Convergence of sequence of means: Define $A_n=\frac{a_1+a_2+...a_n}{n}$, Find $\lim_{n\to\infty} \sqrt{n}(A_{n+1}-A_{n})$ Let $a_n$ be a sequence of real numbers such that $|a_n| \le 1$.    Define $A_n=\frac{a_1+a_2+...a_n}{n}$, Find   $$\lim_{n \rightarrow \infty} \sqrt{n}(A_{n+1}-A_{n})$$   I was thinking of using Stolz Cesaro lemma, but that needs to show that $A_n$ is convergent which means that $a_n$ has to be convergent. But I have no clue how to approach this one.
Compute: $\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)$ where $A_n = \frac{1}{n}(a_1 + a_2 + \cdots + a_n)$ Let $\{a_n\}, {n\geq 1}$, be a sequence of real numbers satisfying $|a_n|\leq 1$ for all $n$. Define  $$A_n = \frac{1}{n}(a_1 + a_2 + \cdots + a_n),$$ for $n\geq 1$. Then find $\displaystyle\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)$ .    I proceed in this way  $$\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)=\lim_{n \rightarrow \infty}\sqrt{n}\left[\frac{1}{n+1}(a_1 + a_2 + \cdots + a_n+a_{n+1})-\frac{1}{n}(a_1 + a_2 + \cdots + a_{n})\right]=\lim_{n \rightarrow \infty}\left[{(na_{n+1}-a_1 - a_2 - \cdots - a_n})\frac{1}{\sqrt{n}(n+1)}\right]$$  Please help me to complete from here
Do paths in Adobe Photoshop have to be closed? I am working to convert a png to a vector graphic.  I am outlining the image using the pen tool.  Photoshop insists that the pen tool produce closed loops in order for me to use the "Stroke path..." tool.  Why is this? Are human bodies programmed to die? Following from this question: :  Is there any evidence that human bodies have systemic self-destruction built into their developmental program? I'm not talking about the cell death response, which I know is an important part of growth, development and cancer prevention.  I've read some things about telomere-shortening but don't know if this is a cause or an effect. I'm working on an app that requires no user input, but I don't want the iPhone to enter the power saving mode.  Is it possible to disable power saving from an app? How to set English as default language instead of Tamil? The default language used in our laptop is Tamil. We don't know how to switch back to English. If we type something we are getting something.   We don't know the font name or version. Even the Menu bar and task bar contents are in Tamil. Reopened questions get a "modified" time stamp in the questions list just showed up on 's front page like this:    ... which makes no sense, because the question hasn't been modified since May 25th, and the person who edited it wasn't CatLord anyway:    Turns out what actually happened was that CatLord cast a fifth reopen vote to reopen the question.    So that stamp in the questions list should probably say this instead:     reopened 4m ago CatLord 10.7k   Or just:     reopened 4m ago If matrix A is invertible, is it diagonalizable as well? If a matrix A is invertible, then it is diagonalizable. Is it true or false? Give an example of a relation R on $A^2$ which is reflexive, symmetric, and not transitive I am just looking for some clarification on this exercise:  Let $A = \{a,b,c,d\}$. Give an example of a relation $R$ on $A^2$ which is reflexive, symmetric, and not transitive.  I understand that if I was only dealing with $A$ and and not $A^2$ I could give an answer similar to:  $R = \{(a,a),(a,b),(a,d),(b,a),(b,b),(c,c),(d,a),(d,d)\}$  Here $R$ is reflexive, symmetric, and not transitive.  What is confusing me is the fact that we have $A^2$. I am not sure what I am missing here. I am looking through my book to see if I can figure this out, but if someone could help shed some light on this it would be appreciated.  Thanks,  Tony Suppose we have two short exact sequences in an abelian category $$0 \to A \mathrel{\overset{f}{\to}} B \mathrel{\overset{g}{\to}} C \to 0 $$ $$0 \to A&#39; \mathrel{\overset{f&#39;}{\to}} B&#39; \mathrel{\overset{g&#39;}{\to}} C&#39; \to 0 $$ and morphisms $a : A \to A&#39;, b : B \to B&#39;, c : C \to C&#39;$ making the obvious diagram commute. The snake lemma states that there is then an exact sequence $$0 \to \ker a \to \ker b \to \ker c \to \operatorname{coker} a \to \operatorname{coker} b \to \operatorname{coker} c \to 0$$ where the morphisms between the kernels are induced by $f$ and $g$ while the maps between the cokernels are induced by $f&#39;$ and $g&#39;$.  It is not hard to show that the morphisms induced by $f, g, f&#39;, g&#39;$ exist, are unique, and that the sequence is exact at $\ker a, \ker b, \operatorname{coker} b, \operatorname{coker} c$. With the use of a somewhat large diagram shown , we can even construct the connecting morphism $d : \ker c \to \operatorname{coker} a$. However, I'm stuck showing exactness at $\ker c$ and $\operatorname{coker} a$. I thought Freyd might have had an element-free proof in his book, but it turns out he proves it by diagram chasing and invoking the Mitchell embedding theorem [pp. 98–99]. Is there a direct proof? Any way to have CPU temperature in the menu bar? I tried  and  but it does not update the temperature and I do not wish to change the fan speed. What is a NullPointerException, and how do I fix it? How does Latex compilers integrate packages into its existing core functions? How do I turn on workspaces? (Why do I only have one workspace?) It appears that Unity on Raring only comes with a single virtual desktop (aka workspaces) by default, and CCSM's Wall plugin has no way to add more. Is this configurable at all? Probability of a 2 pair in poker What is wrong with my method of find the probability of a 2 pair?  For the first card we have 52 choices, one this is picked( call it A) we have 3 choices(different suite than A but this same value) for the second card, ( lets call this card B). Now we have AB, to we have to throw away 2 more cards so we don't have 3 of a kind or a four of a kind. Therefore one our third card we have 48 choices and on our fourth card we have 3 choices again similar to above.  