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Fix $0 < y <1$. Assume that $n$ is a large positive integer. Let $H(z):= \sum_{i=1}^n \ln(z^i+i-1)$. Is it true that $\Re(H(e^{\xi+i\theta}))$ has asymptotically (for large $n$) a peak at $\theta=0$, where $\xi=\frac{\ln(n)+\ln(\alpha)}{\alpha n}$ and $\alpha$ is asymptotically (for large $n$) given by $\alpha=y+O(y/\ln(n)^2)$? Actually $\tilde{z}:=e^\xi$ is the solution of $S'(z)=0$, with $S(z):=\sum_{i=1}^n \ln(z^i+i-1)-\left( \frac{n(n+1)}{2}(1-y^2)+1\right) \ln(z)$. This corresponds to the Saddle point I use in the analysis of the Sum of positions of records in random permutations. I can provide the details to the interested reader.

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This is not a question. – Lee Mosher Dec 12 '12 at 14:46
I've voted to reopen, see also the conversation on meta:… – David Roberts Dec 13 '12 at 3:03
I also voted to reopen, but ask the OP to provide some context for the problem and indicate whether the OP has a proof or is asking whether the statement is true. – Bill Johnson Dec 13 '12 at 4:13

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