利用比值比较法证明收敛级数：若∑yₙ收敛则∑xₙ收敛

Alright, let's wrap up this proof—you're already halfway there by noting that $(y_n) \to 0$ (a standard result for convergent positive-term series). Let's build on that and the given ratio condition to finish the job.

First, let's recap all our given conditions to stay grounded:

• For every natural number $n$, $x_n > 0$ and $y_n > 0$
• There exists some $N \in \Bbb N$ such that for all $n \ge N$, $\frac{x_{n+1}}{x_n} \le \frac{y_{n+1}}{y_n}$
• The series $\sum y_n$ converges (so we know $y_n \to 0$ as $n \to \infty$, which you already leveraged)

Key Step: Transform the Ratio Inequality #

Starting with the inequality we have for $n \ge N$, rearrange terms to get:
$$\frac{x_{n+1}}{y_{n+1}} \le \frac{x_n}{y_n}$$
This tells us the sequence $\left( \frac{x_n}{y_n} \right)$ is non-increasing for all $n \ge N$. Since all $x_n$ and $y_n$ are positive, this sequence is also bounded below by 0.

For non-increasing sequences bounded below, we know the sequence stays less than or equal to its value at the starting index $N$. Let's define a constant:
$$C = \frac{x_N}{y_N}$$
Since $x_N > 0$ and $y_N > 0$, $C$ is a positive real number. And because the sequence is non-increasing, for every $n \ge N$:
$$\frac{x_n}{y_n} \le C \implies x_n \le C y_n$$

Apply the Comparison Test #

Now we use the Comparison Test for positive-term series:

• We know $\sum y_n$ converges, so $\sum C y_n$ also converges (multiplying a convergent series by a positive constant doesn't change convergence).
• For all $n \ge N$, $0 < x_n \le C y_n$.

This means the tail of the series $\sum x_n$ (starting at $n=N$) converges. Adding the finite sum $\sum_{n=1}^{N-1} x_n$ (which is just a fixed finite number, so it can't make the series diverge) tells us the entire series $\sum x_n$ converges.

Quick Note on Your Progress #

You already used $(y_n) \to 0$—while that's a good observation, the real linchpin here is recognizing the ratio inequality gives us a bounded ratio of $x_n$ to $y_n$, which is what lets us apply the Comparison Test directly. That's the missing piece to finish the proof!

内容的提问来源于stack exchange，提问作者John Smith