How do I find using the Squeeze theorem  $$\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{\sqrt{n^2+k}} \;,$$  using the fact that  $$ \lim_{n\to \infty}\frac{n}{\sqrt{n^2+n}}=1.$$  Thank you very much for your help,   C.G
Direct way leads to nowhere.  $$\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} } = \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{n \sqrt {1 +\frac{i}{n^2}} } = $$  $$ = \lim_{n\rightarrow \infty} \frac{1}{n} \cdot \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {1 +\frac{i}{n^2}} } = 0 \cdot \sum^{n}_{i=1} 1 = 0 \cdot n $$  So, I tried to apply squeeze theorem:  $$\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} } \leq \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{n}  = 1$$  Could you help me to figure out the lower bound?
If $f,g : [a,b] \to (0,\infty)$ are continuous functions, prove that there exists $c \in (a,b)$ such that: $$\int_{a}^{c} f(x) \ \mathrm dx = \int_{c}^{b} g(x) \ \mathrm dx$$ What's the easiest way to download all albums from Google Photos? I have around 950 albums at Google Photos.      I want to back them up now, but organised by albums, not by date.   Whenever I create a backup from within Google Photos, the backup splits my photos by date and adds metadata here and there.  What I am doing now is to click on every album, and choose "Download all". But this will take me ages.  Is there a way through Google Drive or another app to have a easier job? Is the axiom of choice used in the proof that every open set is the union of basis elements (Mukres, Lemma 13.1) How to prove that this set is closed? Suppose $Y$ be an ordered set in the order topology. Let $f, g: X \to Y$ be continuous. How to show that the set $\{x| f(x) \leq g(x)\}$ is closed?  This is a excercise from munkres. Maybe trying to show that the set $f(\{x| f(x) &gt; g(x)\})$ is open is suffice. But I could not figure out the connection between this set and the continuity of the two functions.Could you give a hint?  Thanks. Difference between a Structure and a Union On convergence of Bertrand series $\sum\limits_{n=2}^{\infty} \frac{1}{n^{\alpha}\ln^{\beta}(n)}$ where $\alpha, \beta \in \mathbb{R}$ Study the convergence of $$\sum_{n=2}^{\infty}  \frac{1}{n^{\alpha}\ln^{\beta}(n)}$$ where $\alpha, \beta \in  \mathbb{R}$   I have proved that:   This series diverges when $\alpha \leq 0$. This series converges when $\alpha &gt; 1, \beta &gt; 0$ This series diverges when $0 &lt; \alpha &lt; 1, \beta &gt; 0$ This series converges when $\alpha = 1, \beta &gt; 1$      Question: What happens when $\alpha &gt; 0$ and $ \beta &lt; 0$?   There are other questions on MSE which ask about this series, but this question is distinct because   I would like an argument which does not rely on the integral test for series convergence, and this question considers all real $\alpha$ and $\beta$, while other questions ask only about $\alpha, \beta &gt; 0$, where we can apply Cauchy condensation criterion I am trying to monitor multiple JVMs running on different servers through an ssh tunnel.  Thanks to one of the UNIX gurus here, I have managed to get this running for a single server by:   Run jstatd on the target server Set up Visual VM to use 9696 as its socks proxy port. On my local PC, run:  ssh -L 2222:server1:22 bastion-host  On my local PC, run:  ssh -o port=2222 -D 9696 -L 1099:localhost:1099 localhost    This does the trick. Now when I try and tunnel to a second server I try:   On my local PC, run:  ssh -L 3333:server2:22 bastion-host  On my local PC, run:  ssh -o port=3333 -D 9696 -L 2099:localhost:1099 localhost    However, the last step complains with:  bind: Address already in use channel_setup_fwd_listener: cannot listen to port: 9696   Has anyone managed to do something similar?  Update:  The reason this is so complex is that jstatd is an RMI server application:    As with RMI applications, these register with an rmiregistry. Using rmi through a firewall requires me to use SOCKS as described here:    Unfortunately, visualvm only lets me set the SOCKS proxy port once, and the -D option won't allow me to forward the same local port to both servers... Please note: I posted this question about an hour ago on my account (I just made it as I am new to the site). I revisited my post and saw that I had not given definitions of what I was asking (I thought they were common notations) and it would not let me edit the post for some reason so I will ask here now and hopefully someone can delete the other thread.  