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I'm using a recursive function $f: \mathbb{N} \rightarrow \mathbb{N}$, that is defined as \begin{equation} f(n)=\lceil \log(f(n-1)) \rceil +f(n-1) \end{equation}

where $f(1)=F\in \mathbb{N}$, and $\lceil \cdot \rceil$ is the upper integer part of $\cdot$ (that is the cieling function). I would like to either find a close expression or an approximation to $f(n)$, that could help me finding the following limit (or an upper bound) \begin{equation} \lim_{N\rightarrow \infty } \sum_{n=2}^N \log(f(n)) \frac{\lceil \log(f(n)) \rceil - \lceil \log(f(n-1)) \rceil }{\lceil \log(f(n)) \rceil \cdot \lceil \log(f(n-1)) \rceil} \end{equation}

I believe that the limit should exists, since the numerator is almost always $0$ for large $n$. I will appreciate any help, or tip you could give me.

Thanks!

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Perhaps I am misreading something. Don't you have a telescoping sum there? If you do, 1/log F is an upper bound when F>1. –  The Masked Avenger Nov 18 '13 at 17:13
1  
Oh, and the sum needs f(0) to make sense, otherwise the upper bound I state is for the summation with n going from 2 to N. –  The Masked Avenger Nov 18 '13 at 17:20
    
Yes, $F >1$ is needed to make sense, $1/\lceil \log F \rceil$ is the limit, and the whole question is a joke, isn't it? –  Pietro Majer Nov 18 '13 at 22:54
    
thanks. I posted a simpler version of my problem, and I didn't realized it was so obvious. Thanks for pointing it out. –  PolinLnd Nov 19 '13 at 12:01
    
You are allowed to edit the question to make it more challenging, if you haven't solved the less simple version already. Gerhard "Have Someone Proofread The Result" Paseman, 2013.11.19 –  Gerhard Paseman Nov 19 '13 at 17:41

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