I am improving my skill at formal sequence convergence proofs, I find them very tricky. I want to prove that:  $$\frac{n^5}{3^n} \rightarrow0$$  This should be read as "converges to zero", the question is, how large should $n$ be?  I would want a way to compare these two expressions I have trouble picking a big $n$ because I do not quite understand how to compare denominator and enumerator. Could someone drop a small hint so I can continue with my proof, I do not want to use any limit theorems.  What I want is that for $n&gt;n_0$:  $$\left| \frac{n^5}{3^n} \right|&lt; \epsilon$$
In other words, how to prove:  For all real constants $a$ and $b$ such that $a &gt; 1$, $$\lim_{n\to\infty}\frac{n^b}{a^n} = 0$$  I know the definition of limit but I feel that it's not enough to prove this theorem.
Consider this picture of sun beams streaming onto the valley through the clouds.    Given that the valley is only (at a guess) 3km wide, with simple trigonometry and the angles of the beams, this gives the result that the position of the light source is being a few tens of km away at most.   What is wrong with the analysis? Is there a simple way to convert a string to Title Case? E.g. john smith becomes John Smith. I'm not looking for something complicated like , just (hopefully) some kind of one- or two-liner. I was reading the CernCourier, my favorite source of message on Higgs and friends. I was rather shocked, when I saw this:     "The mechanism by which neutrino mass is generated is not known."   What? Not known?  says: In Higgs-based theories, the property of 'mass' is a manifestation of potential energy transferred to particles when they interact ("couple") with the Higgs field, which had contained that mass in the form of energy.  Does this mean that neutrinos don't couple to the Higgs field? Where's the problem in this equation? Resulting in $4 = 5$ I just saw this equation and I can't find out where's the problem:  $$25-45 = 16-36$$  $$25- 2 \cdot 5 \cdot \frac{9}{2} = 16- 2\cdot4\cdot\frac{9}{2}$$  $$25 - 2\cdot 5\cdot \frac{9}{2} + \frac{81}{4} = 16 - 2\cdot 4 \cdot \frac{9}{2} + \frac{81}{4}$$  $$\left( 5-\frac{9}{2} \right) ^2 = \left (4-\frac{9}{2} \right) ^2$$  $$5-\frac{9}{2} = 4 - \frac{9}{2}$$ $$5=4$$ Cannot refer to a non-final variable inside an inner class defined in a different method From what time I've spent with threads in Java, I've found these two ways to write threads:  With implements Runnable:  public class MyRunnable implements Runnable {     public void run() {         //Code     } } //Started with a "new Thread(new MyRunnable()).start()" call   Or, with extends Thread:  public class MyThread extends Thread {     public MyThread() {         super("MyThread");     }     public void run() {         //Code     } } //Started with a "new MyThread().start()" call   Is there any significant difference in these two blocks of code ? Detect which TeX engine is used How can I detect inside a TeX file which engine (pdftex/xetex/luatex) is used to compile the file? Is there a LaTeX package that provides this functionality? How do you define an arbitrary sort sequence for views? Can spells with movable areas of effect be moved out of sight or behind obstacles? With spells like Moonbeam or Flaming Sphere, how should we treat the movement of their areas-of-effect when it comes to passing them behind a thick pillar or around a blind corner?  Should we treat any movement of them as if they are being 're-cast' (and so requiring a clear unobstructed view)? Or could a mage presumably will them somewhere out-of-sight? Let     $$ \mathbb{Z}^{'}_{\mathbb{C}}=\{ z \in \mathbb{C} | \exists f \in \mathbb{Z}[X] \text{ monic such that } f(z)=0\} $$   be the integral closure of $ \mathbb{Z} $ in $ \mathbb{C} $. Prove that $ \mathbb{Z}^{'}_{\mathbb{C}} $ is not finitely generated as a $ \mathbb{Z} $-module.  My attempt: Suppose $ \mathbb{Z}^{'}_{\mathbb{C}}=\mathbb{Z}\alpha_{1}+...+\mathbb{Z}\alpha_{n} $ for some $ \{ \alpha_{i}\}_{i=\overline{1,n}} \subset \mathbb{Z}^{'}_{\mathbb{C}} $. I know that given $ \alpha , \beta \in \mathbb{Z}^{'}_{\mathbb{C}} $ with minimal polynomials of degrees $ m $ and $ n $ respectively, then we can construct a monic polynomial with integral coefficients $ P(z)=\prod(z-(\alpha_{i}+\beta_{j})) $ where $ \alpha_{i},\beta_{j} $ are the conjugates of $\alpha $ , $\beta $ and therefore the minimal polynomial of $ \alpha + \beta $ has degree $ \leq mn $.   