Whilst reading through and attempting some other questions here on Math.SE, I hit upon an idea for testing whether a series is convergent, and I'd like to know:
- If it is a valid test
- How practical this test may be to use?
I'll summarise it briefly here: $$\text{For a function}\space f(n), \text{it's infinite sum }\sum_{n=0}^{\infty}{f(n)}\space\text{is convergent if}$$ $$\exists\rho>0\text{ such that} \space\forall n>\rho; f(n)\ge10f(n+1) $$ In essence, the thought process becomes apparent when we examine the trivial $$\sum_{n=0}^{\infty}{10^{-n}}$$ For which $f(n)=10f(n+1)$ and we have $f(1)=0.1, f(2)=0.01, f(3)=0.001$, and so forth, with obvious convergence to $\frac{10}{9}$.
As a more complicated example, I tried $$\sum_{n=0}^{\infty}{\frac{n^3}{n!}}$$ and upon use of my test I generated: $$\frac{n^3}{n!}\ge\frac{10(n+1)^3}{(n+1)!}\to n^3(n+1)\ge 10(n+1)^3\to(n+1)[n^3-10(n+1)^2]\ge0$$ Iterating $n_{t+1}=\sqrt[3]{10(n_t+1)^2}$ gives us $\rho\approx11.77122803$, and testing $n=12$ confirms the inequality is satisfied for $n>\rho$, thus there is convergence.
I'd like to know your thoughts on this method.
As stated, the test is incorrect without assuming that $f(n)$ is nonnegative. A counterexample is $f(n) = -1$ for all $n$.
With the additional assumption, this is a weak form of the ratio test. Note that the ratio test implies that $10$ may be replaced by $1 + \epsilon$ for any $\epsilon > 0$.