minimum shutter speed for point and shoot camera
How do I use the different shutter speeds my camera offers?
What makes scaling pixel art different than other images? Recently I was coaching a developer who's new to making pixel art games, and realized there's some "conventional wisdom" around scaling pixel art that's not obvious to a newcomer, and might not be well documented. So, here's an attempt to fix that....     I've heard I should only scale pixel art by 200%, 300% etc. not in-between sizes like 120%   Why is that? Shouldn't scaling "just work" the way it does with other images? What if I really need to make my character just a fraction larger or smaller to get the right gameplay or look? How do I fix this?  I'm told it's "bad" to use pixel art at different scales in one game - why?    For example, if I scaled up my character so each sprite texel is 4x4 screen pixels, while my background art is scaled up so each sprite texel is 6x6 screen pixels. Isn't a pixel a pixel at the end of the day? Finding the convergence interval of $\sum_{n=0}^\infty\frac{n!x^n}{n^n}$. I want to find the convergence interval of the infinite series $\sum\limits_{n=0}^\infty \dfrac{n!x^n}{n^n}$. I will use the ratio test: if I call $u_n = \dfrac{n!x^n}{n^n}$, the ratio test says that, if the following is true for some values of $x$, the series will be convergent for these values of $x$: $$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|&lt;1$$ So, I will first calculate the value of $\left|\dfrac{u_{n+1}}{u_n}\right|$: $$\left|\dfrac{\dfrac{(n+1)!x^{n+1}}{(n+1)^{n+1}}}{\dfrac{n!x^n}{n^n}}\right|=\dfrac{(n+1)!|x|^{n+1}}{(n+1)^{n+1}}\times\dfrac{n^n}{n!|x|^n}=\frac{(n+1)n^n|x|}{(n+1)^{n+1}}=|x|\left(\frac{n}{n+1}\right)^n$$ So, $\lim\limits_{n\to+\infty}\left|\dfrac{u_{n+1}}{u_n}\right|$ becomes: $$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|=\lim_{n\to+\infty}|x|\left(\frac{n}{n+1}\right)^n=|x|\lim_{n\to+\infty}\left(\frac{n}{n+1}\right)^n$$ Now I must evaluate the value of $\lim\limits_{n\to+\infty}\left(\dfrac{n}{n+1}\right)^n$. For this, let $y = \left(\dfrac{n}{n+1}\right)^n$; so, instead of calculating $\lim\limits_{n\to+\infty}y$, I will first calculate $\lim\limits_{n\to+\infty}\ln y$: $$\lim_{n\to+\infty}\ln y=\lim_{n\to+\infty}\ln \left(\dfrac{n}{n+1}\right)^n=\lim_{n\to+\infty}n\ln\left(\frac{n}{n+1}\right) =\lim_{n\to+\infty}\frac{\ln\left(\frac{n}{n+1}\right)}{\frac{1}{n}}$$ Applying L'Hôpital's rule: $$\lim_{n\to+\infty}\frac{\ln\left(\frac{n}{n+1}\right)}{\frac{1}{n}} =\lim_{n\to+\infty}\frac{\frac{1}{n(n+1)}}{-\frac{1}{n^2}} =\lim_{n\to+\infty}\left(-\frac{n}{n+1}\right)=-1$$ Now, since we know that $\lim\limits_{n\to+\infty}\ln y = -1$, we have that: $$\lim_{n\to+\infty}y=\lim_{n\to+\infty}e^{\ln y} = e^{-1} = \frac{1}{e}$$. Substituting this back into the expression $\lim\limits_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right| = |x|\lim\limits_{n\to+\infty}\left(\frac{n}{n+1}\right)^n$, we have that the limit of $\left|\dfrac{u_{n+1}}{u_n}\right|$ as $n\to+\infty$ is: $$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|=\frac{|x|}{e}$$ Therefore, the series will certainly be convergent for the values of $x$ for which $\dfrac{|x|}{e}&lt;1$, that is, $|x|&lt;e$. So, I know that the series is convergent for $-e &lt; x &lt; e$, but I have to test whether the series is convergent at $x = e$ or $x = -e$. That is, I have to test whether $\sum\limits_{n=0}^{\infty} \dfrac{n!e^n}{n^n}$ and $\sum\limits_{n=0}^{\infty} \dfrac{(-1)^nn!e^n}{n^n}$ are convergent. Since these limits don't approach zero, I know they are both divergent, but I'm not sure how to find the limits, because of the factorial function. Also, I can't use integral test here, because of the factorial. Probably I should use comparison test, but I haven't found any divergent series to which to compare it. Any hints? Thank you in advance.  Edit: Using the suggestion by Ragib Zaman in the answer below, since the Taylor polynomial $P_n(x)$ of $e^x$ at $a=0$ is $$e^x = 1 + x + \dfrac{x^2}{2!} + \cdots + \dfrac{x^n}{n!}+\cdots,$$ if we substitute $n$ for $x$ we see that $e^n&gt;\dfrac{n^n}{n!}$; therefore, $\dfrac{n!e^n}{n^n} &gt; 1$, and, thus, we show that $\sum\limits_{n=0}^{\infty} \dfrac{n!e^n}{n^n}$ is divergent, because its term doesn't approach zero. $\sum\limits_{n=0}^{\infty} \dfrac{(-1)^nn!e^n}{n^n}$ is also divergent, because, although the absolute value of the ratio between two successive terms, $|e|\left(\frac{n}{n+1}\right)^n$, approaches 1 as $n\to\infty$, it approaches 1 from values bigger than 1; therefore, the absolute value of a term is always greater than the absolute value of the previous term. How do I install Skype? I want to install Skype, but I can't locate it in Software Center or by using Synaptic Package Manager. Can anyone help me? How do you find all subclasses of a given class in Java? How does one go about and try to find all subclasses of a given class (or all implementors of a given interface) in Java? As of now, I have a method to do this, but I find it quite inefficient (to say the least). The method is:    Get a list of all class names that exist on the class path Load each class and test to see if it is a subclass or implementor of the desired class or interface   In Eclipse, there is a nice feature called the Type Hierarchy that manages to show this quite efficiently. How does one go about and do it programmatically? Can I safely charge my laptop with a non-standard, third-party charger? I have a Toshiba Satellite laptop. My charger has stopped working.  