Prove that if $∑_{n=1}^∞b_n$ converges then $∑_{n=1}^∞a_n$ converges. Let $∑_{n=1}^∞a_n$  and $∑_{n=1}^∞b_n$  be series of positive terms satisfying $a_{n+1}/a_n ≤b_{n+1}/b_n$ . Prove that if $∑_{n=1}^∞b_n$ converges then $∑_{n=1}^∞a_n$  converges.
here is my thoughts. I intend to let $c_n =\frac{a_{n+1}}{a_n}$ and $d_n = \frac {b_{n+1}}{b_n}$, from this I have $c_n \leq d_n$. Using the comparison test I have $∑_{n=1}^∞d_n$ converges implies $∑_{n=1}^∞c_n$  converges. However I'm not sure that this can help me, can't it?
 A: Prove by induction that
$$a_{n} \leq b_n \frac{a_1}{b_1}$$
(or rather, just note that $a_n/b_n$ is decreasing to establish the above).
Then use the comparison test.
A: By Cauchy test (i.e. partial sum of a convergent series form a Cauchy sequence), $$\forall\; 0\lt \varepsilon \lt 1, \exists N \in \mathbb N \; \text{such that}\; \forall n\ge N, |b_n+b_{n+1}|\lt \varepsilon$$
Since $b_k \gt 0 \;\forall k, b_{n+1}-b_n\lt |b_n+b_{n+1}|\lt \varepsilon$
Then $\dfrac{b_{n+1}}{b_n}-1 \lt \dfrac{\varepsilon}{b_n}$, so $\dfrac{a_{n+1}}{a_n}\le\dfrac{b_{n+1}}{b_n} \lt 1+\dfrac{\varepsilon}{b_n}\lt 1$ i.e. both {$a_n$} and {$b_n$} are monotonic decreasing from the Nth term onwards. 
Thus we have $a_{n+1}\lt a_n + \dfrac{a_n \varepsilon}{b_n}\lt a_n + \dfrac{a_n }{b_n} \;\forall n\ge N$. Notice that $\dfrac{a_n}{b_n} \ge \dfrac{a_{n+1}}{b_{n+1}} \forall n$, so $a_{n+1} \lt a_n + \dfrac{a_n }{b_n}\le  a_n + \dfrac{a_m}{b_m}\;\forall n\ge N, m\le n$
And therefore, $0\lt \dfrac{a_n }{b_n}\le  \dfrac{a_m}{b_m}\lt 1$  i.e. {$\dfrac{a_n }{b_n}$} is monotonic decreasing and bounded. 
We then have $a_n \le b_n \dfrac{a_N}{b_N}$. Notice that here $\dfrac{a_N}{b_N}$ is a constant with $0 \lt \dfrac{a_N}{b_N} \lt 1$, so by comparison test,  $\sum_{k=0}^{\infty} a_k$ converges.
