# Convergence of $\infty$-accelerating rationals

Suppose $f(1,i)$ is a sequence of rationals above 1. Let $f(k+1,n)=f(k,n+1)-f(k,n)$.

If $f(k,n)$ is for all $k>0$, an increasing function in n, must $\sum_{n>1} 1/f(1,n)<\infty$?

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Is there any particular reason why you assume $f(1,i)$ are rational? – Willie Wong Dec 1 '11 at 12:38
no, reals would have same outcome i suppose – julenøtt Dec 1 '11 at 12:39

## 1 Answer

I assume when you say increasing you mean strictly increasing. Else $f(1,i) = 2$ is a counterexample.

$f(3,n)$ and $f(2,n)$ are increasing in $n$, so we have that $f(3,n) > f(3,1) = f(2,2) - f(2,1) > 0$. So $f(2,n) > f(2,1) + (n-1) f(3,1)$, and so

$$f(1,n) \geq f(1,1) + (n-1) f(2,1) + \frac{(n-1)(n-2)}{2} f(3,1) \geq c (n-2)^2$$

By comparison $\sum_3^{\infty} \left(f(1,n)\right)^{-1} \leq \frac{\pi^2}{6c}$ converges.

(Note that your operation getting $f(k+1,\cdot)$ from $f(k,\cdot)$ is just taking the discrete derivative, so intuitively if the second derivative is everywhere positive, the function must grow at least quadratically, so the reciprocal series must converge.)

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