I was thinking that then the degrees of the minimal polynomials of the elements of $ \mathbb{Z}\alpha_{1}+...\mathbb{Z}\alpha_{n} $ could be somehow bounded by a certain constant, whereas $ \mathbb{Z}^{'}_{\mathbb{C}}$ contains elements whose degrees of the minimal polynomials are arbitrarily large, take $ 2^{\frac{1}{n}} $ for example,$ n\geq 2$. I think this works when we are asked to prove that the set of all algebraic numbers $ \overline{\mathbb{Q}} $ is not a finitely dimensional $\mathbb{Q}$-vector space, but I don't know how to proceed in my case.  I would really appreciate any ideas, hints or solutions. Thank you very much! Understanding minipages - aligning at top I would like to be able to align two minipage at the top, without worrying about what is in the minipage.  I am aware of the optional [t] parameter, but still have some confusion as to how this all works.  I tend to think of a minipage as a regular page in that whatever is placed on it starts at the top and fills the page going across and then down the page, but this does not appear to be the case.  In the MWE example below, I have color coded the sections for reference purposes.   The first case (black) works just as I would expect. The two minipages are aligned exactly as I want with the [t] specifier.  No question here.  The next case (red) only works if the inner minipage also has the [t] specifier. Why is that necessary?  I would have thought that any content would be placed at the top of this page, and since both the outer minipage had a [t] things should have aligned at the top. I would like to place a graphic in the second mini page, and still have the two aligned at the top.   The simple case (blue) of using \includegraphics directly does not result in the content being aligned at the top.  So, then I attempt to put this in a \parbox (green) and this too does not yield the desired results.  I also try to place the graphic inside an inner minipage (brown) and specify the [t] alignment, and even though this worked in the red section, it does not here.  The only difference between the red and brown being that the red section had text, whereas the brown section has a graphic.  I am aware of question regarding  which suggests to place \vspace{0pt} at the top of the minipage, and that does provide the desired alignment (cyan), but adds vertical space at the top.  So, my second question is why is the \vspace{0pt} necessary when a graphic is included, but not when it is text.  This appears to be a hack and wondering if there is a replacement for minipage that works better without having to worry about exactly what is in the minipage?  Also, why did wrapping the figure in a parbox (green) or another minipage (brown) not solve this problem?   Here is the MWE:  \documentclass{article} \usepackage[demo]{graphicx} \usepackage{lipsum} \usepackage{xcolor}  \begin{document} \section{This Works - Aligned at top as desired} \begin{minipage}[t]{0.6\linewidth}     \lipsum[1] \end{minipage} \hfill \begin{minipage}[t]{0.3\linewidth}     \lipsum[2] \end{minipage}  \color{red} \section{Why does inner mini page required the [t]?} \label{Minipage-in-minipage} \begin{minipage}[t]{0.6\linewidth}     \lipsum[3] \end{minipage} \hfill \begin{minipage}[t]{0.3\linewidth}     \begin{minipage}[t]{\linewidth}% Why is [t] required here         \lipsum[4]     \end{minipage} \end{minipage}  \color{blue} \section{Why are These not aligned at top?} \begin{minipage}[t]{0.6\linewidth}     \lipsum[3] \end{minipage} \hfill \begin{minipage}[t]{0.3\linewidth}     \includegraphics[width=\linewidth]{foo} \end{minipage}  \color{green} \section{Why are These not aligned at top?} \begin{minipage}[t]{0.6\linewidth}     \lipsum[3] \end{minipage} \hfill \begin{minipage}[t]{0.3\linewidth}     \parbox{\linewidth}{\includegraphics[width=\linewidth]{foo}} \end{minipage}  \newpage \color{brown} \section{Why does this not work with the [t] as it did in red section?