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With $x_n>0, y_n>0$, $\exists N $ such that $$\forall n>N:\quad \frac{x_n}{ x_{n+1}} \leq \frac{y_n}{ y_{n+1}}$$ Want to prove:

If $\sum y_n$ converges, $\sum x_n$ converges.

I think this is about ratio test, maybe limsup of ratio, but I cannot think it out. Thank you.

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    $\begingroup$ Do you mean $\exists N $ such that $\forall n>N$? $\endgroup$ – Amihai Zivan Apr 21 '15 at 7:20
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    $\begingroup$ @Amihai Zuvan yes. sorry about this mistake $\endgroup$ – k99731 Apr 21 '15 at 7:30
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This is called the Ratio Comparison Test. You can find the proof on page 3 of these notes: http://www.westga.edu/~faucette/research/Raabe.pdf

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Let $\sum_n a_n$ and $\sum_n b_n$ be series having positive terms and satisfying $$ \frac{a_{n+1}}{a_{n}} \le \frac{b_{n+1}}{b_{n}} $$ for all sufficiently large $n$. If we rewrite the inequality as $$ \frac{a_{n+1}}{b_{n+1}} \le \frac{a_{n}}{b_{n}}, $$ we see that the sequence $(a_n/b_n)$ is monotone decreasing. Since $a_n, b_n \ge 0$, this sequence is bounded from below, and it follows that $(a_n/b_n)$ converges by the completeness of the real numbers. In particular, it is bounded from above by some positive number $M$, so $a_n \le Mb_n$ for all $n$ sufficiently large. Since $\sum_n b_n$ converges, so does $\sum_n Mb_n$, and by the Comparison tests, $\sum_n a_n$ converges too$.

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