Let $(x_n)$ be a sequence satisfying   $1 \leq x_n \leq 2$ for all $n \geq 1$, $\lim_{n\to\infty} x_{n+1}/x_n = 1$.   Is this enough to show that $(x_n)$ converges?  This is motivated by  , as well as . In both cases, we can show convergence but making use of the actual definition of the sequence, not simply the two properties listed above.  I have attempted to show that this sequence is Cauchy but the details elude me.
Does the condition $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent?  When I say convergent I mean to $L\in \mathbb{R}$ or to $\infty$ or to $-\infty$  I have a gut feeling this statement is false, just can't find a series which contradicts it.  Thanks a lot
It would help everyone if we provided a mechanism that people could use to flag their post if English is language they are not comfortably fluent in. A lot of non-English speakers have to go to English forums for tech support and find it embarrassing to ask questions in broken Englis.   Worse, people reading a post often cannot tell if a badly-worded post results from a poor English or just laziness and sloppiness on the part of a native English speaker. Personally, I will take the time to puzzle out a poorly-worded post if I know the author is struggling with English, but if I think they're just lazy and careless I won't bother.   A simple check box could flag the author as a non-English speaker and allow everyone to adjust accordingly.   Edit01:  Some answers below have accused me of being 'elitist'. I'm not sure where they got the idea that I was sneering down at anyone because their abilities at either English or coding. That was not my intent.   When I said I would not bother with post that appeared to incoherent owing to laziness or carelessness, I am not being elitist. I'm giving priority in my time budget to people who are respectful enough of my time to spend the their time writing as clear a question as their grasp of English and coding allows. Why should I spend time answering a careless and slapdash post when I could spend that time helping posters who took more of their time writing their questions?   I am perfectly willing to burn a lot of time helping someone struggling to write code in their non-native language. I have a hard enough time coding myself in English that I can't even imagine what a dog's breakfast I would make of it if I had to code in, say, Korean.    I have seen many instances in which questions have been downvoted when a quick glance at the poster's profile gives a strong hint they are not native English speakers. I would like a system that would spare those posters from being downvoted or ignored by using an interface element that clearly communicates in a standardized way whether the poster natively speaks English or not.   I don't see what's elitist about that.    Edit02:   (original unedited) I'm talking about:     Hy, I have a childe view that put it   self over other brother views. The   problem is that when i touch it the   brother also reacts at the touch   event. How can I disable touch   propagation on the underlieng brother   views?   I'm not even exactly sure what he is describing. I can guess that he has a sibling view that extends over its siblings but I'm not sure. If I edited it, I might not end up asking the same question the poster intended.   To answer the question, I will either have to (1) first ask for more detail and then keep checking back until I get it or (2) try to answer what I think the question is and possibly mangle it then have to come back and edit my answer all the while risking a downvote because I answered the wrong question.   I will take the time and the risk of down vote if the poster is not a native English speaker (or is a child) and the confusing question represents their honest best effort. If they're just being lazy, I have other people to help.   