Show that $\sum_{n=1}^{\infty}\left(\frac{1}{(n+1)!}\prod_{k=1}^nf(k)\right)$ diverges.

Question

Let $f:\mathbb{N}\setminus{\{0}\}\to\mathbb{N}\setminus{\{0}\}$ be an injective function. Show that

$$\sum_{n=1}^{\infty}\left(\frac{1}{(n+1)!}\prod_{k=1}^nf(k)\right)$$ diverges.

In the quotient rule I arrived at that $ \frac {a_ {n + 1}} {a_n} = \frac {f (n + 1)} {n + 2} $ with $ a_n: = \frac {1} { (n + 1)!} \prod_ {k = 1} ^ nf (k). $

Can anyone give me a suggestion? It would be a great help, thanks.

Answer 1

$f$ is injective so for any $n$, the minimum of $\prod_{k=1}^nf(k)$ is $1 \cdot2 \cdot3 \ldots n = n!$. So for each $n$ the minimum summand is $\frac{1}{n}$ which diverges.