Prove without induction that $\sum_{r=1}^n {n\choose r}(-1)^{r+1}\dfrac{1}{r}=\sum_{r=1}^n \dfrac{1}{r}$ Prove without induction that $$\sum_{r=1}^n {n\choose r}(-1)^{r+1}\dfrac{1}{r}=\sum_{r=1}^n \dfrac{1}{r}$$ for $n\geq1$.   I have shown the above using induction. However, I am interested in knowing if we can show it directly without using induction.  This is not entirely trivial, I think, if one is armed only with the very basic ideas of combinatorial identities. This can be proved in at least two different ways: one using induction and the other is by evaluating an integral. However, regarding the second approach, I am quite certain that if not known beforehand which integral to compute to get this equality, it is a hard exercise.  EDIT: In the link claiming to be a duplicate of my question, induction was allowed. I am looking for approaches that use no induction or calculus. So this question is different.
Proving Binomial Identity without calculus How to establish the following identities without the help of calculus:  For positive integer $n, $ $$\sum_{1\le r\le n}\frac{(-1)^{r-1}\binom nr}r=\sum_{1\le r\le n}\frac1r $$  and  $$\sum_{0\le r\le n}\frac{(-1)^r\binom nr}{4r+1}=\frac{4^n\cdot  n!}{1\cdot5\cdot9\cdots(4n+1)}$$
Extracting Z data from point layer using ArcGIS Desktop? I have created a point layer with ~150 points. I have a DEM beneath it, which is where I have taken the X and Y values from.   Does anybody know how I can retrieve the Z values, without using the info button?  I am using ArcGIS Desktop 10. I have licences for the spatial analyst etc tools. Ampere's circuital law for finite current carrying wire When I was studying about . Then there comes a question in my mind that "whether this law is applicable for finite current carrying wire or not" How can I get useful error messages in PHP? If a function $f$ defined on $[a,b]$ and differentiable everywhere, does it mean that its derivative $f'(x)$ is continuous everywhere on $[a,b]$?  My understanding is 'yes'. Because if there were 'a gap' in $f'(x)$, we could integrate back to $f(x)$ and show that where the gap is, there would be a sharp "angle" where $f(x)$ is not differentiable. Can I pass arguments to an alias command? I want to know if I can pass an argument with an alias command.  for example:  alias d="dmesg|grep -iw usb|tail -5"    Now d will print the last 5 lines. If I want to use d to print a different number of lines, I have to make change in the alias command declaration of d again.  Is there any way I can modify the declaration of an alias so that I don't have to retype the declaration to change the number of lines. Like incorporating passing the number of lines as an argument while declaring alias for d? Or is there some other method to solve this? "Awesome" vs. "Awful" Almost universal class I so stuck with a problem of set theory. But first a recursive definition:   Define $R_0=\emptyset$ If $R_\alpha$ is defined, then $R_{\alpha+1}=\mathcal{P}(R_\alpha)$ (the power set). For a limit ordinal $\gamma$, if $R_\alpha$ is defined for all $\alpha&lt;\gamma$, then define $R_\gamma=\displaystyle\bigcup_{\alpha&lt;\gamma} R_\alpha$   We define $\text{BF}=\displaystyle\bigcup_{\alpha\in\text{OR}} R_\alpha$ (the class of well founded sets).  Next, my problem     Take $A\subseteq \text{BF}$ a proper transitive class such that $(A,\in)\models\text{ZF}$. Prove that $A$ is almost universal.    I think that the exercise is false because if it was true, then we could conclude that the strongly inaccessible cardinals doesn't exists. This because if $\kappa$ is strongly inaccessible then $R_\kappa$ satisfies the hypothesis but $R_\kappa$ isn't almost universal. I really appreciate any hint or/and suggestion.   Edit: my counterexample is wrong. But, then, how can I solve the exercise? "any other verbs then being located in VP" -- is this acceptable? if the sentence you are analysing contains a modal verb, then it is positioned in ‘I’,any other verbs then being located in VP.      (English Syntax and Argumentation - Bas Aarts)   At first, I think the The bold clause should be "...is located..." but then I realize it might be in the form of a non-finite clause (ing clause with a subject) but I'm not sure. If it isn't, how you describe this usage? Is this common in English? Is there a more complete way of proving this theorem (that I devised)? OR does this theorem already exist and have a name? Last night, I was working on re-solidifying my understanding of parametric equations, and came up with a neat trick for evaluating certain definite integrals. I'm not sure if it exists as a formal theorem, though I refuse to believe I'm the first to notice such a relation. The 'theorem' is a useful trick to evaluate the definite integral of a continuous function over an interval where the function is either strictly increasing or strictly decreasing, given that the inverse of the function is 'nicer' to work with than the function itself. My idea goes as follows:  If $y = f(x)$ is a continuous function over the interval $I$, and either:  $f'(x) &gt; 0$   OR   $f'(x) &lt; 0$   for all values of $x \epsilon I$, then:  $\int_{a}^{b}{f(x)dx} = bf(b) - af(a) - \int_{f(a)}^{f(b)}{f^{-1}(x)dx}$  where $a,b \epsilon I$  Quasi-proof:  Clearly, if a function is strictly increasing or decreasing, it is injective. This is clear because no two values in the domain can correspond to a single value in the range. I feel intuitively that it must also be surjective (though I'm not entirely sure how to prove this part). Assuming it is both injective and surjective over $I$, the function must have an inverse.   