Let $$\sum_{i=1}^\infty c_i^{(0)}$$

be a positive convergent series. Then the series: $$\sum_{i=1}^\infty c_i^{(1)}=\sum_{i=1}^\infty \ln(1+c_i^{(0)})$$

is also a positive convergent series.

Generally let take $c_i^{(N+1)}=\ln(1+c_i^{(N)})$ and define the sum: $$S^{(N)}=\sum_{i=1}^\infty c_i^{(N)}$$

What can we say about the limit:

$$\lim_{N\rightarrow\infty} S^{(N)}.$$

which is certainly exist. Is it zero? Is it nonzero? (given $\{c_i^{(0)}\}$)

  • $\begingroup$ It seems to me that conditions on $c_{i}^{(0)}$ must be very strict to provide any $c_{i}^{(N)}$ being positive $\endgroup$ – cool Aug 27 '14 at 15:35
  • $\begingroup$ $\ln(1+c)>0$ provided that $c>0$ $\endgroup$ – Leaning Aug 27 '14 at 15:36
  • $\begingroup$ @Mr.T, yes, you are right. I was fool:-) $\endgroup$ – cool Aug 27 '14 at 15:38

A start: Use Comparison. Note that if $x\ge 0$ then $0\le \ln(1+x)\le x$.

  • $\begingroup$ Then $S^{(N)}$ is strictly decreasing I think. $\endgroup$ – Leaning Aug 27 '14 at 15:38
  • 2
    $\begingroup$ Yes. If you feel in a formal mood, you can prove it by induction. $\endgroup$ – André Nicolas Aug 27 '14 at 15:41
  • $\begingroup$ Thanks for that point. But I really want a nontrivial estimation, and calculate $S^{(\infty)}$ from $c_i^{(0)}$ if possible :) $\endgroup$ – Leaning Aug 27 '14 at 15:43
  • 1
    $\begingroup$ You are welcome. About the additional detail on rate, at first sight I have nothing particularly useful. If the $c_i$ are small, one can get an easy inequality the other way, using (for small $x$) $\ln(1+x)\gt \frac{x}{2}$. That is much too weak, the rate of decay of the iterated operation is definitely not exponential. $\endgroup$ – André Nicolas Aug 27 '14 at 16:00

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