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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.