Show that the series diverges

I don’t know how to prove that :

Let $$(a_n) \in ( \mathbb R^{+*})^\mathbb{N}$$

We assume that $$\forall n \in \mathbb N ,a_{n+1} and $$\lim(a_n)=0$$.

$$\forall n \in \mathbb N, u_n=\frac{a_n}{a_{n+1}}-1$$

Show that $$\sum u_n$$ diverges.

$$\sum u_n$$ diverges iff $$\prod (1+u_n)$$ diverges. We have that

$$\lim_{N\to\infty}\prod_{n=1}^N(1+u_n)=\lim_{N\to\infty}\prod_{n=1}^N\frac{a_n}{a_{n+1}}=\lim_{N\to\infty}\frac{a_1}{a_{N+1}}=\infty.$$

• has this fact a name? – GhostAmarth Nov 20 at 21:19
• I don´t think so. It is one of sufficient and necessary conditions of convergence of infinite products.. – Pavel R. Nov 20 at 21:46
• Alright thanks anyways – GhostAmarth Nov 20 at 22:02

Note that with $$m \geqslant k > n$$ we have $$a_{k+1} \leqslant a_{n+1}$$ and

$$\left|\sum_{k=n+1}^m u_k \right| = \sum_{k=n+1}^m \frac{a_k - a_{k+1}}{a_{k+1}} \geqslant \frac{1}{a_{n+1}} \sum_{k=n+1}^m(a_k - a_{k+1}) = \frac{a_{n+1} - a_{m+1}}{a_{n+1}} = 1 - \frac{a_{m+1}}{a_{n+1}}$$

Since $$a_{m+1} \to 0$$ we have $$1 - a_{m+1}/a_{n+1} \to 1$$ as $$m \to \infty$$ for any fixed $$n$$.

Hence, there exists a sufficiently large $$m > n$$ such that the RHS exceeds $$1/2$$ and the Cauchy criterion for convergence is violated.