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 Feb 19 comment No Bijection from set $X$ to $X - \{x\}$ @AsafKaragila that is what I thought. thanks. Feb 19 comment No Bijection from set $X$ to $X - \{x\}$ Is a proof possible without some form of cardinality being brought into picture? Feb 12 comment Proving upper bounds? @HagenvonEitzen Wow!! Feb 12 comment Is there an axiom that prevents other axioms from contradicting each other? en.wikipedia.org/wiki/Three_classic_laws_of_thought Feb 5 comment When $dy/dx =0$ for all $x$ in the domain, is $dx/dy$ also zero? +1 For the proper explanation. Feb 1 comment Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Yes. I am Jayesh. Name changed for a month. Feb 1 comment Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Nice solution Chris'ssister! :-) Your question and the corresponding answers provide a decent tool. :-) Feb 1 comment Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Thanks. A different solution. :-) Feb 1 comment Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Okay. No problems. :-) Generally, when people downvote, and the OP asks the reason, it is expected that the person who downvoted leave a comment about it. And hence, my assumption. Feb 1 comment Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Arrrrgh. Thanks. Feb 1 comment Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ @ToddWilcox Counterpoint. meta.math.stackexchange.com/q/2244/14082 Feb 1 comment Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Added the explanation. I hope its correct. Feb 1 comment Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ A few reasons 1. It gives me oppurtunity to verify that the solution is indeed correct. (I self study, so even though I get an answer and I am pretty sure about it, there is no real way to verify the solution completely.) 2. It allows probably other people to offer me better solutions. 3. I can do so on a blog, but then it might not get the same attention on the blog. 4. MSE's editing capabilities are better than almost all other blogging software I have found. Feb 1 comment Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Is the downvote because of self-answering? Jan 4 comment Calculating the integral $\int_0^{\infty}{\frac{\ln x}{1+x^n}}$ using complex analysis Awesome answer. Dec 6 comment Subscript in maximum notation If you found the answer to be correct and helpful, you might want to accept it by clicking the "Right" sign besides the answer. :-) Dec 4 comment Properties about Matrices that can be proved by only using Block Multiplication of Matrices @joriki yes, I rewrote the sentence. I hope this formulates my problem better. Nov 26 comment Compute $\lim_{n\to\infty}(n-(\arccos(1/n)+\cdots+\arccos(n/n)))$ That the error part decreases as $1/N^2$ is decisive for the proof. Nice. :-) Nov 15 comment Information on the sum $\sum_{n=1}^\infty \frac{\log n}{n!}$ @did Sorry, you misunderstood me. I wanted to ask the reason for your comment "What is the meaning of "play with formulas equivalent to the original question and bringing no new information nor understanding to it"? " Nov 15 comment Information on the sum $\sum_{n=1}^\infty \frac{\log n}{n!}$ @did Complete noob here. So may I ask you to explain why the two formulas are equivalent? I have got a suspicion it is so because it is basically just replacing formulas with numbers? Am I correct? For example, is this similar to saying that $\sum_{n=0}^\infty \frac{1}{n!} = e$?