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Proofs involving limsup and liminf I've been working with proofs involving $\limsup$ and $\liminf$, and I'm a bit confused regarding their general methodology. More specifically, I'm unsure about whether my approach to the following problem makes sense.  Problem. Let $(s_n)$ and $(t_n$) be sequences and suppose that there exists $N_0$ such that $s_n \leq t_n$ for all $n &gt; N_0$. Show that $\liminf s_n \leq \liminf t_n$ and that $\limsup s_n \leq \limsup t_n$.   The way I approached it was as follows:   Let $N &gt; N_0$. Then $\limsup_{N \rightarrow \infty} \{ s_n : n &gt; N \} \leq t_n$ as $s_n \leq t_n$, and $\limsup s_n $ is the largest possible limit of a subsequence of $s_n$. As $t_n : n &gt; N$ is (by definition) less than $\limsup_{N \rightarrow \infty} \{ t_n : n &gt; N \}$, the proof is complete.  I'm pretty sure this is incorrect, however, and I'm generally unclear about the method behind such a proof. Any help is appreciated! It seems that I'm missing something about this.  First of all, the series is convergent: $\lim_{n\rightarrow\infty}\frac{2^{-n-1} (n+1)^2}{2^{-n} n^2}=\frac{1}{2}$ (ratio test)  What I tried to do is to find a limit of a partial sum $\lim_{n\rightarrow\infty}S_n$ as follows: $S_n=\frac{\frac{1}{6} n (n+1) (2 n+1)}{\frac{1-\left(\frac{1}{2}\right)^n}{2 \left(1-\frac{1}{2}\right)}}$. Still, the limit is $\infty$ and I'm clearly doing something wrong. Convergence of $U_n=\frac{1}{\sqrt{2n\sigma^2}}\left(\Sigma X_j-\Sigma Y_j\right)$ - central limit theorem Suppose that $U_n=\frac{1}{\sqrt{2n\sigma^2}}\left(\Sigma X_j-\Sigma Y_j\right)$,  where $X_1,X_2,\ldots$ and $Y_i,Y_2, \ldots$ are i.i.d. sequences of random variables with mean $\mu$ and variance $\sigma^2$ and the sequences $\{X_i\}$ and $\{Y_i\}$ are independent.  Then, for all real numbers $x$, show that, as $n\to \infty$,  $$\mathbb{P}(U_n\leq x) \to  \int_{-\infty}^{x} \frac{1}{(\sqrt(2\pi))} \:e^{-\frac 12{u^2}}\mathrm du $$ Approach  I tried to carry out the integration on the right, but this does not integrate nicely. It ends up involving an error function, so I figured this was the wrong was to go.  Should I be looking at convergence in probability to answer this question? Is Chebyshev's inequality relevant here? If so, how and how does that relate to the integral? My office is going to buy a bulk number of desktop PCs (for developmental purposes).  I encouraged them to buy PCs without a pre-installed OS and then install Linux. My colleagues agree with me but they are not regular users of any Linux-based OS, so they want to choose a Linux OS which is best for our purposes and user-friendly and I suggested Ubuntu.  Now I want some data to support my position, what's so good about Linux and specifically Ubuntu? Prove that $a^{2^n}=1 \mod 2^{n+2}$ I would like to prove that  $$a^{2^n}\equiv 1 \pmod {2^{n+2}}$$  I tried induction but could not get it.  Thank you very much! How to remove a filter that is an anonymous object? In my functions.php file I would like to remove the below filter, but I'm not sure how to do it since it's in a class. What should remove_filter() look like?  add_filter('comments_array',array( &amp;$this, 'FbComments' ));   It's on line 88 . Prove inequality $n<3^n$ using mathematical induction Prove that $n&lt;3^n$ where $n \in \mathbb N$,  when $ n=1 $, I have proved it's true. And assumed when $n = p$ , $p&lt;3^p$ is true. Can any body help me in showing that it is true for $n =p+1$ I currently have an Ubuntu-Windows Dual boot. However, I would like to try out a new OS so I burned Linux Mint to a USB stick.   I would like to install Linux Mint and remove Ubuntu while still keeping my windows dual boot. I've already backed up my Ubuntu setup so I can wipe it out.   In the Linux Mint installer, should I choosing the "install alongside windows" option do this? $\mathbb{R}$ with the lower limit topology is not second-countable I am trying to prove that $\mathbb{R}$ with the lower limit topology is not second-countable.  To do this, I'm trying to form an uncountable union $A$ of disjoint, half-open intervals of the form $[a, b)$, $a &lt; b$. Is this possible? I think this would imply the $A$ is open but no countable union of basis elements could coincide with $A$ therefore making the real numbers with the lower limit topology not second-countable.  I think there must exist something like $A$ described above but I am having trouble visualizing it and coming up with a formula to represent it.  Maybe there is some other way to show it is not second-countable. I'm using an old iBook G4 (PowerPC) running Tiger to serve iTunes to the house. For the past week or so, the mac boots fine, and then sometime during the day starts generating   An unexpected error occurred (error code -50).   errors when iTunes attempts to save its library file.  Checking reveals that the boot drive (main drive, in the laptop) behaves as if it is write-protected - I get the same error code when I attempt to create a folder on the desktop, for instance. If I attempt to create files via the terminal, e.g. via     ls > foo.txt I get an "invalid argument" error.   I've set the iBook to run in Target Disk Mode and run Disk Utility to repair the disk (from a Macbook running Snow Leopard) and get "The volume appears to be ok".   