# Sequence and series : Convergence

$\sum_{n=1}^\infty a_n,\sum_{n=1}^\infty b_n$ 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!

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 $1$).
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.