Just wondering, how can I bring back the gray color to my status bar on top of my screen? This black one makes it hard to read the gray fonts (Also it is too dull) ._.  My bad: Please replace orange with gray.
The new top bar layout is fine, but the black color is very visually jarring/intrusive.  Is using a lighter color an available option?
Internet Accounts spontaneously forgets passwords and I can't reauthorize them I've had this problem before, but I was able to resolve it that time. macOS 10.13.4, everything up to date. Yesterday, Internet Accounts forgot my LinkedIn password and would not reauthenticate. Today, it forgot my Facebook password.   I've tried:   App specific password. Removing the account and re-adding it. Turning off 2-factor authentication. Deleting associated entries from Keychain. Logging out of iCloud, deleting Keychain entries, removing and re-adding the account, then logging back in.   First of all, what the hell—why does this keep happening?! Secondly, what else can I do to fix this short of reinstalling macOS? If $\sum_{n=1}^{\infty} a_n^{2}$ converges, then so does $\sum_{n=1}^{\infty} \frac {a_n}{n}$ Let $a_1,a_2,a_3,\ldots$ be reals. Prove that if $\sum_{n=1}^{\infty} a_n^{2}$ converges, then so does $\sum_{n=1}^{\infty} \frac {a_n}{n}$.   For this I have shown the case for when $ a_n^{2} \le\frac {|a_n|}{n}$ $\Rightarrow$ $ |a_n|\le\frac {1}{n}$ $\Rightarrow$ $\frac {|a_n|}{n} \le \frac{1}{n^{2}}$ and we know that $\sum_{n=1}^{\infty} \frac {1}{n^{2}}$ converges and hence $\sum_{n=1}^{\infty}\frac {a_n}{n}$ converges by the comparison test. Now considering $ a_n^{2} \ge\frac {|a_n|}{n}$ $\Rightarrow$ $\frac {|a_n|}{n} \le a_n^{2}$ $\rightarrow$ combining the two cases for any n we have: $\frac {|a_n|}{n}\le\frac{1}{n^{2}}+a_n^{2}$ Hence using the comparion test again we know that $\sum_{n=1}^{\infty} a_n^{2}$ converges and $\sum_{n=1}^{\infty} \frac {1}{n^{2}}$ converges hence the sum converges so we can conclude that $\sum_{n=1}^{\infty} \frac {a_n}{n}$ is absoluetly convergent $\Rightarrow$ convergent.  Not to sure if this is correct, any help would be much appreciated, many thanks. os.path.abspath("..") returns "C://User//" which is false + __file__ is not defined --> execution from Spyder3 (Python 3.6) If $x_1,x_2,\ldots,x_n$ are real numbers larger than $1$, prove that $$\dfrac{1}{1+x_1}+\dfrac{1}{1+x_2}+\cdots+\dfrac{1}{1+x_n} \geq \dfrac{n}{\sqrt[n]{x_1x_2\cdots x_n}+1}$$   Attempt  AM-GM doesn't work here since we will get an upper bound. I don't see Cauchy-Schwarz working either. Thus, I think a substitution might work, but I am unsure of which one to use. Is my proof that $U_{pq}$ is not cyclic if $p$ and $q$ are distinct odd primes correct? Prove that $U_{pq}$ is not cyclic if $p$ and $q$ are distinct odd primes.   I am a self taught person. I just learned this and tried this on my own and came up with this.  $x \equiv 1 \pmod{p}$ and $x \equiv -1 \pmod{q}$ has a solution $[a]_{pq}$, since $p$ and $q$ are relatively prime. Because $q$ is an odd prime, $[-1]_{pq}$ is not a solution, so $[a]_{pq}\neq [-1]_{pq}$. But $a^2 \equiv 1 \pmod{p}$ and $a^2 \equiv 1 \pmod{q}$, so $a^2 \equiv 1\pmod{pq}$ since $p$ and $q$ are relatively prime, and thus $[a]_{pq}$ has order 2.  Can someone please tell me if this proof this correct? Please help with proof as I learned it just now as a self taught person. Ability to update profile from the Stack Exchange Android App It would be nice to update the profile page from within the app.  Perhaps not many people update their profiles more frequently, but when required to do so, I would prefer to do it from the app instead of the mobile browser. The correct omission of the definite and indefinite articles Movie/TV series with a scene of spiky, black goo growing out of a lab container? There was a scene from a sci-fi horror video, not animated, don't know if it's a movie or TV series, where there was this black goo in a glass container and some scientists were looking at it under a microscope.   The pitch black goo grew and grew so fast, eventually breaking out of the inner box and the scientists that were inside the "inner lab" were begging to get out but the scientists outside that inner lab refused to open the door. The black goo was spiky (had spikes).  