Prove by induction that this sum is not a natural number. Prove by induction that $1+\frac{1}{2}+\frac{1}{3} +...+\frac{1}{n}$ for any $1&lt;n$ and $n\in \Bbb{N}$ is not a natural number.
Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer? If $n&gt;1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer.  If you know , then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$.  Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong.
How was this answer posted after this question was closed? Question:   Answer:   The question was closed at 11:57:57 (according to the hover-over-"x-mins-ago" bar) and the answer was posted at 12:16:35. It's a fairly lengthy answer, so it's possible that it was started to be written before it was closed but I swear I've started to answer a question then I got a banner along the top saying "this question has been closed" and the answer was no longer post-able. What is the easiest way to make a backup of an entire hard disk I got myself a dell laptop from the local computer store. Its a used machine with Windows Vista Home Basic on it. I want to load Ubuntu Desktop 10.10 though so I can do perl development. BUT I want to keep a copy of the entire harddrive with the dell utility partition and Windows Vista in case I want to go back. I was thinking I could image the drive but I not sure what to use, I don't have Ghost or anything, Someone had told me about Clonezilla. Would that work for me? Is it hard to use? Also I want to burn the data to a DVD or something more storable than a harddisk. How many solutions for an equation with simple restrictions I'm working on an assignment in which I have to count the number of solutions for this particular equation: $$x_1+x_2+x_3+x_4=20$$for non negative integers with $x_1&lt;8 $ and $x_2&lt;6$   I'm aware that this kind of a task isn't that complicated, but I don't get combinatorics in general that well.  So I've tried two following approaches to get this done.  Firstly I tried to substitute the variable x:  $x_1+x_2+x_3+x_4=20 \Leftrightarrow y_1+y_2+y_3+y_4=34$  in which $y_1=x_1+8$ and $y_2=x_2+6$ (casue $x_1=y_1-8$ and $x_2=y_1-6$) Following this approach the total number of possible solutions would be  $${34+3 \choose 3} $$  But I'm not sure if its the right solution.  The second approach is to sum all of the possible values that $x_1$ and $x_2$ could possibly take, also $x_1=0,1,2,3,4,5,6,7$ and $x_2=0,1,2,3,4,5,6$ And then  count all the possibilities for each of the variables $${20 -x_1-x_2+1\choose 1}$$ and sum them together like this: $${21\choose 1}+{20\choose 1}+{19\choose 1}+{18\choose 1}+... $$ and so on...  I'm sure I'll get the correct number with this one, but I'm not feeling like summing all of this possibilities. There's got to be a better, more elegant way to deal with this.  My professor gave me a hint that I should do it using the complement. How do you subtract Dates in Java? I am considering buying my first DSLR.  How do I know whether I need a 18-200 lens, 17-85, or 18-55? Or something else? Where does the energy required to stop a moving body come from and go? For a long time, I have been thinking about this: In order to stop a moving body, I have to apply impulse to change the existing momentum to $0$. 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Change content of file settings.ini to    [KeyboardShortcuts] Switch=LWin Move=LAlt, Shift, Ctrl MoveAndSwitch=LAlt, Shift Previous=Left Next=Right PlusTen=     Run virtual-desktop-enhancer.exe as administrator. Evaluating this integral : $ \int \frac {1-7\cos^2x} {\sin^7x \cos^2x} dx $ The question :  $$ \int \frac {1-7\cos^2x} {\sin^7x \cos^2x} dx $$  I tried dividing by $\cos^2 x$ and splitting the fraction.  That turned out to be complicated(Atleast for me!)  How do I proceed now? How to change default window background color in Windows 10? In Windows 7 and Windows XP it was very easy to change the default background color of all windows to whatever you want. Staring at white window backgrounds all day is hard on the eyes. Once changed, ALL windows of all apps would conform to the custom color as long as the apps allowed the windows bg to be invoked. So for example it worked great in Notepad and many other apps.   Then the geniuses at M$ decided that this feature was no longer needed in Windows 8, and removed it. There were some reg hacks, but nothing standard. I remember reading that the feature would be coming back in Windows 10, but I'm not finding this to be the case. The only [partial/broken] way I have found to be able to do this is to tweak the following registry keys:  [HKEY_CURRENT_USER\Control Panel\Colors] "Window"="227 221 191"  [HKEY_CURRENT_USER\Control Panel\Desktop\Colors] "Window"="227 221 191"   When I changed those keys to my chosen soothing color and rebooted, it worked ONE TIME. After every reboot thereafter, all the white backgrounds are back AND YET those registry keys still have my custom RGB codes. I really hate Microsoft right now, but that aside, is there any solid way to change windows 10 bg colors and make it stick?  Important: In Settings -> Personalization -> Colors, none of that changes windows backgrounds. Already tried.   I am using Windows 10 Enterprise Version 1511 Build 10586.3.  UPDATE: I rebooted again and the bg colors work correctly...but later in the day all the window bgcolors were back to white again. Windows 8.1 had the exact same bug. My conclusion is that the reg hack "works" but only intermittently. Is this simply a MS bug? Or is there a better answer/solution? So, we know that there is no "natural" isomorphism between a finite dimensional vector space $X$ and its dual $X^{\vee}$. However, given a basis for $X$ $(e_1, \dots, e_n)$ we can always construct a (dual) basis $(e^1, \dots, e^n)$ for $X^{\vee}$ that satisfies $e^i(e_j) = \delta^i_j$. It follows that $V \approx V^{\vee}$ simply because they have the same finite dimension $n$. Now, it seems to me that we can construct an isomorphism by assigning $$ e_i \mapsto e^i $$ and extending by linearity. That is, for an arbitrary vector $v \in X$ given by $v = \sum v^ie_i$ we can specify a linear map   $$\phi:V \rightarrow V^{\vee}$$  by  $$ \phi(\sum v^ie_i) := \sum v^ie^i $$ It is a quick exercise to check that $\phi$ is an isomorphism. Now, although this construction seems correct, there is something that doesn't quite sit right with me about it. For one thing, there is no way to really make sense of the summation convention as the right side of the function will always contain two superscripts thus requiring an explicit summation sign. The other thing that feels rather off is actually taking a lower index and moving it to an upper index, i.e., $e_i \mapsto e^i$ (which, of course, is the reason that the summation convention doesn't work).  So, with this background my questions are   Is there any sense in which the above isomorphism is "favored" or "canonical"? 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