} This worked fine in Section \ref{Minipage-in-minipage}  \begin{minipage}[t]{0.6\linewidth}     \lipsum[3] \end{minipage} \hfill \begin{minipage}[t]{0.3\linewidth}     \begin{minipage}[t]{\linewidth}% Even with [t] here, this does not work         \parbox{\linewidth}{\includegraphics[width=\linewidth]{foo}}     \end{minipage} \end{minipage}   \color{cyan} \section{Adding a \textbackslash{}vspace\{0pt\} works} \begin{minipage}[t]{0.6\linewidth}     \vspace{0pt}% Hack!! But does do the job     \lipsum[3] \end{minipage} \hfill \begin{minipage}[t]{0.3\linewidth}     \vspace{0pt}% Hack!! But does do the job     \begin{minipage}[t]{\linewidth}% Why is [t] required here         \parbox{\linewidth}{\includegraphics[width=\linewidth]{foo}}     \end{minipage} \end{minipage} \end{document} Let $R$ be a simple ring and $T$ be a non-zero ring. Let $f\colon R \rightarrow T$ be a ring homomorphism. Show $f$ is injective.  Proof: $\ker f \lhd R$, so $\ker f=R$ or $\ker f=\{0\}$. If $\ker f=R$ then $R/R \cong \{0\}$, but $T \neq \{0\}$, which is a contradiction. I can't see why does the last part means contradiction.   Thanks for any assistance! Can someone please help me thanks!  this is where I got the image from Why wouldn't the `which` command work for `cd`? I can't find the executable for `cd` either! I tried which cd and it didn't give a path but instead returned the exit code 1 (checked with echo $?). The coreutil cd itself is working, so the executable should be there, right? I also ran a find for cd, but there was no executable file shown. How is it implemented then?  Update:  I don't know if I should ask this in another post but since I think it is good here, I'm expanding (?) the post... So the answer was actually quite simple, there's no executable for that — because it's a builtin — But I've found some builtins (bash shell in Fedora) have the executable files! So builtin -> no executable isn't right I suppose? Maybe an answer explaining what builtins actually are (builtin commands?), which actually is the matter here, rather than focusing more on cd... Some good links posted previously indicate that builtins are not programs... so what are they? How do they work? Are they just functions or threads of the shell? What's the thing with $\sqrt{-1} = i$? Do they really teach this in the US? It makes very little sense, because $-i$ is also a square root of $-1$, and the choice of which root to label as $i$ is arbitrary. So saying $\sqrt{-1} = i$ is plainly false!  So why do people say $\sqrt{-1} = i$? Is this how it's taught in the US? Approximating commuting matrices by commuting diagonalizable matrices Suppose the matrices $A$ and $B$ commute. Do there exists sequences $A_n$ and $B_n$ of matrices such that   $A_n \rightarrow A$, $B_n \rightarrow B$.  Each $A_n$ is diagonalizable  and the same for each $B_n$.  For every $n$, $A_n$ commutes with $B_n$.    Moreover, it would be nice if the following property was additionally satisfied: if $A,B$ are real, then $A_n,B_n$ can be chosen to be real as well. Other word for person who does something to others but does'nt like it done to them This is a duplicate question but has still not been answered for me. I'm so close to the exact word - I know there is one out there - but the closest word I can find is a hypocrite --     a person who claims to have certain moral principles or beliefs but behaves in a way that shows they are not sincere.    The word I'm looking for is a person who does not claim things, but does thing to another and they consider it wrong or immoral to have done to them. Infiniteness of non-twin primes. Well, we all know the twin prime conjecture. There are infinitely many primes $p$, such that $p+2$ is also prime. Well, I actually got asked in a discrete mathematics course, to prove that there are infinitely many primes $p$ such that $p + 2$ is NOT prime. Getting user provided data out of a database and into R Why does instanceof return false for some literals? &quot;foo&quot; instanceof String //=&gt; false &quot;foo&quot; instanceof Object //=&gt; false  true instanceof Boolean //=&gt; false true instanceof Object //=&gt; false false instanceof Boolean //=&gt; false false instanceof Object //=&gt; false  12.21 instanceof Number //=&gt; false /foo/ instanceof RegExp //=&gt; true  // the tests against Object really don't make sense  Array literals and Object literals match... [0,1] instanceof Array //=&gt; true {0:1} instanceof Object //=&gt; true  Why don't all of them? Or, why don't they all not? And, what are they an instance of, then? It's the same in FF3, IE7, Opera, and Chrome. So, at least it's consistent.