> Artin had recourse to an encyclopedia, which he once consulted for help in dealing with the cockroaches that infested the Austrian barracks. At some length, the article described a variety of technical methods, concluding finally with—Artin laughingly recalled in later years—“la caccia diretta" ("the direct hunt"). Indeed, “la caccia diretta” was the straightforward method he and his fellow infantrymen adopted.

Artin survived both war and vermin on the Italian front, and returned late in 1918 to the University of Vienna, where he remained through Easter of the following year.

@saturatedexpo To prove: $\forall q\in\mathbb{N}:\exists n,m\in\mathbb{N}_0:q=f(n,m)=\frac{(m+n)(m+n+1)}{2}+m+1$
Basecase ($q=1$): We find $n=m=0$, such that $f(0,0)=\frac{(0+0)(0+0+1)}{2}+0+1=0+1=1=q$
Induction step: $q\to q+1$
$q+1=(\frac{(m+n)(m+n+1)}{2}+m+1)+1=\left\{\begin{array}{ll}\frac{(m+1+n-1)(m+1+n-1+1)}{2}+m+2=f(m+1,n-1)&\mbox{, if }n>0\\\frac{(m+0)(m+0+1)}{2}+m+2=\frac{(m+1)(m+2)}{2}=f(m+1,0)=f(m+1,n)&\mbox{, if }n=0\end{array}\right.$
Thus, we can find for any arbitrary $q$ at least one pair $(n, m)$ which proves surjectivity.

@NaCl perhaps $$
\begin{align}
\sum^{n+1}_{k=1}{a_kb_k} &= a_{n+1}b_{n+1}+\sum^{n}_{k=1}{a_kb_k} \\
&= a_{n+1}b_{n+1}+A_nb_{n+1}+\sum^{n}_{k=1}{A_k(b_k-b_{k+1})} \\
&= a_{n+1}b_{n+1}+A_nb_{n+1} - A_{n+1} (b_{n+1}-b_{n+2})+\sum^{n+1}_{k=1}{A_k(b_k-b_{k+1})} \\
&= A_{n+1}b_{n+2} + a_{n+1}b_{n+1} - (A_{n+1} - A_n)b_{n+1}+\sum^{n+1}_{k=1}{A_k(b_k-b_{k+1})} \\
&= A_{n+1}b_{n+2} + a_{n+1}b_{n+1} - a_{n+1}b_{n+1}+\sum^{n+1}_{k=1}{A_k(b_k-b_{k+1})} \\
&= A_{n+1}b_{n+2} + \sum^{n+1}_{k=1}{A_k(b_k-b_{k+1})}