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Joshua King came to Cambridge from Hawkshead Grammar School. It was soon evident that the school had produced someone of importance. He became Senior Wrangler, and his reputation in Cambridge was immense. It was believed that nothing less than a Second Newton had appeared. They expected his work as a mathematician to make an epoch in the science. At an early age he became President of Queens’; later, he was Lucasian Professor. He published nothing; in fact, he did no mathematical work. But as long as he kept his health, he was an active and prominent figure in Cambridge, and he maintained his enormous reputation. When he died, it was felt that the memory of such an extraordinary man should not be permitted to die out, and that his papers should be published. So his papers were examined, and nothing whatever worth publishing was found.

comment Calculate limit $\lim_{n\rightarrow\infty}\dfrac{(4n-100)^{4n-100}n^n}{(3n)^{3n}(2n)^{2n}}?$
Hint: Factor $4n$ forcefully outside the parenthesis, and see what you get. :-$)$
comment What is a “hypergeometric series” with differences, not just sums, of indices?
The last expression makes no sense for $n<k$.
comment $ \lim_{n\rightarrow \infty}n^{-\frac{1}{2}\left(1+\frac{1}{n}\right)}\cdot \left(1^1\cdot 2^2…n^n\right)^{\frac{1}{n^2}}$
See hyperfactorial $($ also here $)$.
comment closed form for $I=\int_{0}^{\infty}\frac{x^n}{x^2+u^2}\tanh(x) dx$
It converges only for $n\in(-2,1)$.
revised Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$
Minor Edit.
comment sum of infinite series with telescopes
But what if the sum doesn't telescope ? :-$)$
comment Sum of 2 different irrational logarithms = Irrational?
A related question. See the comment section on @Mookid's now-deleted answer.
comment How to show that $\int_0^1 dx \frac{1+x^a}{(1+x)^{a+2}} = \frac{1}{a+1}$?
A related question.
comment Prove that $\int_0^\infty \frac{\ln x}{x^n-1}\,dx = \left(\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}\right)^2$
$\displaystyle\int_0^\infty\dfrac{x^{^{k-1}}}{1-x^{^n}}dx~=~\dfrac\pi n\cot\bigg(k~\dfrac\pi n\bigg)$
comment integration with pade approximant
This is, of course, the $\Gamma$ function of argument $\dfrac32$, whose value is $\dfrac{\sqrt\pi}2$.
comment To calculate side of the Equilateral triangle
@graydad: He's saying that there's a point inside the equilateral triangle, and that the three distances from this point to each of the three vertices are $3$, $4$, and $5$.
comment q question regarding the numerical zero finding of Riemann zeta function and actual proof of Riemann Hypothesis
The name of Pell's equation arose from Leonhard Euler's mistakenly attributing its study to John Pell. Euler was aware of the work of Lord Brouncker, the first European mathematician to find a general solution of the equation, but apparently confused Brouncker with Pell. :-$)$
comment q question regarding the numerical zero finding of Riemann zeta function and actual proof of Riemann Hypothesis
Neither. ;-$)$
answered Asymptotic behavior of $\sum\limits_{k=1}^n \frac{1}{k^{\alpha}}$ for $\alpha > \frac{1}{2}$
answered How to properly state as to why $\sum_{n=1}^{\infty}\frac{\sqrt{n^3+2}}{n^4+3n^2+1}$ converges.
comment Radius of convergence of $1+3x+\frac{3^2x^2}{2!}+\cdots$
Where am I going wrong? - Your only error is doubting yourself. :-$)$
answered Do there exist integers s and t such that 11s + 9t = 1?
answered An infinite series question.
comment How to evaluate $\int_0^1\frac{\tanh ^{-1}(x)\log(x)}{(1-x) x (x+1)} \operatorname d \!x$?
Not that it matters, but $$I~=~\displaystyle\int_0^1\dfrac{\text{arctanh }x}{1-x^2}\cdot\dfrac{\ln x}xdx~=~\dfrac14\int_0^1\Big(\text{arctanh }^2x\Big)'\cdot\Big(\ln^2x\Big)'dx$$ since $$\quad\text{arctanh }'x=\dfrac1{1-x^2}\quad\text{and}\quad\ln'x=\dfrac1x$$
comment Proof of a sum of positive divisors
@GerryMyerson: I cannot help but wonder whether there might be some connection with the following Faulhaber formula: $\big(\sum n\big)^2=\sum n^3$