robjohn
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960/1000 score
 May 1 comment Prove $\sum\limits_{n=1}^\infty \frac{n!}{3^n\cdot7\times10\times\cdots\times (3n+1)}=\frac{\pi\sqrt3}{2}+\frac32\ln(3)−4$ The answer I get is $$-4+\frac32\log(3)+\frac{\pi\sqrt3}2$$ which agrees numerically with what I gave above. May 1 comment Prove $\sum\limits_{n=1}^\infty \frac{n!}{3^n\cdot7\times10\times\cdots\times (3n+1)}=\frac{\pi\sqrt3}{2}+\frac32\ln(3)−4$ (+1) My answer is a bit different, but along almost identical lines. I scrapped it when your answer appeared, but now I notice that there is a small error here. Your product goes up to $3n+4$ instead of $3n+1$. I didn't notice until I evaluated your integral and got a different answer. May 1 comment Prove $\sum\limits_{n=1}^\infty \frac{n!}{3^n\cdot7\times10\times\cdots\times (3n+1)}=\frac{\pi\sqrt3}{2}+\frac32\ln(3)−4$ This doesn't seem right. The sum is approximately $0.36861747935349131298$ and has a closed form in terms of a hypergeometric function, but the sum given is approximately $4.3122428402397139076$. May 1 comment Prove $\sum\limits_{n=1}^\infty \frac{n!}{3^n\cdot7\times10\times\cdots\times (3n+1)}=\frac{\pi\sqrt3}{2}+\frac32\ln(3)−4$ The previous series was divergent, but would converge if the $(3n+1)$ in the denominator was replaced with $(3n+4)$. May 1 comment Find the summation of the series. Oh... I didn't see it was a finite series. I will leave this answer in case someone is interested in the infinite series, or a sum to a variable upper limit. May 1 answered Find the summation of the series. May 1 comment Why the derivative of $n^{1/n}$ is $n^{1/n} \left( \frac{1}{n^2} - \frac{\log(n)}{n^2}\right)$ @Rinzler: it is not necessary to write it that way; it just makes it easier to see how to apply the chain rule. May 1 comment Evaluating $\lim\limits_{n\rightarrow \infty} \frac1{n^2}\ln \left( \frac{(n!)^n}{(0!1!2!…n!)^2} \right)$ I didn't find a way to do anything with it, but I thought it was interesting that $$\frac{(n!)^n}{(0!1!2!\cdots n!)^2} = \frac1{n!}\prod_{k=0}^n\binom{n}{k}$$ May 1 revised Evaluating $\lim\limits_{n\rightarrow \infty} \frac1{n^2}\ln \left( \frac{(n!)^n}{(0!1!2!…n!)^2} \right)$ remove dependence on Stirling May 1 revised Evaluating $\lim\limits_{n\rightarrow \infty} \frac1{n^2}\ln \left( \frac{(n!)^n}{(0!1!2!…n!)^2} \right)$ explain dependence on Stirling May 1 answered Evaluating $\lim\limits_{n\rightarrow \infty} \frac1{n^2}\ln \left( \frac{(n!)^n}{(0!1!2!…n!)^2} \right)$ May 1 awarded euclidean-geometry May 1 revised Complex integration on circle handle the second integral May 1 comment Complex integration on circle @askazy: I have added a bit to my answer about contour integration and residues (the coefficient of $\frac1{z-a}$ in the Laurent Series at $z=a$). May 1 comment Complex integration on circle I've added a bit more since it seems that contour integration is back on the table :-) May 1 revised Complex integration on circle explain a bit more since it seems that contour integration is on the table. May 1 answered Complex integration on circle May 1 comment Complex integration on circle Then you need to write the circle as $z=2\cos(t)+i(1+2\sin(t))$ and integrate the function in $t$ from $0$ to $2\pi$ Apr 30 revised system of First-Order ODES generalize a bit Apr 30 comment Why the derivative of $n^{1/n}$ is $n^{1/n} \left( \frac{1}{n^2} - \frac{\log(n)}{n^2}\right)$ @Mathemagician1234: in many cases complicated exponential expressions can be simplified by taking logs (as in Essam's answer) or by writing as an exponential. These are both valuable techniques.