Different summation notation? In school, I learned that 
$$\sum_{k=1}^{n}f(k)=f(1)+f(2)+f(3)+ ... +f(n)$$
But in some physics or mathematics book, I saw this kind of representation such like:
$$\lim_{\Delta x\to 0} \sum_{j} f(x_j) \Delta x$$
What is the difference between these kinds of summation notation?
 A: The notion of
$$\lim_{\Delta x\to0^+}\sum_jf(x_j)\Delta x$$
comes from the Riemann sum of an integral.  Another, perhaps easier to understand form:
$$\lim_{\Delta x\to0^+}\sum_{k=1}^{\lfloor1/\Delta x\rfloor}f(x_0+k\Delta x)\Delta x\\\text{or}\\\lim_{n\to\infty}\sum_{k=1}^n\frac{f\left(x_0+\frac kn\right)}n$$
An integral is basically the area under a function.  For example, the area under $y=x$ from $x=0$ to $1$ is shown in this graph.  As you can see, the area is simply a triangle, so we know that it is given by $\frac12bh=\frac12(1)(1)=\frac12$.  That is, the integral of $y=x$ from $0$ to $1$ is equal to $\frac12$.
However, not all integrals can be done with geometry.  This is where the calculus comes in.  The basic idea of a Riemann sum is given in this graph.

The above rectangles add up to what is approximately the integral of $2^x$ from $x=0$ to $x=1$.  We can choose to use more rectangles to get a better approximation of the integral:

Note that the area of a rectangle is base times height.  Here, the base is $1/n$, or $\Delta x$.  The height is given by $f(x_j)$, or $f(x_0+k\Delta x)$, or $f\left(x_0+\frac kn\right)$, all of which mean the same thing.
And then as you take infinitely many rectangles, you will get the exact value for the integral.
A: If you mean why limits of the index are not written, it is because when this happen they are clear from the context.
In particular the formula you have shown can be used to express Riemann sum or upper or lower Darboux sum of $f$ over some interval, depending on how $x_j$'s are chosen. Now in this case the limit is taken as the mesh of the partition of the interval approaches $0$ (not as $\Delta x\rightarrow0$), under the summation $f(x_j)$ is multiplied by $\Delta x_j$ (not $\Delta x$) and $j$ is and integer that runs from $1$ to $n$, where $n$ is the number of the subintervals into which the interval has been partitioned.
A: The    notation 
\begin{align*}
\lim_{\Delta x\to 0} \sum_{j} f(x_j) \Delta x\tag{1}
\end{align*}
is typically used for Riemann sums when introducing integration and adresses more concepts than only finite summation $$\sum_{k=1}^{n}f(k)=f(1)+f(2)+f(3)+ ... +f(n)$$

  
*
  
*Finite sum: One part of (1) is a finite sum given in a somewhat sloppy notation (but quite usual if the context is known):
  \begin{align*}
\sum_{j} f(x_j) \Delta x\tag{2}
\end{align*}
  Here we consider $x$ to be a shorthand of an $(n+1)$-tuple $x=(x_0,x_1,\ldots,x_n)$ with the property
  \begin{align*}
x_0<x_1<\cdots<x_n
\end{align*}
  The symbol $\Delta x$ (aka difference) means the difference of two consecutive elements $x_j-x_{j-1}$ of $x$ with $1\leq j\leq n$. Another more precise notation for (2)    is
  \begin{align*}
\sum_{j} f(x_j) \Delta x&=\sum_{j=1}^n f(x_j) \Delta x_j\\
&=\color{blue}{\sum_{j=1}^n f(x_j) (x_j-x_{j-1})}\tag{3}
\end{align*}
The finite sum (3)  corresponds  to   the finite sum in your question.

But there is another twist, a specific limit of this sum.

  
*
  
*Limit of sum: The notion $$\Delta x \rightarrow 0$$ addresses a specific kind of limit built from a sequence of $(n+1)$-tuples
  $$x^{(n)}=(x_0^{(n)},x_1^{(n)},\ldots,x^{(n)}_n)$$  with $n\geq 1$ requiring that the maximum absolute difference of $\Delta x$ approaches zero when taking the limit.
\begin{align*}
\Delta x\to 0\ &\ \widehat{=}\lim_{n\rightarrow \infty}\Delta x^{(n)}=0\\
&\ \widehat{=}\lim_{n\rightarrow \infty}\left(\max_{1\leq j\leq n}\left(x_j^{(n)}-x_{j-1}^{(n)}\right)\right)=0
\end{align*}

This way we require a nice, convergent behaviour of the Riemann sums necessary for the existence of the Riemann integral which is based upon this principle.
