Aryabhata
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 Mar26 revised Proof : Limit of a sequence edited tags Mar19 comment Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$ +1: Nice....... Mar19 revised Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$ more informative title. Mar19 comment Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$ This is not pedantry. I don't think this is as trivial as you seem to be implying. Unless you have a proof, you are just handwaving. Anyway, you are free not to elaborate, and I am free to leave the downvote intact. Mar19 revised Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$ edited tags Mar19 comment Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$ Please take at least a day or two before accepting an answer. In this case, the accepted answer is incomplete. Mar19 comment Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$ The heuristic is only a first step. Making it rigorous is the hard part. -1 till there is a proof. Sorry. Mar18 comment Quotient of a regular language @AstroNauft: You don't need to determine anything. It is a non-constructive proof. $L$ could be any language. Mar18 comment The limit : $\lim _{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1}$ Mar17 comment How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$? @ADG: It is my name, I will spell it as I want! :-). Just kidding :-). Apparently, it is actually Aryabhata and not Aryabhatta. In fact I had it as Aryabhatta till ShreevatsaR corrected me. Mar15 awarded Nice Answer Mar13 revised Big O notation and limits edited tags Mar11 revised Inequality between real numbers $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$ added 199 characters in body Mar8 comment Inequality between real numbers $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$ If I haven't made any mistake that is... This looks too simple! Mar8 answered Inequality between real numbers $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$ Feb22 awarded Nice Question Feb17 awarded Good Answer Feb16 awarded Enlightened Feb14 awarded Nice Answer Feb11 revised Maths question from an IQ test edited tags