I would like a mechanism that would help me decide how to allocate my time and would protect the non-English speaking posters from unjust downvotes. Spatial query to return multiple values I have a woodland scheme advice layer that has a polygon for each area of advice (brown layer in the picture). I also have a polygon layer of UK local authorities (red boundary layer in the picture). I would like to perform a spatial query which returns the local authority name for each woodland advice scheme. This would work fine if each woodland advice scheme was found in just one local authority boundary. However many are found across a boundary of two different authorities. For instance, one scheme sits across the boundaries of the authorities of Fife and Angus. The scheme is the yellow selected polygon in the picture and the red line is the authority boundary. How do I get the spatial join to return a value of Fife; Angus (or something similar to identify two values) instead of just Fife or Angus. I'm trying to create a command, which calls on an environment, which defines a custom column definition with one argument using \newcolumntype, and defines a second command, containing parameters for a table column.  Then, still within the first command, I begin the tabular environment and call the custom column, with the second command as its argument. I always get an undefined control sequence error, though. Any thoughts on why that might be?  I'm sure that was clear as mud, but I've cut it down to a minimal example, which follows:  \documentclass{article} \usepackage{array} \usepackage[table,dvipsnames]{xcolor} \usepackage{colortbl}  \newenvironment{tablecolor}{     \newcommand{\tlight}{\columncolor{SkyBlue!30}\color{black}}     \newcolumntype{C}[1]{&gt;{##1}c} }{}  \newcommand{\tablegoals}{ \begin{tablecolor} \begin{tabular}{l|C{\tlight}} Goal 7  &amp; \\ \end{tabular} \end{tablecolor} }  \begin{document}     \tablegoals \end{document}     Here's my second attempt. Different approach, same undefined control sequence error with \columncolor.  \documentclass{article} \usepackage{array} \usepackage[table,dvipsnames]{xcolor} \usepackage{colortbl} \usepackage{ifthen}  \newenvironment{tablecolor}{     \newcolumntype{C}[1]{&gt;{         \ifthenelse{\equal{dark}{##1}}{             \columncolor{Cyan}\color{white}             }{             \columncolor{SkyBlue!30}\color{black}         }     }c} }{ }  \newcommand{\tablegoals}[1]{     \begin{tablecolor}         \begin{tabular}{l|C{dark}}             Goal 7  &amp;   #1         \end{tabular}     \end{tablecolor} }  \begin{document}     \tablegoals{Check mark} \end{document} How to start Skype 4 with Ubuntu 12.10? I've installed Skype and it's running all OK but I want it to start automatically when I log with my user.  How can I do that? When a word ends in 's' or 'x', do you add 's or just an '? 1) Alex's house      2) Alex' house     When the noun ends with the letter 's' or 'x', do I need to put 's' after an apostrophe or not?  I remember I read some rules related to this in my school grammar book, but now I've forgotten it. Connected to wifi successfully, but no internet connection Are the rationals a closed or open set in $\mathbb{R}$? There are two checkboxes for Run As Administrator in the Properties dialog for a shortcut in Windows 7. One is under the Shortcut tab in the Advanced Properties dialog, and the other is under the Compatibility tab.  I've seen this under Windows 7 and Windows Server 2008. It might be in Vista and other versions, but I'm not sure.  What is the difference between these two settings? Mixed content on many SE sites Many SE sites (like math.SE) have a problem with mixed content, which causes display bugs in recent browsers. This is particularly evident with math.SE (because of the huge use of MathJax) and with chat sections.  Will this be fixed?  How to reproduce  Visit  or   What happens  Mixed content is blocked, output is missing pieces of code (MathJax):    This picture also includes my main Firefox addons (top right), it relevant. It shows missing styles:    The white on the left is my tab list. Can newsletters take into account a user's ignored tags on that site? I have seen several questions that discuss using your "Favorite" tags to construct a newsletter.   I'm looking for a similar request - but for ignored tags.  I figure it might make sense to split this request from the "favorite" requests because it has different implementation.  Can we exclude questions on the "ignore" tags lists in the newsletter?   For example, I subscribe to the Board and Card Games newsletter.   Currently (and frequently) 6 of the 7 questions are about magic-the-gathering, a tag I have in my ignored list.     I now get a newsletter where at best 1/7 questions is interesting and worth clicking on.  