Let $\sum_{n=1}^{\infty}a_n, \sum_{n=1}^{\infty}b_n$ such that there is $n_0\in \mathbb{N}$ that for all $n_o\leq n$:
a. $$0<a_n,b_n$$
b.$$\frac{a_{n+1}}{a_n}\leq \frac{b_{n+1}}{b_n}$$
Prove: if $\sum_{n=1}^{\infty} b_n<\infty$ then $\sum_{n=1}^{\infty} a_n<\infty$
Let assume $\sum_{n=1}^{\infty} a_n=\infty$ then by the ratio test we get:
$$1\leq q\leq\frac{a_{n+1}}{a_n}\leq \frac{b_{n+1}}{b_n}$$ So
$$
0<a_n\leq a_{n+1}$$
And $a_n$ is monotonic increasing, if $a_n\not \to 0 $ then we are done, else $a_n\to 0$ then
$$0<a_n\leq0$$
Contradiction.
Is it correct?
Best Answer
The ratio test does not guarantee the ratio of terms simply because the series converges or diverges.
Since all numbers are positive, we can rewrite the inequality as $$ \frac{a_{n+1}}{b_{n+1}}\le\frac{a_n}{b_n} $$ Then, by induction, $$ a_n\le b_n\frac{a_1}{b_1} $$ and $$ \sum_{n=1}^\infty a_n\le\frac{a_1}{b_1}\sum_{n=1}^\infty b_n $$