It's not any of the "Alien" movies, "The Blob", "Life", or "The Thing". It's not vintage. It was made after 1990, I'm pretty sure. Some proof question about Fibonacci sequence $\lim\limits_{n \to \infty} \frac{a_n}{b_n}=l$ ; Prove $\sum\limits_{k=1}^\infty a_k$ converges iff $\sum\limits_{k=1}^\infty b_k$ converges. Written, in Ex.8 Ch.9.1 of the book Advanced Calculus by P. M. Fitzpatrick :     Suppose that $\sum\limits_{k=1}^\infty a_k$ and $\sum\limits_{k=1}^\infty b_k$ are series of positive numbers such that $$\lim_{n \to \infty} \frac{a_n}{b_n}=l \ \ \ \text{and} \ l&gt;0.$$ Prove that the series $\sum\limits_{k=1}^\infty a_k$ converges iff the series $\sum\limits_{k=1}^\infty b_k$ converges.   Am I correct by the following (sketch of) proof? :  1- For a given $\epsilon_1$ there is $N_1$ such that $\left|\frac{a_n}{b_n} - l\right| &lt; \epsilon_1$ for all $n \ge N_1$.  2- Since the series $\sum\limits_{k=1}^\infty a_k$ converges, for a given $\epsilon_2$ there is $N_2$ such that $\left|b_{n+1}+\dots+b_{n+k}\right|&lt; \epsilon_1$ for all $n \ge N_2$ any for all natural numbers $k$.  3- Define $N = \max {\{N_1,N_2}\}$.   4- From Step (1), $a_{n+k} &lt; (\epsilon_1+l) b_{n+k}$ for all $n \ge N$ any for all natural numbers $k$. [Also, $a_i$'s and $b_i$'s are all positive]. Thus $a_{n+1}+\dots+a_{n+k} &lt; (\epsilon_1+l) (b_{n+1}+\dots+b_{n+k})&lt; \epsilon_3$. Then the convergence of the series $\sum\limits_{k=1}^\infty b_k$ implies the convergence of the series $\sum\limits_{k=1}^\infty a_k$.  5- For the reverse implication, we use the fact that $\lim\limits_{n \to \infty}\frac{a_n}{b_n}=l \iff \lim\limits_{n \to \infty}\frac{b_n}{a_n}= \frac{1}{l}= l' &gt;0$ and repeat the process this time for a and b exchanged.   6- The Quotient Property For Sequences hold for a nonzero limit in the denominator, but since the limit in the numerator also is zero so we may use the The Quotient Property For Sequences.    Thanks. A matrix and its transpose have the same set of eigenvalues/other version: $A$ and $A^T$ have the same spectrum How do I move the "Show Applications" icon in Ubuntu Dock? Ubuntu 17.10 introduces the GNOME Shell with the  (based on  GNOME Shell extension) and it allows you to put the Dock on the bottom of the desktop:    However, there is no easy way to move the Show Applications icon (shown on the right) to the left of the Dock. There is no such setting in GNOME Control Center. There is already a  filed, but does anybody know any workaround until it gets fixed by Ubuntu Dock developers? If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Cross-site AJAX requests I need to make an AJAX request from a website to a REST web service hosted in another domain.  Although this is works just fine in Internet Explorer, other browsers such as Mozilla and Google Chrome impose far stricter security restrictions, which prohibit cross-site AJAX requests.  The problem is that I have no control over the domain nor the web server where the site is hosted. This means that my REST web service must run somewhere else, and I can't put in place any redirection mechanism.  Here is the JavaScript code that makes the asynchronous call:  var serviceUrl = "http://myservicedomain"; var payload = "&lt;myRequest&gt;&lt;content&gt;Some content&lt;/content&gt;&lt;/myRequest&gt;"; var request = new XMLHttpRequest(); request.open("POST", serviceUrl, true); // &lt;-- This fails in Mozilla Firefox amongst other browsers request.setRequestHeader("Content-type", "text/xml"); request.send(payload);   How can I have this work in other browsers beside Internet Explorer? Seeking methods to solve: $\int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt$ Seeking Methods to solve the following two definite integrals:  \begin{equation} I_(n) = \int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt \qquad J(n) = \int_0^\infty \frac{\ln^2(t)}{t^n + 1}\:dt \end{equation}  For $n \in \mathbb{R},\:n \gt 1$  The method I took was to take the following integral:  \begin{equation}  \int_0^\infty \frac{t^k}{\left(t^n + 1\right)^m}\:dt = \frac{1}{n}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right) \end{equation}  Where: $0 \leq k \lt n$. Here let $m = 1$.   