How to prove for a finite group $G$ that $|\langle g \rangle|=o(g)$?  Here are the definitions:  Let $G$ be a finite group and $g \in G$  $o(g)$ is the smallest $m \in \Bbb{N}$ such that $g^m=e$ (the identity element of the group). (If such an $m$ exists)  $\langle g \rangle=\{g^m : m \in \mathbb{Z}\}$  I really don't know how to show that this is true I tried to say let $o(g)=m$ and then show that $|\langle g \rangle|$ has exactly $m$ elements but I wasn't even sure it did because it would contain the elements $g^0,...g^{m-1}$ so that would mean it's true but what about $g^{-1}$ and the other elements with negative coefficients so it seems like it would actually be bigger than $m$?  Thanks. Restoring default toolbar layout in Blender 2.81 Why does increasing the number of features reduce performance? I'm trying to gain an intuition as to why increasing the number of features could reduce performance.  I'm currently using an LDA classifier which performs better bivariately among certain features but worse when looking at more features.  My classification accuracy is performed using a stratified 10-fold xval.    Is there a simple case of when a classifier would work better univariately than bivaraiately to gain a somewhat physical or spatial intuition of what is happening in these higher dimensions? This question pertains to L1 &amp; L2 regularization parameters in Light GBM. As per official documentation:  reg_alpha (float, optional (default=0.)) – L1 regularization term on weights.  reg_lambda (float, optional (default=0.)) – L2 regularization term on weights  I have seen data scientists using both of these parameters at the same time, ideally either you use L1 or L2 not both together.   While reading about tuning LGBM parameters I cam across one such case: Kaggle official GBDT Specification and Optimization Workshop in Paris where Instructors are ML experts. And these experts have used positive values of both L1 &amp; L2 params in LGBM model. Link below (Ctrl+F 'search_spaces' to directly reach parameter grid in this long kernel)      I have seen same in XGBoost implementations.  My question is why use both at the same time in LGBM/XGBoost.  Thanks. I understand that faster-than-light communication is impossible when making single measurements, because the outcome of each measurement is random. However, shouldn't measurement on one side collapse the wave function on the other side, such that interference effects would disappear? Making measurements on "bunches" of entangled particles would thus allow FTL communication, by making observed interference effects appear or disappear. How does such an experiment not:  1) Clearly imply that faster-than-light communication is possible?  or  2) (if #1 is rejected) Imply that measurement of one half of an entangled pair does not cause the collapse of the other half's wave function.  Why doesn't this thought experiment clearly show that if we maintain that FTL communication is ruled out, we must also rule out "universal collapse" in the Copenhagen interpretation?  EDIT: Here is an example of an explicit experiment (though I think experts could come up with something better):  You can entangle a photon with an electron such that the angle of the photon is correlated with the electron's position at each slit of a double slit experiment. If the photon is detected (it's outgoing angle measured), then which-path information is known, and there is no interference. If the photon is not detected, the interference remains.   The experiment is designed such that the photon and electron go in roughly opposite directions, apart from the tiny deflection which gives which-path information. You set up a series of photon detectors 100 ly away on one side, and your double slit experiment 100 ly away in the opposite direction. Now you produce the entangled pairs in bunches, say of 100 entangled pairs, each coming every millisecond, with a muon coming between each bunch to serve as a separator.   