Show if the following series $$ \sum_{n=2}^{\infty}\frac{(-1)^n}{(-1)^n + n} $$ are absolutely convergent, conditionally convergent or divergent. I think I succeeded in proving that it is NOT absolutely convergent. My attempt: $$ \sum_{n=2}^{\infty}\left|\frac{(-1)^n}{(-1)^n + n}\right| = \sum_{n=2}^{\infty}\frac{(-1)^n}{\left|(-1)^n + n\right|} $$ Then we have $$ \lim_{n\to\infty}\frac{\frac{(-1)^n}{\left|(-1)^n + n\right|}}{\frac{1}{n}} = \lim_{n\to\infty}\frac{n}{(-1)^n+n} = 1 $$ Therefore $$ \sum_{n=2}^{\infty}\frac{(-1)^n}{\left|(-1)^n + n\right|}\ \text{is convergent}\ &lt;=&gt; \sum_{n=2}^{\infty}\frac{1}{n}\ \text{is convergent.}\ $$ The last series are NOT convergent, so our series are NOT absolutely convergent. I'm still stuck on proving if it is conditionally convergent or divergent. intuitively, I think It's divergent, but I need a formal proof. It crossed my mind that I can show $a_n = \frac{1}{(-1)^n +n}$ is NOT monotonously decreasing so It's divergent(Leibniz). But I think Leibniz's theorem doesn't work in this direction. So I'm still stuck. Looking forward for any ideas!
How do I check the series $\sum_{n=2}^{\infty} \frac{(-1)^n }{n+(-1)^n}$ for absolute convergence/conditional convergence ?  How can I solve this problem?
I have a mouse with lots of buttons, but it's not a mainstream make like Logitech. For Windows, I have a driver that lets me assign actions like close-window (Ctrl+W) or next-tab (Ctrl+Tab), but I don't have a Linux driver. Since Linux is so flexible, I thought perhaps there is a general way to do this, regardless of brand?  Update: Based on input from Cyrex, I installed and ran sudo apt-get install btnx which found several but not all mouse buttons. Found: left, right, wheel, wheelclick, thumb fwd, thumb back. Not found: wheel left, wheel right, thumb middle button. Vendor ID is 0x04d9, Model ID is 0xa015.  Update 2: In System>Prefs>Mouse there's a lightbulb icon for testing double-click speed. Every working button can turn the bulb on&amp;off, but the missing buttons can't. It would seem that Ubuntu isn't aware of these buttons and thus doesn't register their clicks. I guess I need to hunt for a driver, though a mainstream mouse is probably the easier way. What is the expected minimum euclidean distance between $N$ points uniformly and independently chosen on the shell(sphere) of 3-D ball of radius $R$?   Note that the expected minimum distance might be difficult to compute, so a good lower bound is also fine.   My approach  I take the approach similar to this   .   Suppose we put $N$ circles of radius $r$ uniformly on the surface of the ball. Therefore we could compute the expected minimum distance by the following expression \begin{align} E[D]=\int_0^\infty P(D&gt;t) dt. \end{align}  So, it remains to compute  $P(D&gt;t)$. Actually, since we are after a lower bound on  $E[D]$ it is enought to give a lower bound on $P(D&gt;t)$.  Let $S_i$ be the event that the pair of circles does not intersect which should be given by \begin{align} P(S_i) =  \frac{ Surf(R)-Surf(r)}{Surf(R)}= 1-\frac{  \pi r^2}{ 4\pi R^2}=1-\frac{ r^2}{ 4 R^2} \end{align}  Now the probability that euclidean minimum distance $D$ between $N$ balls is bigger than $t$ should be similar to  \begin{align} P[ D \ge t] \stackrel{?}{=}  P[ \cap S_i] \end{align} However, I think we somehow need to converte from the geodesic distance to euclidean distance but I am not very sure how to do that? Lie algebra and tangent space at the identity are isomorphic Define the Lie algebra $\mathfrak{g}$ of a Lie group G to be the set of all left-invariant vector fields of $G$. I want to prove that   $$          f: \mathfrak{g} \rightarrow T_eG \\              X \mapsto X(e) $$ is a linear isomorphism.  The only thing that I have left to prove is surjectivity.Letting $v\in T_eG$, we set $X\in \mathcal{X}(G)$ such that $\forall x\in G,\,X(x)=L_{x_{*,e}}(v)$ (I still have to show that such an $X$ is indeed a vector field on $G$). We get that  $$ f(X)=X(e)=L_{e_{*,e}}(v)=v $$  where the last equality holds because  $$       L_{e_{*,e}}(v)f = v(f \circ L_e) = v(f \circ id_G) = v(f) $$ for any $f\in C^\infty(G)$. (I am thinking of the tangent vectors as derivations).  Now all we have left to do is prove that such an $X$ is left-invariant, and it is therefore an element of $\mathfrak{g}$, but I do not know how. I tried to prove that  $$            \forall f\in C^\infty(G),\, L_{{x}_*,y}(X(y))f=X(L_x(y))f $$ but with no success. Does $1^{\infty}=e$ or $1^{\infty}=1$? How do I scale the enormously sized desktop icons on Ubuntu 18.04? When should we use "can", "could", "will", "would"? Most people say I wish I could, I wish you would.    