How would I derive the partial sum formula for $\sum _{ n=1 }^{ \infty }{ (-1)^{ n }\cdot n } $? I was solving a practice competitive programming question when I realized that this seems very familiar to the things I have done in a Calculus class. So, I tried to figure out the closed form partial sum for $\sum _{ n=1 }^{ m  }{ (-1)^{ n }\cdot n } $, but I realized that I am in over my head. I entered "$-1 + 2 + -3 + 4 + -5 +...$" into Wolfram Alpha, and I got that the partial sum formula for $\sum _{ n=1 }^{ m  }{ (-1)^{ n }\cdot n } = \frac { 1 }{ 4 } (2(-1)^{ m }m+(-1)^{ m }-1)$ 
I feel guilty for blindly applying a formula to solve a problem, and I have not been able to find any wikipedia article, or any other source that provides an explanation for this closed form partial sum formula. I would appreciate if someone can provide an explanation for this.

You are computing the sum of the even numbers minus that of the odd numbers.
Assuming $m=2k+1$,
$$\sum_{n=0}^k2k-\sum_{n=0}^k(2k+1)=\sum_{n=0}^k(2k-(2k+1))=-\sum_{n=0}^k1=-k-1.$$
Or assuming $m=2k$, undo the last odd term and get
$$-k-1+2k+1=k.$$

To get a closed formula, notice that
$$m=2k\to S=\frac m2,\\
m=2k+1\to S=-\frac{m+1}2,$$
which can be summarized as
$$(-1)^m\frac{m+\dfrac{1-(-1)^m}2}2$$ or
$$(-1)^m\frac{m+m\bmod2}2.$$
