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I just learnt character theory and I am reading the paper https://academic.oup.com/jlms/article-abstract/66/3/623/811347 which is a quite beautiful paper. Section 4.3, on p. 631 contains an inequality which confuses me for a while:
Let $P=\{\lambda \vdash n, \lambda \neq (1^n), (n)\}$ be the set of partition of the natural number $n$ except for the two trivial ones. We know irreducible characters of $S_n$ correspond to partitions of $n$. Then third equality of section 4.3 is $$O\left(\left[\sum_{\lambda\in P}(\chi_{\lambda}(1))^{-1}\right]\left[\max_{\lambda\in P} (\chi_{\lambda}(1))^{-(d-3)}\right]\right)=O(n^{-1}2^{-(d-3)}). $$ Now except for the trivial characters we have $\chi_{\lambda}(1)\geq 2$ which bound the term with max by a negative power of $2$. But why is the sum of the reciprocal bounded by $O(n^{-1})$? Here the context is that the $n$ is fixed, but $d$ later tends to infinity. But the number of partition grows like exponential of square root of $n$, so it is not obvious to me that the sum of reciprocal is small as there might be many nonzero terms.

It took me a while but I have not found any related identity regarding the character of symmetric group. Thank you very much for any help.

Han
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    Good news is that this estimate seems to follow from p. 636, formula (i), (ii) of https://academic.oup.com/jlms/article-abstract/66/3/623/811347 by letting $R=s=1$ and summing (i)+(ii) and apply symmetry to get the sum of the reciprocals of the dimension of the irreducible characters. – Han Feb 23 '24 at 04:05

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