On our last card we only have 44 choices left. So the probablity of getting a 2 pair should be (52*3*48*3*44)/(52 choose 5).  But this isn't correct.     Note: I edited this to "2 pair"; the original version was wrong. Why (and how) did using cat on binary files mess up the terminal? Request of a domain with no virtualhost in apache ends to default virtualhost If I ask Apache for domain.com that has no Virtual Host setup in Apache configuration, Apache will show the default Virtual Host. How can I prevent this from happening or simply drop the connection? Improving Review Queues - Design overview II: Changing review bans and other new features This is one of three project announcements for Improving Review Queues. We’ve summarized the project objectives and goals . As a reminder, this project is still in the . In this post, we are sharing proposed changes to review bans and other new features. We are asking for your feedback before we begin implementation. After we collect community feedback, we will be open to including changes into the next design iteration. Suspension of privileges  Most users learn how to be good reviewers by actually reviewing posts, but sometimes find themselves making wrong decisions. This can lead to a less than obvious suspension of privileges, with too little guidance as to what they did wrong and how they can improve in the future. In the event of a suspension, a user will be notified in the Review Queue dropdown. A post notification will be available with reason and time remaining on the Review Queue main page, along with guidance for continued learning in this area. We’re also proposing softening the language by renaming “review bans” to “review suspensions.” Open questions:  Do you have suggestions of other ways we can notify reviewers of a ban? What information would be most helpful to you in this situation?  New features  My tasks We’re considering introducing a new, additional workflow and means of task discovery. My Tasks is a curated experience that’s currently based on your Watched Tags. This page includes an overview of pending tasks that may require more immediate attention or pique interest in a user’s subject-matter expertise.  Queue filtering We’d also like to add more useful, robust filtering options on all queues. Right now, the filter function is difficult to discover. We want to make this feature more noticeable and add queue-specific options. For example, you will be able to filter by your Watched Tags and sort tasks that are soon to expire out of each queue. Badges Currently, the gold Steward badge is awarded once at 1,000 reviews per queue regardless of whether you’ve performed 1,000 or 10,000 reviews. We propose that reviewers can now earn this badge multiple times, adding more opportunities to earn badges while reviewing - this change is intended to be retroactive. Open questions:  How would you want to be able to customize in My Tasks? Are there other review-centric badges you would like to see added?  In case you missed it... We propose changes to Onboarding and Updated workflows/pathways here: . I have two questions. First, which command lists files and directories, but lists directories first?  Second question: I want to copy a list of files into a single directory, but make the target directory the first filename in the command. This thing came up in a combinatorics course I am taking.  Choose a fixed set of primes $p_1,p_2,\dots,p_k$ and let $A_n$ be number of integers in $\{1,2,\dots,n\}$ which are not divisible by any of the $p_i$'s. $A_n$ is given by $ n - \sum_{1\leq i_1 \leq k} \lfloor \frac{n}{p_{i_1}} \rfloor + \sum_{1\leq i_1 &lt; i_2 &lt; k} \lfloor \frac{n}{p_{i_1}p_{i_2}}\rfloor - \dots $. Now, if $n$ is divisible by each of the $p_i$'s then we have the simpler expression : $A_n = n \prod_{i=1}^{k}(1-\frac{1}{p_i})$(added later : note I am not assuming $n = \prod_{i=1}^{k}p_i$ here, this holds true whenever $n = c \prod_{i=1}^{k} p_i $ for some integer c, since each $\lfloor \frac{n}{p_{i_1} \dots p_{i_j}} \rfloor = \frac{n}{p_{i_1} \dots p_{i_j}}\ )$.  Another student pointed out that even if $n$ does not have $\prod_{i=1}^{k} p_i$ as a factor, the approximation $B_n = n \prod_{i=1}^{k}(1-\frac{1}{p_i})$ to $A_n$ is quite close in some specific cases. It is easy to see that $\lim_{n\to\infty}\frac{A_n - B_n}{n}=0$ as $\lim_{n\to\infty}\frac{1}{n}\lfloor \frac{n}{p_{i_1}p_{i_2}\dots p_{i_j}} \rfloor \to \frac{1}{p_{i_1}p_{i_2} \dots p_{i_j}},$ however that is not strong enough to imply that $A_n - B_n$ will be always close. Is there any way to analyze how well this approximation will perform in general? I am interested in the worst case for small to moderately sized $n$.  Just to give a feel of how close $A_n$ and $B_n$ can get, assuming the set of primes is $\{2,3,7\}$ we have (assuming my program is correct):  n     A(n)     B(n) 17    5    4.86   27    8    7.71   37    11      10.57   107   31      30.57   1111  318     317.43   3001  858     857.43   4007  1145    1144.86   5000  1429    1428.57 . I recently installed an SSD and AFAIK Ubuntu makes a weekly FSTrim on SSD's that support it.  Is there a log file of this trims? Question about Coulomb's law I am just very confsed about how to do these sort of questions where they give you a point in the middle of two charges. The farthest I've gotten in questions relating to coulomb's law, is using the formula \$F = {kq_1q_2 \over d^2}\$ to work out one of the unknowns when given two variables. Can someone please explain how you would do a question like this:  Two point charges of charge -2 µC and -3 µC respectively are separated by a distance of 15 cm. Point X is positioned at an equal distance away from each of the charges.      a. Determine the net electric field at X.     b. Find the point of equilibrium between the charges.