= Solution
The definition of a <convergent sequence> is
$$
\boxed{\forall\epsilon>0\ \exists N\ \forall n\ge N:\ |u_n-l|<\epsilon.}
$$
To prove the product rule, convergence first gives an eventual bound $|u_n|\le |l|+1=:M$. Then
$$
|u_nv_n-lk|=|u_n(v_n-k)+k(u_n-l)|\le M|v_n-k|+|k||u_n-l|.
$$
Choose $n$ sufficiently large that $|v_n-k|<\epsilon/(2M)$ and $|u_n-l|<\epsilon/[2(|k|+1)]$, as well as the bound on $u_n$. The right side is less than $\epsilon$, proving \b[$u_nv_n\to lk$] directly from the <limit of a sequence> definition.
If $l\ne0$, eventually $|u_n-l|<|l|/2$, so the <reverse triangle inequality> gives $|u_n|>|l|/2$. Consequently
$$
\left|\frac1{u_n}-\frac1l\right|=\frac{|u_n-l|}{|u_n||l|}\le\frac{2|u_n-l|}{|l|^2}\longrightarrow0.
$$
Thus \b[$1/u_n\to1/l$]. The condition $u_n\ne0$ ensures the reciprocal sequence is defined at every index; the nonzero limit supplies the eventual uniform bound away from zero. Finally $\boxed{u_n=1/(n+1)\to0}$ has every term nonzero, showing that this termwise property alone does not prevent a zero limit.
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