If we were to draw two rectangles out of our bounds, they will have areas:  $af(a)$ and $bf(b)$, so that   $bf(b) - af(a)$ will be the area of the region we are observing.  This area is equivalent to the sum of the areas connecting the curve $f(x)$ to the $x$-axis and to the $y$-axis. Thus:  $bf(b) - af(a) = \int_{a}^{b}{f(x)dx} + \int_{f(a)}^{f(b)}{f^{-1}(x)dx}$  Solving for $\int_{a}^{b}{f(x)dx}$, we get:  $\int_{a}^{b}{f(x)dx} = bf(b) - af(a) - \int_{f(a)}^{f(b)}{f^{-1}(x)dx}$  A particularly illuminating example is evaluating:  $\int_{a}^{b}{ln(x)dx}$  $ln(x)$ is strictly increasing over any interval, so the 'theorem' applies.  $\int_{a}^{b}{ln(x)dx} = bln(b) - aln(a) - \int_{ln(a)}^{ln(b)}{e^xdx}$  $\int_{a}^{b}{ln(x)dx} = ln(b^b/a^a) - (e^{ln(b)} - e^{ln(a)})$  $\int_{a}^{b}{ln(x)dx} = ln(b^b/a^a) + a - b$  and so we've evaluated the integral without the need of integrating the function itself (by method of integration by parts), but rather by integrating its inverse, which is trivial to evaluate.!  Response to Jason DeVito's answer: [This image was taken from   () My DOM looks like this:  &lt;div id="d1"&gt;    &lt;div class="c1"&gt;             &lt;a href="#"&gt;&lt;img src="img1_on.gif"&gt;&lt;/a&gt;             &lt;a href="#"&gt;&lt;img src="img2_on.gif"&gt;&lt;/a&gt;    &lt;/div&gt; &lt;/div&gt;   When someone clicks on an image, I want the image src to change to &lt;img src="imgx_off.gif"&gt; where x represents the image number 1 or 2.  Is this possible or do I have to use CSS to change the images? How does vendor restocking work in Fallout 4? Solve this integral for free WiFi Vertical uniform circular motion.. can it really be uniform? This is the picture in my mind:    For centripetal force, I learned that: $T-mg\cos\theta= \frac{mv^2}{r}$  In vertical circular motion, the velocity is naturally going to decrease as kinetic energy is converted to potential energy as the particle moves up the circle- and thus resulting into a non-uniform circular motion.  However, I was told that if tension changes accordingly, the velocity will remain constant(as from the equation). What I do not get about this part of the explanation is that which force will balance mgsinθ (tangential force in the diagram  )? My website's google cache copy looks broken.   Below is link to cache.  http://webcache.googleusercontent.com/search?q=cache:http://paintcollar.com  Can this affect the SERPs and crawling by google?  What can I do to rectify it?  PS. Few pages are being cached properly though.Below is URL to one such page. http://webcache.googleusercontent.com/search?q=cache:http://paintcollar.com/aneel VMware won't work after Ubuntu Upgrade I upgraded to Ubuntu 15.10 and now if I want to open VMware Player it doesn't do anything. I also uninstalled and installed it again, still nothing.  What can I do? Definite article — “on television” vs. “on the radio” Why are these different?        We heard the news on the radio.   We watched the news on television.      In this , the author says we must use television without the. Why? It makes me crazy. Is there a valid reason? My sister, cousins, and other relatives speculate that my mom's grandmother was Jewish (for various reasons). I've heard people say at times things like, for example, "I'm 50% German" or "I'm 25% Swedish." This has to do with their ancestry. But, how exactly is this calculated? Say for example that my mother's paternal grandmother was indeed 100% Jewish, what would that make me and my siblings, and how is it calculated? Does Wasserstein distance require the source and target distributions to have the same mass? If we minimize the Wasserstein loss, $$W_1 (P_S, P_T) = \underset{\gamma \in \Pi}{\text{min }} \sum_{x,y} |x-y|\gamma(x,y)$$ which means we are looking for the coupling $\gamma(x,y)$ that minimizes the cost of moving mass from the source distribution $P_S$ to the target distribution $P_T$,  Do the masses of the source and target distributions have to be identical? If so or not, how are the masses of $P_T$ and $P_S$ accounted for mathematically in, or when minimizing, the shown formula, or what might be missing/overlooked by the Wasserstein approach overall due to the mass differences? How to kill The Twins In Terraria In Terraria I often try to defeat spazmatism first, but when he transformed, his flamethrower killed me. What is easiest way to kill The Twins? Filling cells with sequential numbers in an Excel (2003) macro I need to fill an excel column with a sequential series, in this case from -500 to 1000.  I've got a macro to do it, but it takes a lot of lines for something that seems like it should be a single function [something like FillRange(A2:A1502, -500, 1000, 1)].  But if that function exists, I can't find it.  Is the following as simple and elegant as it gets?  'Draw X axis scale Cells(1, 1).Value = "mV" Cells(2, 1).Value = -500 Cells(3, 1).Value = -499 Cells(4, 1).Value = -498  Dim selection1 As Range, selection2 As Range  Set selection1 = Sheet1.Range("A2:A4") Set selection2 = Sheet1.Range("A2:A1502")  selection1.AutoFill Destination:=selection2