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I don’t know how to prove that :

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

We assume that $\forall n \in \mathbb N ,a_{n+1}<a_n$ 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.

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

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

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

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