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$\sum_{n=1}^\infty a_n,\sum_{n=1}^\infty b_n$ with $a_n, b_n >0 $ such that $\frac{a_{n+1}}{a_n} <= \frac{b_{n+1}}{b_n}, n>=\mathrm{some\space integer}$.

Suppose $ \sum_{n=1}^\infty b_n$ converges,then $\sum_{n=1}^\infty a_n$ converges.


My solution: Since $\sum_{n=1}^\infty b_n$ converges, by ratio test $\lim_{n\to\infty} |\frac {b_{n+1}}{b_n}|<1$.

Then for large $n$, we know $\frac{a_{n+1}}{a_n} \leq \frac{b_{n+1}}{b_n}$.Thus, $$\lim_{n\to\infty} |\frac {a_{n+1}}{a_n}| \leq \lim_{n\to\infty} |\frac {b_{n+1}}{b_n}| <1$$ So by ratio test again, $\sum_{n=1}^\infty a_n$ converges

My question : Is my solution correct? thanks!

If it is not, Can you please suggest other way of doing it?thanks!

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Your solution is not correct. Convergent sequences do not necessarily have ratio limits less than $1$. For instance,

$$ \sum_{n \geq 1} \frac{1}{n^2} $$

is a reasonable, convergent sequence. But the ratio of consecutive terms approaches $1$ (and so doesn't converge to a number less than $1$).

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  • $\begingroup$ Can guide me for this question? thx, what is ur way of doing? $\endgroup$ – UnusualSkill Apr 25 '15 at 7:48
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By your hypothesis, the sequence $(a_n/b_n)$ is monotonically decreasing and bounded below (by $0$). Hence it is bounded, so $\exists M > 0$ such that $$ a_n \leq M b_n $$ for $n$ large enough. Now the comparison test applies.

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