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Dec
31
awarded  Favorite Question
Dec
31
answered Double Euler sum $ \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} $
Dec
31
revised Prove that $ x_1+ \dotsb + x_k=n, \frac1{x_1}+ \dotsb + \frac1{x_k}=1$
improved title, wording, grammar
Dec
31
answered Prove that $ x_1+ \dotsb + x_k=n, \frac1{x_1}+ \dotsb + \frac1{x_k}=1$
Dec
28
comment How this operation is called?
@IgorRivin: The "binomial convolution" reference is on page 365 of the second edition of Concrete Mathematics. Perhaps you are talking about Generatingfunctionology?
Dec
22
reviewed Approve construct an equilateral triangle with out knowing its scale
Dec
20
comment Curious graph: expected number of balls in the $i$th ordered bin
This is an interesting and, I think, fairly difficult question. The expected value of the largest element (with the roles of $n$ and $k$ switched) is given in exact and approximate form by my answer here. Finding an expression for the $N$th largest will be much harder, I think.
Dec
20
answered Applications of functions of the form $f(x)^{g(x)}$
Dec
17
revised Evaluate $\int_0^1\ln(1-x)\ln x\ln(1+x) \mathrm{dx}$
referenced Robjohn's answers to my other question
Dec
17
accepted Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$
Dec
17
revised Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$
gave explanation for acceptance switch
Dec
17
comment Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$
Thanks, Rob. I appreciate you taking the time to add this answer.
Dec
17
revised Approximating Distribution of a Data Set
fixed speling in title
Dec
17
comment $\sum_{n=1}^{\infty}\frac{H_n}{n^2 2^n}=\zeta(3)-\frac{1}{2}\log(2)\zeta(2)$
+1. Nice, Rob! The factor of $2^n$ in the denominator made me think of the Euler series transformation, too, but I couldn't make it go through. Also, your work starting from equation (13) is an evaluation of $A(2,1)$ from my question here, which means you've got a derivation for the last of the three sums from that question. I'd be happy if you were willing to finish off your series of answers to my question with your derivation for $A(2,1)$ here.
Dec
17
comment Infinite Series $\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}$
@RaymondManzoni: You're welcome. It was one of those questions where the quality of the answers far exceeded my expectations!
Dec
16
comment Infinite Series $\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}$
+1. For a different evaluation of $S(-1)$ that relies solely on manipulation of the summation, see robjohn's answer here.
Dec
16
revised The $n^{th}$ root of the geometric mean of binomial coefficients.
edited tags
Dec
15
awarded  Necromancer
Dec
3
revised Number of elements of $3n$ binary tuples, where the ordinates add up to $2n$.
fixed grammar
Dec
3
answered Number of elements of $3n$ binary tuples, where the ordinates add up to $2n$.