# Strong and weak Ratio test?

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.

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

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 $$n$$.