I tried to do a change of variable with $t=\frac{1}{n}$ in order to use Taylor expansion but the factor $n!$ becomes a fractional number (during the lesson it was define only for $\mathbb{Z}$ set number).  Another attempt I made was to write the limit as $\lim_{n \to \infty} \sqrt[n]{\frac{n!}{n^n}} \rightarrow e^{\frac{1}{n}log{\frac{n!}{n^n}}}$ but it doesn't work. After that I tried to use the Stirling formula and yes... it works! But I would find something that doesn't use this formula. Some advice?
Show:$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \frac{1}{e}$$  So I can expand the numerator by geometric mean. Letting $C_{n}=\left(\ln(a_{1})+...+\ln(a_{n})\right)/n$. Let the numerator be called $a_{n}$ and the denominator be $b_{n}$ Is there a way to use this statement so that I could force the original sequence into the form of $1/\left(1+\frac{1}{n}\right)^n$
There is a common story that airline ticket prices rise significantly if you buy them less than two weeks before departure.  And that any time before that will yield standard prices.  Is that advice valid for flying during holiday seasons?  As an example: if I want to fly coast to coast USA on December 7th, must I buy my ticket ASAP, or would I get a similar price buying on November 21st? Put your shoes on and Take your coat off I have often heard the following statements when someone talks about shoes, cloths, etc. I don't know which one is more appropriate or grammatically correct.   Hey, put your shoes on.  Hey, put on your shoes.  Take off your coat, please.  Take your coat off, please.    Any thoughts? Can we also say &quot;take out your coat&quot;? What is the difference between Clamav and Clamtk? I know that both of them are antivirus. VMware Linux Server -- how can you tell if you are a vm or real hardware? An interesting question.  I have logged into a Linux (most likely SuSE) host.  Is there some way that I can tell programmatically that I am a VM host or not?    Also assume that the vmtools are not installed. Prove that if the identity is written as the product of $r$ transpositions, then $r$ is an even number Theorem. If the identity is written as the product of $r$ transpositions, $id=τ_1τ_2\dots τ_r$, then $r$ is an even number. Proof. We will employ induction on $r$. A transposition cannot be the identity; hence, $r &gt; 1$. If $r = 2$, then we are done. Suppose that $r &gt; 2$. In this case the product of the last two transpositions, $τ_{r−1}τ_r$, must be one of the following cases: $(ab)(ab) = id\\ (bc)(ab) = (ac)(bc)\\ (cd)(ab) = (ab)(cd)\\ (ac)(ab) = (ab)(bc),$ where $a$, $b$, $c$, and $d$ are distinct. The first equation simply says that a transposition is its own inverse. If this case occurs, delete $τ_{r−1} τ_r$ from the product to obtain $id=τ_1τ_2···τ_{r−3}τ_{r−2}$. By induction $r − 2$ is even; hence, $r$ must be even. In each of the other three cases, we can replace $τ_{r−1} τ_r$ with the right-hand side of the corresponding equation to obtain a new product of $r$ transpositions for the identity. In this new product the last occurrence of $a$ will be in the next-to-the-last transposition. We can continue this process with $τ_{r−2} τ_{r−1}$ to obtain either a product of $r − 2$ transpositions or a new product of $r$ transpositions where the last occurrence of $a$ is in $τ_{r−2}$. If the identity is the product of $r − 2$ transpositions, then again we are done, by our induction hypothesis; otherwise, we will repeat the procedure with $τ_{r−3}τ_{r−2}$. At some point either we will have two adjacent, identical transpositions canceling each other out or $a$ will be shuffled so that it will appear only in the first transposition. However, the latter case cannot occur, because the identity would not fix $a$ in this instance. Therefore, the identity permutation must be the product of $r − 2$ transpositions and, again by our induction hypothesis, we are done. Question: That is the proof written in the book, but unfortunately after more than $10$ times reading I don't understand the last two paragraph (i.e., from &quot;In each of the other three cases&quot;). Is there another simpler proof or could someone please clarify what is going on? Thank you. How to use measured step response to tune control system How does one use a measured step response to tune either a PID or convolution control scheme? Inspired by , I'm interested a more detailed explanation of how to implement a control system based on the measured step response.  