Can we use I wish I can, I wish you will?    I'd like to know what the main differences between the usage of can/will and could/would are when wishing.  Is it related to the subjunctive? Because one says “I wish I were”, not “I wish I am”, so are would and could the subjunctive forms of can and will?  Or is it something else entirely? Quickest way to rename files, without retyping directory path How to prove $\,x^a-1 \mid x^b-1 \iff a\mid b$ How to prove $x^a-1\mid x^b-1 \iff a \mid b$, where $x \ge 2$ and $a,b,x \in \Bbb Z$.  I've tried the following in attempting to solve this:  $$a\mid b \Rightarrow aq=b \Rightarrow x^{aq}=x^b \Rightarrow x^ax^q=x^b$$  Because $x^q \in \Bbb Z$, it follows that $x^a\mid x^b$.  This is as far as I have gotten; any help getting further is appreciated.  Note: It may be that this identity is not true at all? ++ is dangerous for C programmers Smoothing/interpolating across images in an ImageCollection to remove missing data I have some code that is working properly to calculate the monthly NDVI averages for the previous two years from a date but I sometimes have missing data in some of the months and I would like to get an average based on the mean of the previous two months. I suppose I could also interpolate from the closest geographic data point but it seems like it would be better to interpolate across time than distance given that sometimes the clouds cover reasonably large areas.   But I'm certainly open to other opinions, I just want to fill these holes as accurately as possible. Is there an efficient way to do this.  and here is the code itself:  // pick a landsat tile footprint to use as my geometry var wrs2_descending = ee.FeatureCollection('ft:1_RZgjlcqixp-L9hyS6NYGqLaKOlnhSC35AB5M5Ll'); // use a manually defined point to pick the WRS2 tile var wrs2_filtered = wrs2_descending.filterBounds(roi); var layer1 = ui.Map.Layer(wrs2_filtered, {}, 'WRS2 filtered'); Map.layers().set(1, layer1);  var imageCollection = ee.ImageCollection('LANDSAT/LC08/C01/T1_SR').filterBounds(wrs2_filtered); var monthCount = ee.List.sequence(0, 11);  // Function to cloud mask from the pixel_qa band of Landsat 8 SR data. function maskL8sr(image) {   // Bits 3 and 5 are cloud shadow and cloud, respectively.   var cloudShadowBitMask = 1 &lt;&lt; 3;   var cloudsBitMask = 1 &lt;&lt; 5;    // Get the pixel QA band.   var qa = image.select('pixel_qa');    // Both flags should be set to zero, indicating clear conditions.   var mask = qa.bitwiseAnd(cloudShadowBitMask).eq(0)       .and(qa.bitwiseAnd(cloudsBitMask).eq(0));    // Return the masked image, scaled to reflectance, without the QA bands.   return image.updateMask(mask).divide(10000)       .select("B[0-9]*")       .copyProperties(image, ["system:time_start"]); }  // run through the image collection and generate monthly NDVI median images var composites = ee.ImageCollection.fromImages(monthCount.map(function(m) {   var startMonth = 1; // thinking that I should always start from Jan so the series are similar   var startYear = ee.Number(2017-1); // number is one year before the current one    var month = ee.Date.fromYMD(startYear, startMonth, 1).advance(m,'month').get('month');   var year = ee.Date.fromYMD(startYear, startMonth, 1).advance(m,'month').get('year')    // filter by year and then filter by month to get monthly mosaics   var filtered = imageCollection.filter(ee.Filter.calendarRange({     start: year.subtract(1), // we want an average of the last two years     end: year,     field: 'year'   })).filter(ee.Filter.calendarRange({     start: month,     field: 'month'   }));   // mask for clouds and then take the median   var composite = filtered.map(maskL8sr).median();   return composite.normalizedDifference(['B5', 'B4']).rename('NDVI')       .set('month', ee.Date.fromYMD(startYear, startMonth, 1).advance(m,'month')); })); print(composites);  Map.addLayer(composites, {min: 0, max: 1}, 'check');  // stack the ImageCollection into a multi-band raster for downloading var stackCollection = function(collection) {   // Create an initial image.   var first = ee.Image(collection.first()).select([]);    // Write a function that appends a band to an image.   