Strong and weak Ratio test?

Question

Is this a valid test for convergence of $\sum_{n=1}^\infty a_n$ where $a_n$ are all positive? Define:

$\rho_n=a_n/a_{n+1}$

The series converges if $\rho_n>1$ for all n>N

The series diverges if $\rho_n \le 1$ for all n>N

where $N$ is some positive integer. Note this is not the same as the usual ratio test which states that the series converges if $\lim_{n \to \infty}\rho_n>1$ and diverges if $\lim_{n \to \infty}\rho_n<1$ with no conclusion for 1.

I ask this because Kummer's test has been stated as: $\rho_n=D_n a_n/a_{n+1}-D_{n+1}$ where $D_n$ is a positive term series, with convergence for $\rho_n>0$ and divergence for $\rho_n \le 0$ and $D_n$ divergent, for some $n>N$. It has also been stated in the limit form where $\lim_{n \to \infty}\rho_n<0$ and $D_n$ divergent for divergence. Substituting $D_n=1$ into Kummers test gives the above statement (no limits), along with the usual ratio test involving limits.

Answer 1

No. For example, your test predicts that $$ \sum_{n \geq 1} \frac{1}{n^2} $$ diverges.

Answer 2

Ok, thanks, now I see the problem. Kummer's $\rho_n$ is $D_n a_n/a_{n+1}-D_{n+1}$ and for convergence, there must be a $c>0$ such that $\rho_n \ge c$ which is NOT the same as $\rho_n > 0$, which I mistakenly supposed. If $a_n=1/n^2$, then $\rho_n=(1+1/n)^2$ which, although it is greater than zero for all $n\ge 1$, there is no $c>0$ that it is greater than or equal to for any $n$.