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 Feb19 comment No Bijection from set $X$ to $X - \{x\}$ Is a proof possible without some form of cardinality being brought into picture? Feb12 comment Proving upper bounds? @HagenvonEitzen Wow!! Feb12 comment Is there an axiom that prevents other axioms from contradicting each other? en.wikipedia.org/wiki/Three_classic_laws_of_thought Feb5 comment When $dy/dx =0$ for all $x$ in the domain, is $dx/dy$ also zero? +1 For the proper explanation. Feb1 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. Feb1 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. :-) Feb1 comment Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Thanks. A different solution. :-) Feb1 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. Feb1 comment Limit of $s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx$ as $n \to \infty$ Arrrrgh. Thanks. Feb1 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 Feb1 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. Feb1 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. Feb1 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? Jan4 comment Calculating the integral $\int_0^{\infty}{\frac{\ln x}{1+x^n}}$ using complex analysis Awesome answer. Dec26 comment How to check if a point is inside a rectangle? Excellent answer. Dec6 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. :-) Dec4 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. Nov26 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. :-) Nov15 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"? " Nov15 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 $\sum_{n=0}^\infty \frac{1}{n!} = e$?