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the mean square The Mean Square
(with one standard deviation and several unusual ones)

aka Rob Johnson

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Jan
8
comment bounded functions supported on a set E
I think we have the same function represented two ways, except I haven't handled the infinite part... except there is no infinite part since $f$ is bounded by $M$. +1 in any case :-)
Jan
8
answered bounded functions supported on a set E
Jan
8
revised Equivalent of an improper integral
make the relative sizes more explicit
Jan
7
answered What's the explanation for why n^2+1 is never divisible by 3?
Jan
7
comment Sum of reciprocals of binomial coefficients
It is interesting that your asymptotic expansion of two terms gives $O(n^{-6})$ error (and it does). My expansion, while simple to continue, requires five terms.
Jan
7
revised Sum of reciprocals of binomial coefficients
increase the asymptotic expansion to match Peter Košinár
Jan
7
awarded  Nice Answer
Jan
7
comment Equivalent of an improper integral
I think that to get votes, you might need to give more detail.
Jan
7
comment Equivalent of an improper integral
No, I am not using that. I am using the exact formula $$\frac{\Gamma(x+n)}{\Gamma(x+1)}=(x+1)(x+2)\cdots(x+n-1)$$ which follows inductively from the relation $x\Gamma(x)=\Gamma(x+1)$.
Jan
7
revised Sum of reciprocals of binomial coefficients
add an asymptotic expansion
Jan
7
comment Equivalent of an improper integral
@Julien: The Gamma function is convex on $(0,\infty)$. The bounds follow from the fact that $\Gamma'(1)=-\gamma$ and $\Gamma(1)=\Gamma(2)=1$.
Jan
7
revised Equivalent of an improper integral
improve the exposition
Jan
7
comment Equivalent of an improper integral
My better lower bound now matches yours which is $\frac1{n!\log(n)}\left(1-\frac{\gamma}{\log(n)}\right)$ if you use $H_n\sim\log(n)+\gamma$.
Jan
7
revised Equivalent of an improper integral
improve asymptotic approximation
Jan
7
comment Equivalent of an improper integral
I like your method for the lower bound (+1). I will improve my answer to match yours better by using a better lower bound for $\Gamma(1+x)$ on $[0,1]$ than $\frac78$.
Jan
7
answered Equivalent of an improper integral
Jan
7
revised If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$
add missing end of proof
Jan
7
comment Sum of reciprocals of binomial coefficients
Now your two expressions are the same, only differing by a change of index $k\mapsto n-k$.
Jan
7
comment Sum of reciprocals of binomial coefficients
Your first update expression matches my $(2)$. Your second update expression would be the same as your first if you divide by $2^n$ instead of $n$ (as it stands, it is not correct).
Jan
6
answered Sum of reciprocals of binomial coefficients