There is a popular claim going around the internet that the sum of the positive integers is $-1/12$. There are proofs to this statement and I am not going to try and refute them. But I could not understand at what point the sum starts becoming more negative, since when I add positive numbers, the sum just gets more positive.
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(I was requested to edit the question to explain why it is different that a proposed duplicate question. This seems counterproductive to do here, inside the question it self, but that is what I have been asked by the site and moderators. There is no way for me to vote against their votes. So, here I go: Please stop voting this as a duplicate so quickly, which will eventually lead to this question being closed off. Yes, the other question linked to asks the same math, but any newcomer to the problem who was exposed to it via physics, as I was, will prefer this question instead of the one that is purely mathematically. I beg the moderators to not be pedantic on this one. This question spills into physics, which is why I did the cross post to the physics forum as well.)  How does the sum of the series “1 + 2 + 3 + 4 + 5 + 6…” to infinity = “-1/12”, in the context of physics?  I heard Lawrence Krauss say this once during a debate with Hamza Tzortzis (). I found a transcript of another debate between Krauss and William Lane Craig which has the same sum. Here is the paragraph in full:     Let’s go to some of the things Dr. Craig talked about. In fact, the   existence of infinity, which he talked about which is   self-contradictory, is not self-contradictory at all. Mathematicians   know precisely how to deal with infinity; so do physicists. We rely on   infinities. In fact, there’s a field of mathematics called “Complex   Variables” which is the basis of much of modern physics, from   electro-magnetism to quantum mechanics and beyond, where in fact we   learn to deal with infinity; without the infinities we couldn’t do the   physics. We know how to sum infinite series because we can do complex   analysis. Mathematicians have taught us how. It’s strange and very   unappetizing, and in fact you can sum things that look ridiculous. For   example, if you sum the series, “1 + 2 + 3 + 4 + 5 + 6…” to infinity,   what’s the answer? “-1/12.” You don’t like it? Too bad! The   mathematics is consistent if we assign that. The world is the way it   is whether we like it or not.   -- Lawrence Krauss, debating William Lane Craig, March 30, 2011  Source:   CROSS POST: I'm not sure if I should post this in mathematics or physics, so I posted it in both. Cross post:   EDIT: I did not mean to begin a debate on why Krauss said this. I only wished to understand this interesting math. He was likely trying to showcase Craig's lack of understanding of mathematics or logic or physics or something. Whatever his purpose can be determined from the context of the full script that I linked to above. Anyone who is interested, please do. Please do not judge him out of context. Since I have watched one of these debates, I understand the context and do not hold the lack of a full breakdown as being ignorant. Keep in mind the debate I heard this in was different from the debate above.
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I have found two formulas. I want to connect them!  The number of ways in which a given positive integer $n≥3$ can be expressed as a sum of three positive integers $x,y,z$ (i.e. $x+y+z=n$) , subject to $x≤y≤z$ is the integer closest to $\frac{n^2}{12}$. If $x+y+z=n$ where $x,y,z$ are postive integers then the number of ways is ${n-1 \choose 2}$ .  The difference between these two formula is the condition $x≤y≤z$. Can we connect these two ?? Please help me! I mainly wants to know how one implies the other.
