Let $\sum a_n$ be the positive termed series. Prove that if $\sum \frac{a_{n+1}}{a_n}$ converges, then so is $\sum a_n$.
My try:
Since $\sum \frac{a_{n+1}}{a_n}$ is convergent, we have $$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=0$$, then $\forall \epsilon >0$, $\exists n_0 \in N$ such that
$$\left|\frac{a_{n+1}}{a_n}-0\right|<\epsilon,\:\forall n \geq n_0$$
So we have
$$\frac{a_{n_0+1}}{a_{n_0}}<\epsilon$$
$$\frac{a_{n_0+2}}{a_{n_0+1}}<\epsilon$$
$$\frac{a_{n_0+3}}{a_{n_0+2}}<\epsilon$$
$$\frac{a_{n_0+4}}{a_{n_0+3}}<\epsilon$$
$\vdots$
$$\frac{a_{n_0+n-n_0+1}}{a_{n_0+n-n_0}}<\epsilon$$
Multiplying all we get
$$\frac{a_{n+1}}{a_{n_0}}<\epsilon^{n-n_0+1}$$
$\implies$
$$a_{n+1}<a_{n_0} \epsilon^n \epsilon^{1-n_0}$$
Now if we choose $\epsilon <1$, by comparision test we get $\sum a_n$ convergent. But how about when $\epsilon >1$?
Best Answer
The reasoning is correct, only that the $\epsilon$ you have is from the definition of the limit (which you already know exists and is true for any value of $\epsilon$), not from the convergence of the series that you want to be convergent (which must meet for all $\epsilon$), as they mention in a comment, just take the value that suits you best, in this case you need $\epsilon$ to be some value between zero and one, so that the series that is formed on the right is convergent. If it is still not clear, tell me and I will try to give you a more detailed explanation.