Related questions about using "Favorites": Is errno thread-safe? If $x,y,z\in\mathbb R\setminus \{1\}$ and $xyz=1$, prove that $\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1$. If $x,y,z\in\mathbb R\setminus \{1\}$ and $xyz=1$, prove that $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1$$  Without using calculus. There are a few ways I've tried solving this:  $1)$ We could try using the Cauchy-Schwarz inequality: $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge \frac{(x+y+z)^2}{(x-1)^2+(y-1)^2+(z-1)^2}$$  But it's apparent that nothing's useful here.  $2)$ We could use AM-GM as well: $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 3\sqrt[3]{\frac{x^2y^2z^2}{(x-1)^2(y-1)^2(z-1)^2}}=\frac{3}{\sqrt[3]{(x-1)^2(y-1)^2(z-1)^2}}$$  So we only have to prove that: $$\sqrt[3]{(x-1)^2(y-1)^2(z-1)^2}\le 3$$  We could raise both sides to the power of 3: $$(x-1)^2(y-1)^2(z-1)^2\le 27$$  But this inequality doesn't hold.   $3)$ We could try cleaning the denominators by multiplying both sides by $(x-1)^2(y-1)^2(z-1)^2$. After a bunch of expanding and simplifying we get that: $$x^2y^2+y^2z^2+x^2z^2-6(xy+yz+xz)+2(x+y+z)+9\ge 0$$  I can't tell so easily whether the inequality is true or not. You could help me out on this one. Just don't forget that $x,y,z\in\mathbb R\setminus\{1\}$ and so we can't just simply use AM-GM, unless we're using it for squares that have to be non-negative. A formal name for "smallest" and "largest" partition Consider a set $A=\{1,2,3,4,5\}$, is there any terminology for the following partitions of $A$ ?  (1) $A=\{ \{1\},\{2\},\{3\},\{4\},\{5\} \}$  (2) $A=\{\{1,2,3,4,5\}\}$. booting windows 10 spends too long In august 2015, I have upgrade my laptop from windows 8.1 to windows 10. But it spends always 10 minutes to boot it. It is not normal that it spends so long. What can I do for boot him faster? Below some information about my laptop.   age: +/- 2 year (on august 2015) edition: Windows 10 Home   File trace are removed.  Update:    I must also say that always there comes a windows update, I have the same problem again form slow booting. Is it good that I turn off the updates? If yes, how can I do that? Also it is in periods that my pc booting to slow. Some time it is like normal (a few seconds) and other times a few minutes. I don't know how it comes that it is fast or slow. Which parts are from Eminems real life? In "" we saw how Bunny Rabbit got into the rapping gang and won the heart of millions of rap lovers. Now in Eminem's original life too similar thing happened. We know how Dr. Dre helped him getting into an industry which is dominated by mainly black people in USA. In fact,  called the film semi-autobiographical. So I want to know which parts are direct picturisations of his real life? Why are there no animals in "The Matrix"? What is the difference between Reduce and Solve? Can I gift a game via the PlayStation Network? Is it possible to "gift" a game (e.g., Castle Crashers) to someone through the PlayStation Network?  I don't have a PS3 but was considering buying Castle Crashers as a gift for a friend so that we could play co-op.  Is this possible? Let the ring of fractions be denoted as $S^{-1}R$.       Proposition (from Robert Ash's textbook "Basic Abstract Algebra")      Define $f: R \rightarrow S^{-1}R$ by $f(a) = a/1$. Then f is a ring homomorphism. If S has no zero divisors then f is a monomorphism, and we say that R can be embedded in $S^{-1}R$. In particular:      (i) A commutative ring R cna be embedded in its complete (or *full) ring of fractions ($S^{-1}R$, where S consists of all non-divisors of zero in R).       (ii) An integral domain can be embedded in its quotient field.    Why does the proposition say "If S has no zero divisors..."? If S really had zero divisors, then this wouldn't even be properly defined, right? Because we would have a/0 for some a in R. Maximum of $|\sin(z)|$ as $\{z: |z| \leq 1 \} $  So according to the Maximum Principle, the maximum is when $|z|=1$. I tried using the fact that $\sin z= \dfrac {e^{iz}-e^{-iz}}{2i}$, but didn't know how to continue from there. Setting $z=e^{it}$ doesn't help much.  I saw this question was answered before, however the answer uses hyperbolic functions ($\sinh, \cosh$) which I haven't actually studied.  Thanks in advance for any assistance!