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I have been wondering about this question for a while (I am almost sure that I read it in some contest mathematics book a while ago)     Determine which positive integers cannot be written as a sum of consecutive positive integers.   All numbers will be considered positive integers in the following unless stated otherwise.   I was thinking as follows: odd numbers will always work since if we take an odd $m$, we have  $$ m=2\cdot k+1=k+(k+1). $$   Then I realised that all numbers having at least one odd prime factor will also work. Take $n$ as a number having an odd prime factor and let this odd prime factor be $m$ and let $n=m\cdot k$. Then $$ n=\underbrace{k+k+\ldots+k}_{m \ \text{times}} $$ and since $m$ is odd, say $m=2\cdot l+1$ we have $$ n=\underbrace{k+k+\ldots+k}_{l \ \text{times}}+k+\underbrace{k+k+\ldots+k}_{l \ \text{times}} $$ and we can start moving $1$s one by one from the left side of the middle "$k$" to the right side of it $$ n=\underbrace{k+k+\ldots+(k-1)}_{l \ \text{times}}+k+\underbrace{(k+1)+k+\ldots+k}_{l \ \text{times}} $$ $$ n=\underbrace{k+k+\ldots(k-2)+(k-1)}_{l \ \text{times}}+k+\underbrace{(k+1)+(k+2)+\ldots+k}_{l \ \text{times}} $$ this way we end up in a sum of consecutive integers. It may happen that some integers on the left will be negative which is undesired but since these numbers are consecutive integers we always be able to cancel out the negative ones without disturbing the structure of the uncancelled ones. To see this in action let us take $22=11\cdot 2$. This would give $$ 2+2+2+2+2+2+2+2+2+2+2 $$ so $$ 2+2+2+2+1+2+3+2+2+2+2 $$ the next step gives $$ 2+2+2+0+1+2+3+4+2+2+2 $$ it seems that now we are out of numbers but we just continue moving an increasing number of $1$s to the right $$ 2+2+(-1)+0+1+2+3+4+5+2+2 $$ $$ 2+(-2)+(-1)+0+1+2+3+4+5+6+2 $$ finally $$ \underbrace{(-3)+(-2)+(-1)+0+1+2+3}_{=0}+4+5+6+7 $$ so we get $$ 22=4+5+6+7. $$  Now we have a method to generate the desired expression for all odd numbers and all the ones having at least one odd prime factor. Which numbers are left? The ones having ONLY even primes in their prime factorisation. These are of the form $2^n$. I was thinking what do these have in common and I thought that they are separated by a factor of $2$. The first of these are $1$ for $n=0$ and $1$ cannot be expressed in the desired for so if I could show that  $$ m \ \text{cannot be expressed}\Rightarrow 2m \ \text{cannot be expressed} $$ I am done. So I was trying to show the contrapositive $$ 2m \ \text{can be expressed}\Rightarrow m \ \text{can be expressed}. $$ So assume that $2m$ can be expressed. Now, eihter $m$ is odd in which case we are done since odds were easy to express or it is even, that is $m=2l$. From here the argument repeats. Eihter $l$ is odd in which case we are done or it is even, so $l=2k$ and so on. After a while we factored out all of the $2$'s and we end up with and odd number which we can express and we are done.  So integers of the form $2^n$ cannot be expressed in this form.  Is my reasoning correct?  If not please point out where I made a mistake. If it turns out to be correct then we are all happy I can sleep well tonight but it would be nice to see a more rigorous proof of this so if someone can provide one I am more than happy to read it. Since this is a contest question I expect it to have a really elegant solution and I do not consider mine an elegant one.
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Why subtraction of a negative number from a positive number is addition? Eg: $a - (-b) = a + b$ When looking through the number scale I am unable to relate this, please kindly clarify ?
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Supposing that a real number $c$ is given, is the following true?  "If $n^c$ is a natural number for every natural number $n$, then $c$ is a non-negative integer."  Though this seems true, I can't prove that. Can anyone help?
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Let $\alpha$ be a positive number. Calculate $$\lim_{n\to\infty} \sum_{k=1}^{n} \Big(\frac{k}{n}\Big)^{\alpha k}$$  Edit: I have deleted my attempt, it didn't seem to lead me anywhere and I discovered a couple of mistakes. :)  Please give me a hint!
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I am looking to prove the following statement false:  Let $k$ be a positive integer, then $k\mid n$ if and only if $k\mid n^2$.  So I am trying to find a $k$ where this does not hold but after aimlessly trying numbers I was curious if there is a methodical way to find $k$.
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I need help!  I need to prove that for any $2 \le n,k$ positive integers $(n - 1)^2 \mid n^k -1$ if and only if $(n - 1) \mid k$  Thanks!