var appendBands = function(image, previous) {     return ee.Image(previous).addBands(image);   };   return ee.Image(collection.iterate(appendBands, first)); }; var stacked = stackCollection(composites); print('stacked image', stacked);  // Display the first band of the stacked image. Map.addLayer(stacked.select(0).clip(wrs2_filtered), {min:0, max:1}, 'stacked'); Meaning of "stackexchange" What does stackexchange mean? I have checked out this  to read about the Stack Exchange Network. Story about a deadly alien creature on another planet that has some chemical in its body that makes it faster I can not remember if it was a short story or a novel, but I am leaning towards it being a novel. There is a colony of humans and I can't remember if they crash-landed or not.   There is a creature that they come to discover after it has killed a few people. The creature has some kind of chemical in its body that is like a super-adrenaline. One part is I remember is one of these creatures getting very hot from using up its reserve of this chemical and it dived into a river or pond and it started to steam. Something like that.  It has to be at least 10 years since I read the story. After upgrading to macOS Sierra, I'm getting the notification that "Your disk is almost full. Save space by optimizing storage.":    The options appear to be to store my files in iCloud, automatically delete files, or manually delete files:    My problem/irritation is that I have 80GB free of my 440GB volume.  The question: is there a way to silence this notification in a (semi-)permanent fashion, or to change the threshold it uses for the notification? Using OR comparisons with IF statements When using IF statements in Python, you have to do the following to make the "cascade" work correctly.  if job == "mechanic" or job == "tech":         print "awesome" elif job == "tool" or job == "rock":         print "dolt"   Is there a way to make Python accept multiple values when checking for "equals to"? For example,  if job == "mechanic" or "tech":     print "awesome" elif job == "tool" or "rock":     print "dolt" Proof that no polynomial with integer coefficients can only produce primes Doing a discrete math review and am trying to solve problem 1.6 in the text found here:  - I believe I've gotten parts (a) and (b) correctly, but (c) is a bit tricky for me. Would appreciate a review of a/b and a hint for c if possible. Restatement here:  For n = 40, the value of polynomial $p(n)::=n^2 + n + 41$ is not prime, as noted in Section 1.1. But we could have predicted based on general principles that no nonconstant polynomial can generate only prime numbers.  In particular, let $q(n)$ be a polynomial with integer coefﬁcients, and let $c::=q(0)$ be the constant term of q.  (a) Verify that $q(cm)$ is a multiple of c for all $m \in Z$  Proof: Since q is a polynomial, it will be of the form: $q(x) = a_nx^n+a_{n-1}x^{n-1} + ... + a_1x + c$  If $x = cm$, then notice that cm will be inside every term of the polynomial, and since the final term is c, c can be factored out of every term. Hence, q(cm) is a multiple of c for all $m \in Z$.  b) Show that if q is nonconstant and c > 1, then as n ranges over the nonnegative integers, $N$, there are infinitely many $q(n) \in Z$ that are not primes.  Proof: Continuing from the result found in (a), it is easy to see that for all $m \in Z$, there will be an infinite number of multiples of c that will be generated by q. Since c > 1, these multiples are guaranteed not to be prime.  c) Conclude that for every nonconstant polynomial, q, there must be an $n \in N$ such that $q(n)$ is not prime. Hint: Only one easy case remains.  c = 0 is trivially easy to prove using the (b) above.  I assume the "easy" case is referring to c = 1, but in this case I'm not sure how to continue, since the result of (a) and (b) don't apply: I can't use them since adding 1 to any even number may make it prime. If the result of the terms not including c is odd, then adding 1 to that result makes it even and it is not prime. However, if the terms add up to an even number, I don't see a way of using the knowledge I have so far to prove conclusively that that number + 1 will NOT in fact be prime. How do I find out the number, type, resolution &amp; refresh rate of additional monitors I can attach to my MacBook or MacBook Pro?  This is an attempt at a canonical QA, as per the Meta QA -   to try prevent the myriad "same but different" questions we get on every possible combination of computers, cable &amp; monitors There are perfectly good resources out there that Ask Different should not have to duplicate the efforts of. How do I get my C# program to sleep for 50 milliseconds?  This might seem an easy question, but I'm having a temporary brain failure moment! Express the complex numbers given in Problem in the form $a+ib$ How to increase label size of OpenLayers plugin basemap layers? New breaker and panel: What's the brown goo where they connect?