I would not try to guess all the effects.  There are probably some non-obvious things going on, and you can't know all the parameters. I would measure the step response.  Find two pump settings that both result in the ball being within the measurable range within the tube.  Then have the controller suddenly switch from one setting to the other open loop.  Meanwhile measure what the ball does over time.  That is the step response.  You can take the derivative of that and get the impulse response.  From the impulse response you can predict the motion of the ball for any history of pump settings, assuming this is a linear system.  It is probably linear enough over the small range of settings to keep the ball within its normal range. You can use this as a simulation base to find the parameters for old fashioned PID control.  Or you can use the impulse response directly to do a convolution control.  You have to low pass filter the control input enough so that convolution kernel doesn't go negative unless your pump is actually reversible and can suck the ball back down.  How exactly does this work? ; I assume &quot;convolution control&quot; is the use of the pole-zero or transfer function, but don't see exactly how to get the parameters. RELATED:  * Accessing class members on a NULL pointer Ephemeral error message: "cryptsetup: unknown fstype, bad password or options?" Let $m$ be Lebesgue measure and $a \in R$. Suppose that $f : R \to R$ is integrable, and $\int_a^xf(y) \, dy = 0$ for all $x$. Then $f = 0$ a.e. This is a corollary to a proof in Bass, but I don't understand why it follows from the proof he gives. I follow everything up until the last statement. Why is it that proving that the integral is $0$ for all Borel measurable sets guarantees it is true for all Lebesgue measurable sets? Am I missing something vital about the relationship between Lebesgue measurable sets and Borel measurable sets? Statement: Let $m$ be Lebesgue measure and $a \in\mathbb R$. Suppose that $f : \mathbb R \to\mathbb R$ is integrable, and $\int_a^xf(y) \, dy = 0$ for all $x$. Then $f = 0$ a.e. Proof: We prove the conclusion first for intervals, then for unions of intervals, then for open sets, then for Borel measurable sets. For any interval $(c,d)$, we have $$\int_c^df = \int_a^d f - \int^c_a f = 0.$$ By linearity, if $G$ is the finite union of disjoint open intervals, then $\int_G f = 0$. Now, if $G$ is open, then $G$ is the countable union of disjoint, open intervals, $G = \bigcup^n_{i=1}I_i$. We have $$\int_G f = \int \lim_{n\to\infty} f \cdot \chi_{\bigcup^n_{i=1} I_i}$$ and since $f_n$ is integrable for all $n$, and $|f_n| \le f$ for all $n$, by the dominated convergence theorem, $$\int_G f = \lim_{n\to\infty} \int_{\bigcup^n_{i=1}I_i} f =0.$$ Now, if $G_n$ is a decreasing sequence of open sets converging to $H$, since $|f \cdot \chi_{G_n}| \le f \cdot \chi_{G_1}$ for all $n$, we again have by the DCT that $\int_H f = 0 = \lim_{n\to\infty} \int_{G_n}f = 0$. Finally, if $G$ is a Borel measurable set, then we know that there exists a sequence $G_n$ of decreasing, open sets, that converge to some set $H$, where $H \setminus G$ is a null set. Thus, $$\int_G f = \int_H f = 0.$$ We have found that for every Borel measurable set $G$, $\int_G f =0$. Since $f$ is real valued and measurable, $f = 0$ a.e. Show that $(2,0,4) , (4,1,-1) , (6,7,7)$ form a right triangle What I tried:  Let $A(2,0,4)$, $B(4,1,-1)$, $C(6,7,7)$ then  $$\vec{AB}=(2,1,-5), \vec{AC}=(4,7,3), \vec{BC}=(2,6,8)$$  Then I calculated the angle between vectors:  $$\begin{aligned} \alpha_1 &amp;= \cos^{-1}\left(\frac{(2,1,-5)(4,7,3)}{\sqrt{2^2+1^2+(-5)^2}\sqrt{4^2+7^2+3^2}}\right) \\ &amp;= \cos^{-1}(0)=90° \\ \alpha_2 &amp;= \cos^{-1}\left(\frac{(4,7,3)(2,6,8)}{\sqrt{4^2+7^2+3^2}\sqrt{2^2+6^2+8^2}}\right) \\ &amp;= \cos^{-1}\left(\frac{74}{\sqrt{74}\sqrt{104}}\right)=32.49\\ \alpha_3 &amp;= \cos^{-1}\left(\frac{(2,6,8)(2,1,-5)}{\sqrt{2^2+6^2+8^2}\sqrt{2^2+1^2+(-5)^2}}\right) \\ &amp;= \cos^{-1}\left(\frac{-30}{\sqrt{104}\sqrt{30}}\right)=122.5° \end {aligned}$$  As you can see, these angles don't even form a triangle, what am I doing wrong, any thoughts? Can I safely uninstall games purchased from Steam? Can I uninstall purchased games from my computer and then later download them through Steam without having to re-purchase if I'd like to?   