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There exists a sequence of strictly decreasing real positive numbers $x_n$ such that its series converges, but the quantity $$\frac{x_n^2}{x_n-x_{n-1}}$$ doesn't converge to zero?

All the famous ones aren't good, but I tried also something different:

$x_n-x_{n-1}=-x_n^2$ is a recurrence formula that describe a sequence, but I don't know if the series converges or not.

Also, setting $z_n$ the partial sum of the $x_k$, we obtain $$\frac{x_n^2}{x_n-x_{n-1}}=\frac{(z_n-z_{n-1})^2}{z_n-2z_{n-1}+z_{n-2}}$$ that looks like the discrete derivation, so probably it's sufficient to find a real crescent function with $$\lim_{x\to\infty}f(x)=c>0 \qquad \lim_{x\to\infty}\frac{(f'(x))^2}{f''(x)}\ne 0$$

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We can make things real bad with $$1,1,\frac{1}{2},\frac{1}{2},\frac{1}{4},\frac{1}{4},\frac{1}{8}\dots.$$ (This answers the question as it was put originally. The small modification below answers the revised question.)

To make the sequence strictly decreasing, replace the first $\frac{1}{2^n}$ in each pair by $\frac{1}{2^n}+\frac{1}{2^{2n}}$.

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  • $\begingroup$ ok, my bad, I forgot an hypotesis (in reality I thought it was useless): the $x_i$ are strictly decreasing $\endgroup$ – Exodd May 16 '15 at 22:35
  • $\begingroup$ I made a minor modification to make it decreasing. $\endgroup$ – André Nicolas May 16 '15 at 22:42
  • $\begingroup$ It seems to work! Thanks! Markov chains will always be grateful to you! $\endgroup$ – Exodd May 16 '15 at 22:50
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas May 16 '15 at 22:51
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How about $$ x_n = \begin{cases} 2^{-n} & n\text{ odd} \\ 2^{-(n+1)}+2^{-3(n+1)} & n\text{ even} \end{cases} $$

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