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I came across this sequence as part of my work. Could someone indicate me the methodology I should follow to solve it? I guess it involves harmonic numbers and/or the digamma function?

I tried to express $U_{n}$ as a function of $n$, I tried expressing it as a function of $U_{n-1}$, I tried looking at $U_{n+1} - U_{n}$, all without success. I built an Excel macro to look at what the sequence looks like. With that, I can confirm that the sequence does have a limit (and different from zero, but depending from $a$ and $b$), after having tried several values for $a$ and $b$. I tried inferring the value of the limit from the Excel calculations, but it is not obvious.

Let $a$ and $b$ be natural numbers, with $1\leq a< b$
We define $U_{n}$ by:

$$ U_n := \begin{cases} \displaystyle \frac{1}{b+n} \left [ \frac{1}{a+n} + \sum_{i=0}^{n-1}\left ( U_{i} \sum_{k=n-i}^{b+n-1} \frac{1}{k} \right ) \right ] & \text{for $n\geq 1$,} \\[1ex] \displaystyle \frac{1}{ab} & \text{for $n=0$.} \end{cases} $$

I want to find $\lim_{n \to +\infty}U_{n}$.

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  • $\begingroup$ Also posted to MO, mathoverflow.net/questions/332539/… $\endgroup$ – Gerry Myerson May 27 at 0:25
  • $\begingroup$ There's no indication, Christophe, of how you came across this question, what progress you've made on it, where you get stuck, whether there's any reason to think the limit exists, whether there are any values of $a,b$ for which you can prove anything or have even done any calculations. This doesn't bother me, but apparently it does bother those who voted to close your question, so perhaps you should attend to it. Also, it's considered an abuse of the system to post the same question to two different sites without at the very least linking each post to the other. $\endgroup$ – Gerry Myerson May 27 at 0:29
  • $\begingroup$ Sorry about that, I'm new here... I tried to express $U_{n}$ as a function of n, I tried expressing it as a function of $U_{n-1}$, I tried looking at $U_{n+1} - U_{n}$, al without success. I built an Excel macro to look at what the sequence looks like. With that, I can confirm that the sequence does have a limit (and different from zero), after having tried several values for a and b. I tried inferring the value of the limit from the Excel calculations, but it is not obvious. $\endgroup$ – Christophe May 27 at 11:07
  • $\begingroup$ And I didn't know if this question was more appropriate for MO or for MSE... $\endgroup$ – Christophe May 27 at 11:16
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    $\begingroup$ All of this extra information should be edited into the body of the question, not buried in the comments. And not knowing where the question was appropriate does not respond to the complaint that you didn't notify each site of what you were doing. $\endgroup$ – Gerry Myerson May 27 at 11:54

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