Convergence: infinite series $\sum_{n=1}^\infty a_n,\sum_{n=1}^\infty b_n$ with $a_n, b_n >0 $ such that $\frac{a_{n+1}}{a_n} \leq \frac{b_{n+1}}{b_n}, n\geq\text{some integer}$.
Suppose $ \sum_{n=1}^\infty b_n$ converges,then $\sum_{n=1}^\infty a_n$ converges.

Anyone can guide me for this question? Appreciate your help!thx!
 A: from $
\frac {a_n} {a_{n-1}} \le 
\frac {b_n} {b_{n-1}}
$
you get
$$
a_N = \frac {a_0} {b_0} \times b_0 \times 
\prod_{n = 1}^{N} \frac {a_{n}} {a_{n-1}}
\le \frac {a_0} {b_0} \times b_0 \times 
\prod_{n = 1}^{N} \frac {b_{n}} {b_{n-1}} = \frac {a_0} {b_0} \times b_N
$$
so
$$\sum b_N<\infty \implies \sum a_N<\infty
$$
A: An answer to expand on my comment.
First, there is a $n_0$ such that $a_nb_n\neq0$ for $n\geq n_0$, otherwise your quotients are not well defined.
Assume $n_0=1$, otherwise you start the summation from $n_0$, and you can rename the variable $k$ with $n=n_0+k-1$.
Assume further that $a_1=b_1=1$, otherwise you can divide all $a_n$ by $a_1$, and all $b_n$ by $b_1$, without changing convergence or divergence.
With these assumptions, you can prove by induction that $a_n\leq b_n$, hence $S_N\leq T_N$, with
$$S_N=\sum_{n=1}^N a_n$$
$$T_N=\sum_{n=1}^N a_n$$
Then, since $\sum_{n=0}^\infty b_n$ converges, that means (it's the definition) that the sequence $T_N$ converges, thus $S_N$ is bounded, and since it's inceasing (because the $a_n$ are positive), it's also convergent.
Start the induction: you have obviously $a_1\leq b_1$ since $a_1=b_1=1$.
Induction step:
If $0<a_n\leq b_n$, then since
$$\frac{a_{n+1}}{a_n}\leq\frac{b_{n+1}}{b_n}$$
You have then
$$a_{n+1}\leq b_{n+1}\frac{a_n}{b_n}$$
And $\frac{a_n}{b_n}\leq 1$, so $a_{n+1}\leq b_{n+1}$.
So by induction, $a_n\leq b_n$ for all $n\geq 1$, then conclude.