I'm running short on space on my laptop, looking to free up some space and wanted to be sure that I can safely re-download anything at a later time similar to how you can with the Xbox Live marketplace. How did we 'discover' dark matter? I'm an astrophysics student and I've been researching this topic and there is one point that keeps eluding me.  How did the scientific community realize that there had to be  in the Universe? How to Run an app as Administrator without the prompt? When I start SQL Server Management Studio 2008, I get an error that's been around forever and is documented .  One of the suggestions to fix it that actually worked was to start SSMS with Run as Administrator. So I modified the properties of the shortcut to always run it as the administrator.  The only annoyance is that I get a prompt every single time asking whether I really want to do it. Is there a way to tell just this one program (SSMS) to not pop up the prompt, while keeping this behavior for other apps? Is there a case where including the same header twice is actually helpful? Finding the limit $\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}$ Currently using Ubuntu 18.04, and my wireless driver is not being detected.  I've included a .  $ dmesg | grep iwl [    6.212724] Loading modules backported from iwlwifi [    6.212724] iwlwifi-stack-public:master:6988:0d8892f2 [    6.224583] iwlwifi 0000:00:14.3: enabling device (0000 -&gt; 0002) [    6.229811] iwlwifi 0000:00:14.3: Direct firmware load for iwl-dbg-cfg.ini failed with error -2 [    6.230540] iwlwifi 0000:00:14.3: Direct firmware load for iwlwifi-9000-pu-b0-jf-b0-39.ucode failed with error -2 [    6.230547] iwlwifi 0000:00:14.3: Direct firmware load for iwlwifi-9000-pu-b0-jf-b0-38.ucode failed with error -2 [    6.230552] iwlwifi 0000:00:14.3: Direct firmware load for iwlwifi-9000-pu-b0-jf-b0-37.ucode failed with error -2 [    6.230557] iwlwifi 0000:00:14.3: Direct firmware load for iwlwifi-9000-pu-b0-jf-b0-36.ucode failed with error -2 [    6.230562] iwlwifi 0000:00:14.3: Direct firmware load for iwlwifi-9000-pu-b0-jf-b0-35.ucode failed with error -2 [    6.236670] iwlwifi 0000:00:14.3: loaded firmware version 34.0.0 op_mode iwlmvm [    6.248314] iwlwifi 0000:00:14.3: Detected Intel(R) Dual Band Wireless AC 9560, REV=0x318 [    7.264081] iwlwifi 0000:00:14.3: SecBoot CPU1 Status: 0x3, CPU2 Status: 0x2477 [    7.264084] iwlwifi 0000:00:14.3: Failed to start INIT ucode: -110 [    7.276165] iwlwifi 0000:00:14.3: Failed to run INIT ucode: -110   sudo modprobe iwlwifi returns no output. Skype window problem in Ubuntu 14.04 At what point does a caster define the target of a spell? The new School of Invention tradition for Wizards () while still in playtesting (and indeed a bit silly in my opinion, more of a "chaos wizard"), got me thinking about its reckless casting ability. This allows the caster to cast magic they do not have prepared and get a random result. The random spells have different targeting criteria (Wall of Stone, Fireball, Feign Death , Mass Cure Wounds) and so it is not possible to decide the target of the spell until it has been cast, as you don't know what you are going to get until it is cast.  This suggests that the designer of the subclass has made the assumption that you don't actually target until you have cast the spell and release it, which makes sense from one point of view, strongly supported by the way a prepared action casting works, PHB p.193 where you cast the spell and hold the energy until the condition is met for the readied action at which point you target the spell.  While the Unearthed Arcana is not RAW or finalized and released, my question is not about this subclass, but about the official rules as a whole.  Is there was any direct RAW, or other supporting or opposing RAW, that states at what point a caster defines the target of a spell?  As a sub-question: are there any circumstances where it actually matters, other than the already defined Readied Action spell casting, in the official rules? Why the principal value of complex number argument is taken within $(-\pi,\pi]?$ Why the principal value of complex number argument is taken within $(-\pi,\pi]?$ What is the harm in considering similar other intervals like $[0,2\pi)$ for the purpose? Single service account running all SQL Servers Is there any negative performance impact for using the same service account and agent account to run SQL Server and SQL Agent respectively for all SQL server instances running at a small company of say 35 servers? The biggest database is 0.5 GB. any suggestions are welcome. Thank you.