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I just read somewhere that there are as many even numbers as natural numbers! Is there an intuitive way of coming to terms with this fact? Because honestly, it's quite baffling. Also, does the concept of odd/even apply to negative integers as well? By definition, it should. But I have never seen a negative multiple of two being used as an example of an even number.
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Given a positive integer $n$. How to prove that $\sum\limits_{k=0}^{n}(-1)^k \dbinom{n}{k} = 0 $ ?   I've tried using mathematical induction, then:  $$p(1)=\sum_{k=0}^{1}(-1)^k \binom {1}{k}=0 $$  And my induction hypothesis is: $$p(n)=\sum_{k=0}^{n}(-1)^k \binom {n}{k}=0 $$ So, i need prove: $$p(n+1)=\sum_{k=0}^{n+1}(-1)^{k} \binom {n+1}{k}=0 $$ Let $\sum_{k=0}^{n+1}(-1)^{k} \binom {n+1}{k}$, $$\begin{align} \sum_{k=0}^{n+1}(-1)^{k} \binom {n+1}{k} &amp;= \sum_{k=0}^{n}(-1)^{k} \binom {n}{k}+(-1)^{k} \binom {n+1}{n+1} \\ &amp;=0 + (-1)^{k} \end{align}$$ What am i doing wrong?
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If $l$ and $n$ are any positive integers, is there a  proof of the identity     $$\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(l-k)^n=n!\;$$   which uses the Inclusion-Exclusion Principle?  (If necessary, restrict to the case where $l\ge n$.)    This question is closely related to   and also
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Can anyone help me with this?     If 738 consecutive integers are added together, where the 178th number in the sequence is 4,256,815, what is the remainder when this sum is divided by 6?
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Prove or disprove: If a, b belong to the set of positive integers, and if a divides b and b divides a, then a=b. Does this hold if if a,b are not necessarily positive? Why or Why not?   Here is what I have:  If a and b are integers, we say a divides b if there exists an integer k so that b=ak. Thus:  $$b=ak$$  $$a=bj$$   $$a=akj$$ $$kj=1$$   Since a divides b and b divides a, we know that k,j must both be integer, which means k and j must both equal 1 or -1. This means:   $$a=b$$  This proves that a=b when we assume that a,b are both positive integers.   For the second part of the question, I am asked if I can still say a=b if the assumption about a, b is relaxed so that a,b belong to the integers (not the positive integers.) I think The answer is no, I cannot say that a=b for every case when we allow a,b to also be negative integers.   Here is my reasoning (I use the same idea that a=bj and b=ak here)   $$+/-a=akj$$  $$+/-1=kj$$   So $ |a|=|b|$ but not $a=b$  I'm very new to proofs so any suggestions or thoughts are greatly appreciated.
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This is the question I've run into: Show that given a set of  positive n integers, there exists a non-empty subset whose sum is divisible by  n  I'm having trouble understanding how they came to the conclusion    the part that I'm having trouble understanding is how subtracting the two subsets results in a sum that's divisible by n given that the two subsets have the same remainder
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How do I prove that any even integer $n \neq 2^k$ is expressible as a sum of positive consecutive integers (more than 2 positive consecutive integer)? For example:   14 = 2 + 3 + 4 + 5 84 = 9 + 10 + ... + 15  n = sum (k + k+1 + k+2 + ...)  n ≠ 2^k
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Find the value of $\sum_0^n \binom{n}{k} (-1)^k \frac{1}{k+1}$.  Writing out several terms, I think the answer is $1/(n+1)$, but I'm struggling to prove this. I would greatly appreciate it if someone could show how to prove this.
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Find all positive integers $a, b, c$ such that $1/a + 1/b + 1/c = 4/3999$.  The contest is just ended, so you may freely answer. (I did not attend the contest: it is an Italian contest for schools and the groups may ask other people for hints. This is how I got involved, but of course asking on MSE would have been cheating)  I found three solutions: I am explaining below what I did. I would like however to know if there is a more general approach.