I don't have the option in Disk Utility to Repair Permissions in Target Disk Mode. When I ran Repair Permissions on the iBook itself, it came back with no errors, no permissions corrected, but did not in fact fix the problem.  Rebooting always solve the problem - for the next 5-10 hours or so.   Would Disk Warrior help, or some other software fix, or is it time to buy a new drive? (And, in any case, what's going wrong?) Retrieve Preview Image URL of Salesforce Image Files Do indexes work with "IN" clause If I have a query like:  Select EmployeeId  From Employee  Where EmployeeTypeId IN (1,2,3)   and I have an index on the EmployeeTypeId field, does SQL server still use that index? How to enable exFAT in Ubuntu 14.04 So i have this $3\times 3$ matrix  $$     \begin{pmatrix}     -2 &amp; 0 &amp; 0 \\     3 &amp; 1 &amp; -6 \\     0 &amp; 0 &amp; -2 \\     \end{pmatrix} $$  I want to check if the matrix is diagonizable. First thing i do is find the roots : namely $\lambda_1 = −2$ of multiplicity $2,$ and $\lambda_2 = 1$ of multiplicity $1.$  In order to ensure that the matrix is diagonalizable we need to find a basis of $\mathbb{R}^3$ whose elements are eigenvectors for $A.$ In particular, since we have the eigenvalue $\lambda_1=-2$ having multiplicity $2$ we need to check that the dimension of the eigenspace $E(\lambda_1)$ is equal to $2.$ We compute the eigenspaces of $A.$  $\DeclareMathOperator\Nul{Nul}$On the answer sheet is says that we need to find $\Nul(A+2Id)$... but why $2Id?$ Isn't is supposed to be $-2,$ so $\Nul(-2Id-A)$??   I do not understand this $\Nul(A+2Id).$ I always thought that we should compute $\Nul(-2Id-A),$ so what do i need to do ?   $\Nul(-2Id-A)$ gives me a different result from $\Nul(A+2Id).$  Does that mean that since $-2$ has a multiplicity of $2,$ we need to work with $2$ and not $-2$?  Somebody please help with a clear and simple explanation ! Thanks! Make WCF service be accessible through Internet Can a function be made to accept a variable amount of inputs? Which Distribution Does the Data Point Belong to? I have two distributions which are derived from 2 separate sets of data. These distributions are not normal, and it is not clear at this point if they belong to any family of known pdfs (they are not symmetric either). Given a data point, I need to decide which distribution it is most likely to belong to. If these were normal distributions, I could do usual parametric tests and go from there, but in this case I am not sure how to proceed. I searched around a bit, but I couldn't find anything, probably because I am not using the right keywords. Any help is much appreciated.  Edit to clarify: I should have mentioned that it was univariate. I might as well explain the actual problem here as well. We have time-spent data for users on websites. We also have information on whether a user liked or disliked some of those sites (about 4% of all the sites). So we know the distribution of time spent for like and dislike. An obvious question, then, is for a random user who spent x amount of time, are they more likely to have liked the page or dislike the page. Our time spent information is based on seconds, so the distributions are really discreet, but in real life, they are continuous. Prove that $\bigcap_{k = 1}^\infty C_k$ is also compact and connected. Let $X$ be a Hausdorff space and let $C_0 \supset C_1 \supset ...$ be a decreasing sequence of compact connected subsets of $X$.  Prove that $\bigcap_{k = 1}^\infty C_k$ is also compact and connected.  I have the compact part.  Just don't know how to show it is connected. Ambiguity of "quite" The adverb "quite" has the following meanings :     1: wholly, completely ("not quite   finished")      2: to an extreme : positively "quite   sure" —often used as an intensifier   with a "quite a swell guy" "quite a   beauty"      3: to a considerable extent : rather   ("quite near")   My understanding is that #1 and #2 conflict with #3. For example, when you say "She's quite right", do you mean "She's completely right" or "To a considerable extent, she's right"? Notice that in the latter case you're not saying that she is completely right.  Another example is "I'm quite sure". Does it necessarily mean "I'm completely sure"?   Does the word really have conflicting meanings that can possibly cause ambiguity? Or am I missing something? Is there a nice example of integral domain $R$ and $s,t \in R$, with $Rs \cap Rt$ not principal but lcm($s,t$) exists? I am currently searching for an integral domain $R$ which allows lcm, but where the intersection of two principal ideals is not necessarily a principal ideal (or at least an example with two certain elements).   As converse example (where the sum of two ideals is not principal but $R$ allows gcd) is $\mathbb Z[X]$. It allows gcd, but obviously $\left&lt;X+1,2\right&gt;$ is not principal.  An example of an integral domain where the intersection of two principal ideals is not principal would be $\mathbb Z[\sqrt{-5}]$ with the ideal $\left&lt;2\right&gt; \cap \left&lt;1+\sqrt{-5}\right&gt;$, but this doesn't allow lcm.  Since I can't find a nice example, I put the question here. Maybe someone has already seen this? (I am not completely sure, one even exists, but pretty certain there is an example. In the case my hunch is wrong, I would be happy, if someone could proof me wrong.)