Understanding this summation identity I'm currently reading a book in which part of the solution to the problem involve this identity:
$$\sum_{j=i+1}^{n}j = \sum_{j=1}^{n}j-\sum_{j=1}^{i}j$$
Which I cannot derive myself. The only thing I can do with it is this:
$$\sum_{j=i+1}^{n}j = \sum_{j=1}^{n}j+i = \sum_{j=1}^{n}j + \sum_{j=1}^{i}i$$
Which seems to me completely useless. 
Any help in understanding this (as I am unaccustomed to summation manipulation in general) would be greatly appreciated. 
I know it's related to " Calculate integer summation when lower bound is a variable " but I still don't see the why. 
 A: We have 
$$\sum_{j=1}^{i}j+\sum_{j=i+1}^n j=\sum_{j=1}^n j.\tag{$1$}$$
The result you are looking for follows by subtraction.
If $(1)$ seems unclear, let us take particular numbers, say $i=7$ and $n=19$.
We have
$$\sum_{j=1}^{i}j =\sum_{j=1}^7j=1+2+\cdots+7$$
and 
$$\sum_{j=i+1}^{n}j =\sum_{j=8}^{19} j=8+9+\cdots +19.$$
If you add them, you get
$$1+2+\cdots+7+8+9+\cdots +19.$$
This is equal to
$$\sum_{j=1}^{19} j.$$ 
Remark: Exactly the same argument shows that if $a_1,a_2,\dots$ is any sequence, then
$$\sum_{j=i+1}^n a_j=\sum_{j=1}^n a_j -\sum_{j=1}^i a_j.$$
A: Note that
\begin{align}
\sum_{j=1}^{n} j & = 1 + 2 + \cdots + i + (i+1) + (i+2) + \cdots + n\\
& = \left(1 + 2 + 3 + \cdots + i \right) + \left((i+1) + (i+2) + \cdots + n \right)\\
& = \sum_{j=1}^{i} j + \sum_{j=i+1}^{n} j
\end{align}
Hence, we get that $$\sum_{j=i+1}^{n} j = \sum_{j=1}^{n} j - \sum_{j=1}^{i} j$$
A: \begin{align}
6+7+8 = \Big(1+2+3+4+5+6+7+8\Big) - \Big(1+2+3+4+5\Big)
\end{align}
That's all there is to it.  Your way of phrasing the question makes me wonder if you think there's something more to it than that.
