Show that $$\sum_{k=2}^n (-1)^k \sum_{j=1}^{k-1} a_ja_{k-j} = \Bigl(\sum_{k=1}^n (-1)^ka_k\Bigr)^2 - \sum_{k=n+1}^{2n} (-1)^k \sum_{j=k-n}^n a_ja_{k-j}$$ Source: Determine whether $\sum_{k=1}^\infty\sum_{j=1}^{k-1}\frac{(-1)^k}{[j(k-j)]^p}$ converges.
I'm not quire sure how to prove this. Does anybody have an idea or suggestion? Thanks.
Since I'm "the author", I think I have to explain it myself. Start with $$\left(\sum_{k=1}^{n}b_k\right)^2=\sum_{i=1}^{n}\sum_{j=1}^{n}b_i b_j=\sum_{\substack{1\leqslant i,j\leqslant n\\i+j\leqslant n}}b_i b_j+\sum_{\substack{1\leqslant i,j\leqslant n\\i+j>n}}b_i b_j.$$ In the first sum on the RHS, we collect the terms with fixed value of $k:=i+j$ (which runs from $1+1=2$ to $n$); the possible values of $j$ are from $1$ to $k-1$, and for a fixed value of $j$, the (single possible) value of $i$ is then equal to $k-j$. Thus, the first sum is equal to $\sum_{k=2}^{n}\sum_{j=1}^{k-1}b_j b_{k-j}$.
Similarly, if we fix $k=i+j$ in the second sum (the possible values of $k$ run from $n+1$ to $2n$), then the smallest possible value of $j$ is $k-n$ (and the greatest one is $n$ of course), and the single possible value of $i$ (for a fixed $j$) is then again $k-j$.
This explains $$\left(\sum_{k=1}^{n}b_k\right)^2=\sum_{k=2}^{n}\sum_{j=1}^{k-1}b_j b_{k-j}+\sum_{k=n+1}^{2n}\sum_{j=k-n}^{n}b_j b_{k-j}.$$ Now put $b_k=(-1)^k a_k$.