Link between Infinite coin flipping and Uniform distribution on unit interval
Limit distribution of infinite sum of Bernoulli random variables
Probability that the sum of 6 dice rolls is even I remember that a couple of years ago a friend showed me and some other people the following expression:  $$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$  As shown below, I can prove that this limit exists by the monotone convergence theorem. I also remember that my friend gave a very dubious "proof" that the value of the limit is $\frac{1}{e-1}$. I cannot remember the details of the proof, but I am fairly certain that it made the common error of treating $n$ as a variable in some places at some times and as a constant in other places at other times. Nevertheless, numerical analysis suggests that the value my friend gave was correct, even if his methods were flawed. My question is then:  What is the value of this limit and how do we prove it rigorously?  (Also, for bonus points, What might my friend's original proof have been and what exactly was his error, if any?)  I give my convergence proof below in two parts. In both parts, I define the sequence $a_n$ by $a_n=\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ for all integers $n\ge 2$. First, I prove that $a_n$ is bounded above by $1$. Second, I prove that $a_n$ is increasing.   (1) The sequence $a_n$ satisfies $a_n&lt;1$ for all $n\ge 2$.  Note that $a_n&lt;1$ is equivalent to $1^n+2^n+\cdots+(n-1)^n&lt;n^n$. I prove this second statement by induction. Observe that $1^2=1&lt;4=2^2$. Now suppose that $1^n+2^n+\cdots+(n-1)^n&lt;n^n$ for some integer $n\ge 2$. Then  $$1^{n+1}+2^{n+1}+\cdots+(n-1)^{n+1}+n^{n+1}\le(n-1)(1^n+2^n+\cdots+(n-1)^n)+n^{n+1}&lt;(n-1)n^n+n^{n+1}&lt;(n+1)n^n+n^{n+1}\le n^{n+1}+(n+1)n^n+\binom{n+1}{2}n^{n-1}+\cdots+1=(n+1)^{n+1}.$$  (2) The sequence $a_n$ is increasing for all $n\ge 2$.  We must first prove the following preliminary proposition. (I'm not sure if "lemma" is appropriate for this.)  (2a) For all integers $n\ge 2$ and $2\le k\le n$, $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We observe that $k^2-1\le kn$, so upon division by $k(k^2-1)$, we get $\frac{1}{k}\le\frac{n}{k^2-1}$. By , we may find:  $$\frac{k+1}{k}\le 1+\frac{n}{k^2-1}\le\left(1+\frac{1}{k^2-1}\right)^n=\left(\frac{k^2}{k^2-1}\right)^n.$$  A little multiplication and we arrive at $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We may now first apply this to see that $\left(\frac{n-1}{n}\right)^n\le\left(\frac{n}{n+1}\right)^{n+1}$. Then we suppose that for some integer $2\le k\le n$, we have $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$. Then:  $$\left(\frac{k-1}{n}\right)^n=\left(\frac{k}{n}\right)^n\left(\frac{k-1}{k}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}\left(\frac{k}{k+1}\right)^{n+1}=\left(\frac{k}{n+1}\right)^{n+1}.$$  By backwards (finite) induction from $n$, we have that $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$ for all integers $1\le k\le n$, so:  $$a_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n\le\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}&lt;\left(\frac{1}{n+1}\right)^{n+1}+\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}=a_{n+1}.$$  (In fact, this proves that $a_n$ is strictly increasing.) By the monotone convergence theorem, $a_n$ converges.  I should note that I am not especially well-practiced in proving these sorts of inequalities, so I may have given a significantly more complicated proof than necessary. If this is the case, feel free to explain in a comment or in your answer. I'd love to get a better grip on these inequalities in addition to finding out what the limit is.  Thanks! ISP blocking port 25 Dirac delta function as a limit of sinc function How can I mute the YouTube app while playing music on another player? In this  to a question about a series, a theorem was stated:     If $A= \{a_i \}$ is a set such that $\sum_{i = 1}^{\infty} \frac{1}{a_i}$ converges, then $d(A) = 0$, where $d(A)$ is the  of the set.   My background in number theory is basically zero and all my attempts to prove this have been utterly unseccessful; would anyone please help me, or at least provide me with a hint to prove it? In order to show $e^{x+y}=e^{x}e^{y}$ by using Exponential Series, I got the following:   $$e^{x}e^{y}=\Big(\sum_{n=0}^{\infty}{x^n \over n!}\Big)\cdot \Big(\sum_{n=0}^{\infty}{y^n \over n!}\Big)=\sum_{n=0}^{\infty}\sum_{k=0}^n{x^ky^n \over {k!n!}}$$  But, where should I go next to get $e^{x+y}=\sum_{n=0}^{\infty}{(x+y)^n \over n!