The following words which end with an ean are pronounced as /*i:n/ whereas Sean is pronounced as /ʃɒn/:      bean, dean, lean, mean...   Why is Sean not pronounced as /si:n/ but as /ʃɒn/? Where does this irregularity come from? Are there known rules in English for such cases?
I've always had this question about the pronunciation of Sean.  Is Sean a word from another language? Is it actually not pronounced Shawn and instead it's some sound between Shawn and Seen? Also, why isn't it pronounced as Sawn instead of Shawn, when the word Sean doesn't even have the letter 'h' in it to get that -sh sound?  Bottom line is, why is Sean pronounced Shawn instead of Seen?
This question is related to a couple of other questions ( and ) relating to some material that I teach in sampling theory.  Consider the critical points of the  defined as follows.  The value $t_{n-1, \alpha/2}$ is a function of $n$ defined by the implicit equation: $$\frac{\alpha}{2} = \frac{1}{\sqrt{(n-1) \pi}} \cdot \frac{\Gamma(\tfrac{n}{2})}{\Gamma(\tfrac{n-1}{2})} \int \limits_{t_{n-1, \alpha/2}}^\infty \Big( 1+ \frac{r^2}{n-1} \Big)^{-n/2} dr.$$ Normally we treat the input $n&gt;1$ as an integer, but it is possible to extend the treatment to consider it a real value (since the T-distribution is well-defined for non-integer degrees-of-freedom).  I'm trying to come up with a (relatively simple) proof that this critical point value is decreasing in $n$, but I seem to be making quite a mess.  Differentiation using Leibniz rule gets part way, but it quickly becomes a bit of a mess.  Consequently, I would like to solicit proofs of this result --- this simpler the better. 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