Answer by Thomas Andrews for Proof for an identity (from Ramanujan written)
Assume $(m-1)^2<9n+17<m^2$. (We never get equality because $9n+17\equiv -1\pmod{3}$.) So we get: $$(m-1)^2+1\leq 9n+17\leq m^2-1$$or $$\frac{(m-1)^2-16}{9}\leq n\leq\frac{m^2-18}{9}.$$So...
View ArticleAnswer by Khosrotash for Proof for an identity (from Ramanujan written)
if $$|\sqrt n+\sqrt {n+2}+\sqrt{n+4} - \sqrt {9n+17}|<\epsilon \\\Rightarrow \lfloor \sqrt n+\sqrt {n+2}+\sqrt{n+4} \rfloor=\lfloor \sqrt {9n+17}\rfloor $$ so ,I will show $|\sqrt n+\sqrt...
View ArticleProof for an identity (from Ramanujan written)
I saw an identity by Ramanujan$$\forall n \in \mathbb{N} ,n>1 :\lfloor \sqrt n+\sqrt {n+2}+\sqrt{n+4} \rfloor=\lfloor \sqrt {9n+17}\rfloor$$ I tried to prove it by limit definition . I post my trial...
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