I have access to a Lenovo charger. Can I use this charger on my laptop? PHP in_array() / array_search() odd behaviour Check if one list contains element from the other why does fdisk add one additional sector when creating partitions? I'm playing around with fdisk and I tried to create a partition on an empty USB drive. Here's how that went: Command (m for help): n Partition number (1-128, default 1): 1 First sector (34-61187102, default 2048): 2048 Last sector, +/-sectors or +/-size{K,M,G,T,P} (2048-61187102, default 61187102): +30000000           Created a new partition 1 of type 'Linux filesystem' and of size 14,3 GiB.  so just one partition with 30 million sectors. However, when I check how many sectors it contains I get:  I'm checking with a different program and I see again that additional sector:  Sorry for the pettiness :) I'm learning. Wifi losing connection, weak signal, Intel 7260 adapter I yesterday upgraded from Windows to Ubuntu 15.04 but I have problem with my wireless adapter Intel Corporation Wireless 7260 (rev 73). It randomly loses internet connection. Restarting adapter always helps. (same it did with Windows, except that sometimes restarting adapter didn't help).   Additionally, I have weaker connection then I had with Windows.  Thank you! I tried to log into Minecraft today to find that my password isn't working. I tried to change it, but I can't login to my email because I forgot the answer to my security question. I tried changing the email to a different one, but that didn't work for some reason (it kept saying "invalid email"). I don't have my transation ID because I forgot which credit card I used.  Does this mean that I have to buy a new Minecraft account? Changing affiliation on publication Where do I ask questions about the Windows operating systems? Be it Win 8, 7 or XP.  The questions that I want to ask may not be strictly related to a problem I am facing, but related to files and features of the OS itself or asking just because of my curiosity. For example, I want to know the purpose of the file "Bootvid.dll" in a Win 8 OS.  Or, for example, I want to what "hal.dll" is.  Are these questions appropriate for Stack Overflow or Super User? If not, please recommend some SE sites in which I can post such questions, because I am interested in getting into the guts of the Windows OS. Open to all suggestions. Sending a simple TCP message using Netcat I am trying to use netcat to send a simple message over TCP, e.g. I want to send the digit 1.  So I understood that I write the following in the terminal, after installing netcat  netcat [ip-address] [port]  But how do I specify the message to be sent? i.e. where to I write "1" ? How to check who owns the Slack's workspace? I'd like to check who owns (or at least some kind of public profile page) some workspace based on its subdomain (e.g. xyz.slack.com).  Is it possible? I admit this feature request is probably somewhat limited in useful scope, but I'm throwing it out there anyway. Inspired by , and because I want to use it on , I'm requesting that name be supported on a tags in posts.  On very long answers, such as the closing/migration guidance answer, this would allow direct linking to the specific closure reason. This would then allow us, when someone , to link directly to the appropriate reason and description thereof.  I recognize the limited scope of this, however, I have seen other long answers that could stand to have that kind of "deep" linking ability as well.  (The original incarnation of this post had either name or id, but preferenced name. Per Koper's answer, which I agree with, I took out the idea of supporting id, because Koper's right -- too dangerous.) I just noticed . Hey! I'm there! But that's beside the point.  The point is that Nick Craver is there too. Nick seems to like really long tag names, so his profile shows up like this:    Akk! Diamonds everywhere!  It seems that the page is optimized for only two lines of tags per user. One tag per line means three lines. Three lines means that Nick's reputation (tag) is cut in half.  Let's not cut Nick's rep (tag) in half on the Users page, no? Which SUSY models are affected by the recent LHCb result? Can you be sued for using publicly used images on a blog without copyright permission? Is there a "sentence stretch" that acts horizontally in a similar way as \setstretch acts vertically? It confuses/blows my mind that the Binomial has variance proportional to $p(1-p)$. Equivalently, the Fisher information is proportional to $\frac{1}{p(1-p)}$. What is the reason for this? Why is the Fisher Information minimized at $p=0.5$? That is, why is inference most difficult at $p=0.5$?  Context:  I'm working on a sample size calculator, and the formula for $N$, the sample size needed, is an increasing factor of $p(1-p)$, the result of a variance estimation in the derivation.