# Limit of sum of series

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 $\{c_i^{(0)}\}$)

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

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

• Then $S^{(N)}$ is strictly decreasing I think. – Leaning Aug 27 '14 at 15:38
• Yes. If you feel in a formal mood, you can prove it by induction. – André Nicolas Aug 27 '14 at 15:41
• Thanks for that point. But I really want a nontrivial estimation, and calculate $S^{(\infty)}$ from $c_i^{(0)}$ if possible :) – Leaning Aug 27 '14 at 15:43
• 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. – André Nicolas Aug 27 '14 at 16:00