with respect to $k$ and taking the limit as $k \rightarrow 0^+$ (via the ), i.e.   \begin{align} \lim_{k \rightarrow 0+} \frac{d}{dk} \left[ \int_0^\infty \frac{t^k}{t^n + 1}\:dt \right] &amp;= \lim_{k \rightarrow 0+} \frac{d}{dk} \left[\frac{1}{n}B\left(1 - \frac{k + 1}{n}, \frac{k + 1}{n} \right) \right] \\ \lim_{k \rightarrow 0+}  \int_0^\infty \frac{t^k \ln(t)}{t^n + 1}\:dt  &amp;= \lim_{k \rightarrow 0+}  \left[\frac{1}{n^2} B\left(1 - \frac{k + 1}{n}, \frac{k + 1}{n}\right)\left[\psi^{(0)}\left(\frac{k + 1}{n} \right) - \psi^{(0)}\left(1 - \frac{k + 1}{n} \right)\right] \right] \\ \int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt&amp;= \frac{1}{n^2} B\left(1 - \frac{1}{n}, \frac{1}{n}\right)\left[\psi^{(0)}\left(\frac{1}{n} \right) - \psi^{(0)}\left(1 - \frac{1}{n} \right)\right] \\ &amp;=- \frac{\pi^2}{n^2}\operatorname{cosec}\left(\frac{\pi}{n}\right)\cot\left(\frac{\pi}{n} \right) \end{align}  Which is our expression for $I_n$. Taking the same approach but differentiating twice with respect to $k$ we arrive at our expression for $J_n$:  \begin{equation}  J(n) = \int_0^\infty \frac{\ln^2(t)}{t^n + 1}\:dt = \frac{\pi^3}{n^3}\operatorname{cosec}\left(\frac{\pi}{n} \right)\left[\operatorname{cosec}^2\left(\frac{\pi}{n}\right) + \cot^2\left(\frac{\pi}{n}\right) \right] \end{equation}  And in fact we may generalise:  \begin{equation}     \int_0^\infty \frac{\ln^p(t)}{\left(t^n + 1\right)^m}\:dt = \lim_{k\rightarrow 0}\frac{d^p}{dk^p}\left[\frac{1}{n} B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right)\right] \end{equation}  Where $p \in \mathbb{N}$  This method however was just an extension of another integral. I'm curious, if I had just started with $I_n, J_n$ what alternative methods could be used? Why is the Earth so fat? I made a naive calculation of the height of Earth's equatorial bulge and found that it should be about 10km.  The true height is about 20km.  My question is: why is there this discrepancy?  The calculation I did was to imagine placing a ball down on the spinning Earth.  Wherever I place it, it shouldn't move.  The gravitational potential per unit mass of the ball is $hg$, with $h$ the height above the pole-to-center distance of the Earth (call that $R$) and $g$ gravitational acceleration.  Gravity wants to pull the ball towards the poles, away from the bulge.  It is balanced by the centrifugal force, which has a potential $-\omega^2R^2\sin^2\theta/2$ per unit mass, with $\omega$ Earth's angular velocity and $\theta$ the angle from the north pole.  This comes taking what in an inertial frame would be the ball's kinetic energy and making it a potential in the accelerating frame.  If the ball doesn't move, this potential must be constant, so  $$U = hg - \frac{(\omega R \sin\theta)^2}{2} = \textrm{const}$$  we might as well let the constant be zero and write  $$h = \frac{(\omega R \sin\theta)^2}{2g}$$  For the Earth,   $$R = 6.4*10^6m$$  $$\omega = \frac{2\pi}{24\ \textrm{hours}}$$  $$g = 9.8\ m/s^2$$  This gives 10.8 km when $\theta = \pi/2$, so the equatorial bulge should be roughly that large.    , the Earth is 42.72 km wider in diameter at the equator than pole-to-pole, meaning the bulge is about twice as large as I expected.  (Wikipedia cites diameter; I estimated radius.)  Where is the extra bulge coming from?  My simple calculation uses $g$ and $R$ as constants, but neither varies more than a percent or so.  It's true the Earth does not have uniform density, but it's not clear to me how this should affect the calculation, so long as the density distribution is still spherically-symmetric (or nearly so).  (Wikipedia also includes , without derivation, that agrees with mine.) Error message "sudo: unable to resolve host (none)" If A is a finite set and B is an infinite set, then B\A is an infinite set Prove: If $A$ is a finite set and $B$ is an infinite set, then $B \setminus A$ is an infinite set.   I've tried proving the contrapositive, that if $B \setminus A$ is a finite set then $A$ is infinite or $B$ is finite, but I don't think this is the way to go. javascript not working, working on fiddle but not on my webserver How to calculate 95% CI for cumulative incidence rate using R?