}$.  Thanks in advance. Proving Pascal's identity So I came across Pascal's identity: Prove that for any fixed $r\geq 1$, and all $n\geq r$, $$ \binom{n+1}{r}=\binom{n}{r}+\binom{n}{r-1}. $$  I know you can use basic algebra or even an inductive proof to prove this identity, but that seems really cumbersome. I was wondering if anyone had a "cleaner" or more elegant way of proving it. For example, I think the following would be a decent combinatorial proof.   Proof: Let $S$ be a set with $n+1$ elements, and consider some fixed $x\in S$. There are $\binom{n+1}{r}\;\; r$-subsets of $S$--count them according to whether or not they contain $x$: there are $\binom{n}{r}$ not containing $x$, (each formed by choosing $r$ of the remaining $n$ elements in $S\setminus\{x\}$), and there are $\binom{n}{r-1}\;\; r$-sets containing $x$, (each formed by selecting an additional $r-1$ elements in $S\setminus\{x\}$).   Is that right? Are there any other efficient ways of doing it? What are all the correct answers to the Squirrel's questions, and what is the reward when I get them all correct? Book about a teenager programmed to be an assassin and trying to get back at the people responsible I read this book a few years back (around 5?). It was part of a multibook series but I have only read one book.  It's about this teenager who I think had been programmed to be an assassin but then somehow retained his conscience and was out to get the people responsible for that.  I briefly remember a girl, and this one part where he realizes he can ride a horse and then there's a chase too. I think the guy's name was Jim or Jimmy, but not quite. I'm also quite sure he was either part computer, or had a chip or something in his brain which had been programmed to give him a host of mental and physical abilities which he retained and used as well as the series progressed. How can I get free or trial ExactTarget instance Assigning Multiple Cores to a Python Program If $m^4+4^n$ is prime, then $m=n=1$ or $m$ is odd and $n$ even I have been stuck on this one for months, really simple to state, really giving me trouble.     Show that if $m^4 + 4^n$ is prime, $m&gt;0$, $n&gt;0$, then $m$ is odd and $n$ is even, except when $m=n=1$.   The case $m=n=1$ gives $5$, so that is why it is excluded. Clearly $m$ is odd, for if it was even, the number would be divisible by at least $4$, but I can't seem to get rid of the case where $n$ must be odd.   The only thing I've managed to try is to write $m$ as an odd number, say $m = 2k+1$, and $n = 2\ell + 1$, then get $$ m^4 + 4^n = (2k+1)^4 + 4^{2\ell+1} = 16k^4 + 32k^3 + 24k^2 + 8k + 1 + 4^{2 \ell + 1} $$ and this is congruent to $0 \mod 5$ unless $k \equiv 2 \mod 5$, and now it gives you an even more ugly polynomial if I write $k = 5j+2$, and it's not working. I've tried other inconclusive approaches, like trying to see if two numbers of some form generate numbers of the same form.. (for instance primes of the form 8k+1 and 8k+7 always generate primes of the same form when multiplied together, stuff like that).  Any ideas? Even just ideas you haven't gave a thought... they're welcome! Printer not recognized on Ubuntu Libre Office I am running 14.04 and have a HP Deskjet 2545 wireless printer which is supported by Ubuntu. I have run HpLIP, the latest version and installed the printer.  The printer is recognised by Ubuntu, printer and network, but LibreOffice do not see the printer. Only a Generic printer.  I don't know what to do anymore and would like to get some good advice Backup GNOME-Terminal For any prime $p > 3$, why is $p^2-1$ always divisible by 24? Let $p&gt;3$ be a prime. Prove that $24 \mid p^2-1$.   I know this is very basic and old hat to many, but I love this question and I am interested in seeing whether there are any proofs beyond the two I already know. Why is the time limit for editing comments 5 minutes? What is the reasoning behind this? Could someone with experience in using these exercise typesetting packages please offer a quick compare and contrast of these packages to help me choose?  Packages under consideration are:         Having read briefly the documentation, they all seem to do the same thing generally, so could you please focus on the differences?  Some differences will not be obvious until I've learnt and used all the different packages but I want to better appreciate the differences now, a priori. Can Ubuntu really run all Windows programs? 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?