= Solution
A real <sequence> $x_n$ converges to $l$ if, for every $\varepsilon>0$, there is an <integer> $N$ such that $|x_n-l|<\varepsilon$ for every $n\ge N$. A <convergent series> $\sum_{n=1}^\infty x_n$ is one whose partial sums $s_N=\sum_{n=1}^Nx_n$ converge to a finite real limit.
If $s_N\to s$, then $x_N=s_N-s_{N-1}\to s-s=0$. This necessary condition is the <term test for divergence>. Its converse fails: the <harmonic series> has terms $1/n\to0$ but diverges. For a direct proof, each block from $n=2^{j-1}+1$ to $2^j$ contains $2^{j-1}$ terms each at least $2^{-j}$, so contributes at least $1/2$. Infinitely many such blocks make the partial sums unbounded.
For the positive <sequences> in the final request, the inequality makes $y_n$ decreasing and bounded below by zero. The <bounded monotone sequence theorem> gives $y_n\to l\ge0$. Summing the decrements gives
$$
\frac12\sum_{n=1}^N\min(x_n,y_n)\le y_1-y_{N+1}\le y_1.
$$
Suppose $l>0$. Since $y_n\ge l$, this implies
$$
\sum_{n=1}^N\min(x_n,l)\le2y_1\quad\text{for every }N.
$$
But <clipping preserves divergence of a positive series>: if infinitely many $x_n\ge l$, the clipped series has infinitely many terms equal to $l$; otherwise its tail equals the divergent tail of $\sum x_n$. Either case contradicts the displayed bound. Hence the only possible limit is
$$
\boxed{y_n\longrightarrow0.}
$$
This proves the <minimum-decrement convergence criterion> without assuming that the divergent positive <sequence> $x_n$ itself tends to zero.
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