If $\sum a_n=0$, what can we say about $\sum n a_n$? Suppose we know that 
$$\sum_{n=0}^{k}(-1)^na_n=0$$
for even $k$, and that $\{a_n\}$ is decreasing after $n/2$. What conditions can we set on $a_n$ to make 
$$\sum_{n=0}^k(-1)^na_n(-n)$$
either positive or negative?
 A: I reckon you meant $\{a_n\}$ is decreasing after $\frac{k}{2}$.
In that case, let $\displaystyle a_n=e^{-\left(n-\left(\frac{k+1}{2}\right)\right)^2}$, then $\displaystyle \sum_{n=0}^{k}(-1)^na_n\approx0$ and $\displaystyle \sum_{n=0}^k(-1)^na_n(-n)$ is negative when $\frac{k}{2}$ is odd and positive when $\frac{k}{2}$ is even.
A: Ok, it seems to me that the original sum is
$$S_1:=a_0-a_1+a_2-....\pm a_k=0$$
Our second sum is therefore:
$$S_2:=a_1-2a_2+3a_3-4a_4+...\pm ka_k$$
I guess that if we can find an $M$ such that $a_{i+1}/a_i=M(i+1)$, and we define $M(1)=1$,  then we can say that 
$$S_2=a_1-2\Big( M(2)a_1 \Big)+3\Big( M(3)M(2)a_1\Big)-...\pm k\Big(a_1 \prod_{i=1}^{k} M(i) \Big)$$ 
$$=a_1 \sum_{j=1}^{k}(-1)^{j+1}j\Big( \prod_{i=1}^{j}M(i)  \Big)$$
Now, we can safely say that if $\{b_n\}$ is an increasing sequence, then
$$\sum_{i=1}^k(-1)^{i+1}b_n n$$
is positive if $k$ is odd and negative otherwise.  Therefore, if we have that 
$$V(j)=\prod_{i=1}^{j}M(i)$$
is an inreasing function of $j$, then we can govern and control the positivity of $S_2$ as is evident in the example given by Maazul.  Of course, $V(j)$ will only be increasing if $\{a_n\}$ is increasing.  So this is all pretty useless.
