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I'd like to know if this proposition is true:

$$ \left\{ x_n \right\} | x_{n+1}>x_n \wedge \lim_{n \to \infty}x_n = \infty \Rightarrow $$ $$\sum_{n \geq 1} \left( 1 - \frac{x_n}{x_{n+1} } \right) = \infty$$ Can anyone help me?

Ludox
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1 Answers1

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Yes, this is true. You have $$ \sum_{k=1}^n \ln\frac{x_{k+1}}{x_k} = \ln x_{n+1} - \ln x_1 \to \infty. $$ If $(x_{k+1}/x_k)$ does not converge to $1$, then the series trivially diverges because the terms do not converge to $0$. Otherwise you have $x_{k+1}/x_k \le 2$ eventually, and so $$ \ln \frac{x_{k+1}}{x_k} \le \frac{x_{k+1}}{x_k}-1 \le 2 \left(1 - \frac{x_k}{x_{k+1}}\right), $$ implying your claim by comparison of series.

Lukas Geyer
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  • Hi! Thanks for the answer, but I am quite sure that the last inequality is false! That's because (a/b - 1 < 1- b/a ; a/b + b/a < 2 ; aa + bb < 2ab) and that's false! – Ludox Sep 29 '15 at 19:57
  • Or you can, for example, consider x_k=2 and x_k+1=3 and you obtain 1/2<=1/3 – Ludox Sep 29 '15 at 20:00
  • @Ludox: Oops, that is true, but it is easy to fix. – Lukas Geyer Sep 29 '15 at 20:26
  • This doesn't improve the answer. Now you can consider x_k=2 and x_k+1= 6 and you obtain 2<4/3 – Ludox Sep 29 '15 at 20:34
  • @Ludox: Read the statement just before the inequality. Either $x_{k+1} \le 2 x_k$ (for all sufficiently large $k$), or the series diverges by an easier argument. – Lukas Geyer Sep 29 '15 at 20:37
  • You are right, my fault, sorry! – Ludox Sep 29 '15 at 20:43