Then the idea is that someone at the photon detector side can send information to someone watching the double-slit experiment, by selectively detecting all of the photons in some bunches, but not in others. If all of the photons are detected for one bunch, then the corresponding electron bunch 200 ly away should show no interference effects. If all of the photons are not detected for one bunch, then the corresponding electron bunch 200 ly away would show the usual double-slit interference effects (say on a phosphorus screen). (Note that this does not require combining information from the photon-detector-side with the electron-double-slit side in order to get the interference effects. The interference effects would visibly show up as the electron blips populate the phosphorus screen, as is usual in a double-slit experiment when which-path information is not measured.)  In such a way the person at the photon detectors can send '1's and '0's depending on whether they measure the photons in a given bunch. Suppose they send 'SOS' in Morse code. This requires 9 bunches, and so this will take 900 milliseconds, which is less than 200 years. The point is that such an experiment would only work if you assume that the measurement of the photon really does collapse the wave function nonlocally. Evaluating this integral $ \small\int \frac {x^2 dx} {(x\sin x+\cos x)^2} $ The question:      Compute$$ \int \frac {x^2 \, \operatorname{d}\!x} {(x\sin x+\cos x)^2} $$   Tried integration by parts. That didn't work.  How do I proceed? With the last update of Fallout Shelter, Bethesda has added these "limited time" quests.  Question is, how limited is that time exactly? What happens when the quests "end", will it still be possible to get the benefits? How do I add a .sty file to my MacTeX/TeXShop installation? I've searched online, but I have not found any clear instructions how to permanently add .sty files to my TeXShop installation. I'm using a Mac with OSX (TeXShop is a mac only program).  What is the best way of doing this? The one comprehensive set of instructions I found .   I've got as far as extracting the files (running LaTeX on the .ins and . dtx files) but after that I don't know what to do. I also don't understand how to update the index of my distribution, and whether there are particular OSX specifics (the instructions are for Windows). How to get a function name as a string? In Python, how do I get a function name as a string, without calling the function?  def my_function():     pass  print get_function_name_as_string(my_function) # my_function is not in quotes   should output "my_function".  Is such function available in Python? If not, any ideas on how to implement get_function_name_as_string, in Python? Solve for $f(x)$ if $f(f(x))=6x-f(x)$ If $f: [0,\infty) \rightarrow [0,\infty)$ and   $f(f(x))=6x-f(x)$ $f(x)&gt;0$ $  \forall x \in (0,\infty) $   Find f(x) What is the name of the black&white movie about a mirror Earth (universe)? Astronauts in a spaceship have an accident and fall back to the Earth. They are tried in court. It turns out to be a mirror Earth. As I remember (though maybe I'm not totally right) only one professor knows something about that and in the final scene he crashes into a mirror with his wheelchair. In another scene an astronaut realizes that his name can be read from the mirror directly.  Does anybody remember the movie? computing the orbits for a group action I'm trying to install the flashybrid but when I enter the apt-get install flashybrid command, I get this error:  root@user-desktop:/home/user# apt-get install flashybrid Reading package lists... Done Building dependency tree Reading state information... Done flashybrid is already the newest version. 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded. 1 not fully installed or removed. After this operation, 0B of additional disk space will be used. debconf: DbDriver "config": /var/cache/debconf/config.dat is locked by another process: Resource temporarily unavailable Setting up flashybrid (0.15+nmu2) ... debconf: DbDriver "config": /var/cache/debconf/config.dat is locked by another process: Resource temporarily unavailable dpkg: error processing flashybrid (--configure): subprocess post-installation script returned error exit status 1 Errors were encountered while processing: flashybrid E: Sub-process /usr/bin/dpkg returned an error code (1)   I also tried this one to find what is already using that file:  root@user-desktop:/home/user# lsof | grep /var/cache/debconf/config.dat lsof: WARNING: can't stat() fuse.gvfs-fuse-daemon file system /home/user/.gvfs Output information may be incomplete. dpkg-prec 4672 root 4rW REG 8,6 39658 1697827 /var/cache/debconf/config.dat   But I don't know how to fix this.