Mike Spivey
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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 $2n$.