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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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6h
comment How to show a sequence converges
(+1) I think this is in essence the same as what I did, though I had a longer answer that I overwrote that implies that $$\frac23=\frac12\frac54\frac{17}{16}\frac{257}{256}\cdots$$
7h
revised How to show a sequence converges
simplify argument
7h
answered How to show a sequence converges
8h
comment Is the $\sum_{n=1}^{\infty} \frac{(2n+1)^{1/2}}{n^2}$ convergent or divergent?
@user138069: probably since $2n+1\le3n$ for $n\ge1$.
8h
comment How to calculate the following sums?
@user4140: other than summing the finite sum, I don't know of a simpler formula for the first sum.
8h
revised How to calculate the following sums?
give an asymptotic expansion of the first sum
8h
answered How to calculate the following sums?
9h
comment Different methods of calculating $\zeta(s)$'s Laurent series.
No one has answered. Change the question. It's only been 10 minutes, so I don't think that anyone has spent too long working on this yet.
9h
comment Different methods of calculating $\zeta(s)$'s Laurent series.
Since the result gives the coefficients of the Laurent series at $z=1$, I would be surprised if there were.
11h
revised Show $\displaystyle\int_0^af(x)g(x)dx\ge\int_0^af(a-x)g(x)dx$
give better explanation
11h
answered Show $\displaystyle\int_0^af(x)g(x)dx\ge\int_0^af(a-x)g(x)dx$
11h
revised Prove this identity: $ \tan(2x)-\sec(2x) =\tan(x-\pi/4)$
add a bit of explanation
11h
answered Prove this identity: $ \tan(2x)-\sec(2x) =\tan(x-\pi/4)$
12h
comment Probability question with dice
@RB: I noticed that you got the right answer, so I figured it was a typo.
13h
comment Show $\lim\limits_{a \rightarrow + \infty} \int_0^{\infty} \frac{y}{1+y^2}e^{-ay} dy =0 $
would the downvoter care to comment?
14h
comment Probability question with dice
$\frac{2!}{6^3}=\frac{1}{72}$?
1d
answered Integral Inequality
1d
comment Show $\lim\limits_{a \rightarrow + \infty} \int_0^{\infty} \frac{1}{1+y^2}e^{-ay} dy =0 $
@5xum: Your answer is quite nice (I've upvoted it), but I was simply replying to the question. The LDC question is answered in the "duplicate" question.
1d
comment Show $\lim\limits_{a \rightarrow + \infty} \int_0^{\infty} \frac{1}{1+y^2}e^{-ay} dy =0 $
@5xum: Lebesgue Dominated Convergence. You need one dominating function as $a\to\infty$ (e.g. $e^{-y}$ for $a\ge1$)
1d
revised Show $\lim\limits_{a \rightarrow + \infty} \int_0^{\infty} \frac{y}{1+y^2}e^{-ay} dy =0 $
note the difference from the previous question