TenaliRaman
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 Aug5 revised Prove that $e^{\sum 1/p_k^2} > \pi/2$ added 216 characters in body Aug5 comment Prove that $e^{\sum 1/p_k^2} > \pi/2$ @TheoJohnson-Freyd I was bounding the series $\sum_{k = 1}^{\infty} \frac{\mu(k)}{k} \ln(\zeta(2k))$ below with just the first term in the series. The first term gives $\ln(\pi^2/6)$ which as observed in the comments is slightly larger the final sum and hence the bound as used clearly isn't true. Maybe there is a way to salvage it yet by considering few more terms from the series and showing that it does bound the series below, but my math-fu skill are lacking for that :-). Aug4 revised show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ Minor typo in the final result of the second technique Aug4 comment show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ (+1) The second technique is amazing and makes the whole integral brilliantly simple! Aug4 suggested approved edit on show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ Aug4 comment Prove that $e^{\sum 1/p_k^2} > \pi/2$ (+1) for your answer. And it was my answer that used Prime Zeta Function and it wasn't correct, so I retracted it before it got any undeserved attention -_-; sheepish grin Aug4 comment Prove that $e^{\sum 1/p_k^2} > \pi/2$ Yes, realized that a while ago. Proving things without sleep is a bad idea. Aug4 answered Prove that $e^{\sum 1/p_k^2} > \pi/2$ Aug1 comment Prove: $\int_0^{\infty}\left(\frac{\sin x}{x}\right)^2dx=\pi/2$ @PeterTamaroff is correct (+1). Put $\delta = \frac{1}{m}$ for some $m$. What does that give you? Jul31 comment Prove: $\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$ @BarryCipra Agreed. Jul31 comment Prove: $\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$ +1 Shouldn't that be <= because it is equal for n = 1? Jul29 comment A minimization problem Taking derivative gives $w - u + 2\beta\frac{w}{x^{\circ 2}}$ and second derivative is semi-definite. Therefore, setting the derivative to zero should give you a local minimum. Jul29 comment Find the sum : $\sin^{-1}\frac{1}{\sqrt{2}}+\sin^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\cdots$ +1 This is a very beautiful answer! Jul29 revised Show if $\|\cdot\|$ is a norm then $\|f(\cdot)\|$ is a norm where $f$ is linear and invertible Minor correction to notation Jul29 suggested approved edit on Show if $\|\cdot\|$ is a norm then $\|f(\cdot)\|$ is a norm where $f$ is linear and invertible Jul27 revised show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ added 95 characters in body Jul27 answered show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ Jul26 comment How to show $\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$? @EugeneBulkin True. I apologize for the confusion caused. Jul25 comment How to show $\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$? Your definition of f(x) should not contain that n. Jul23 comment Given $\{\log_ab \mid a,b\in \mathbb N, \mathrm{gcd}(a,b)=1,a,b≥3\}$ does the sum of any two $\log_ab$ form an irrational or rational number? Try verifying the conditions for the group. Is it closed? Is it associative? Does it have an identity? What about inverses?