a discrete point in time (obviously, this isn't the case for actual
  rainstorms). Making this assumption can greatly simplify the analysis
  involved in constructing a unit hydrograph, and it is necessary for
  the creation of a geomorphologic instantaneous unit hydrograph.

It seems that I have successfully reinvented the wheel.
 A: This looks to me like a convolution of the two functions $f(t), g(t)$.
$$h(t)  = \lim_{n\rightarrow\infty}_{m\rightarrow-\infty}_{\Delta t\rightarrow 0}\sum_{i=m}^n f(t-i\ \Delta t)\ g(i\ \Delta t)\ \Delta t=\int^\infty_{-\infty}f(t-\tau)g(\tau)\, d\tau$$
Using the Hyetograph as $f(t)$, and defining $g(t)$ as $$g(t)=\begin{cases} 0 & t<0, t>1 \\ 1 & 0 \le t \le 1 \end{cases}$$
Convolving your Unit Hyetograph gives me a graph similar to yours:

However, convolving your Arbitrary Hyetograph gives me this constructed Hydrograph instead, and I'm wondering whether I understood your question wrongly.

My arbitrary Hyetograph as I define it is as follows:
$$f(t)=\begin{cases} 0 & t<0, t>6 \\ 2 & 0 \le t<2 \\ 3 & 2 \le t<4 \\ 1 & 4\le t<6 \end{cases}$$
Edit: As requested, here is my code in Mathematica. I use version 9.0.
hyetograph = 
  Function[t, 
   Piecewise[{{0, t < 0}, {2, 0 <= t < 2}, {3, 2 <= t < 4}, {1, 
      4 <= t < 6}, {0, t >= 6}}]];
Plot[{hyetograph[t], g[t]}, {t, 0, 10}, AxesLabel -> {"Time", ""}, 
 PlotLegends -> {"hyetograph[t]", "g[t]"}]
g = Function[t, HeavisidePi[t - 1/2]];
hydrograph = Convolve[hyetograph[t], g[t], t, y];
Plot[hydrograph, {y, 0, 10}, AxesLabel -> {"Time", ""}, 
 PlotLegends -> {"hydrograph[t]"}]

A: You probably want the $\frac 1n$ inside the limit.  This makes $n$ into a dummy variable.  Then $\lim_{n \to \infty}\frac 1n \sum f(t+dt)=f(t+dt)$.  Unless you have a term subtracted from this, you can just let the $dt$ be zero and get $f(t)$
