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 Oct 20 comment Rate of convergence of $\left[ \left( \sum\limits_{i=j}^n {2i+1}\right)^{\frac{1}{2}}\right]$ $[\ ]$ is for integer parts, $\{\ \}$ for fractional parts. Oct 20 answered Rate of convergence of $\left[ \left( \sum\limits_{i=j}^n {2i+1}\right)^{\frac{1}{2}}\right]$ Oct 20 comment Eigenvectors of $\left( \begin{array}{ccc} a & 0 \\ 0 & -b \end{array} \right)$ "Assuming $a\neq 0 \neq b$" Why? Oct 20 comment Verifying solution of difference equation? Maybe you can show that $h_x=t_1^x$ solves the difference equation, then that $h_x=t_2^x$ solves it as well, then the full case? What is stopping you? Oct 20 comment Ladder against a wall. "In this form" refers to the fact that the sequence of coefficients of the polynomial is symmetric with respect to its midpoint, here (a,b,c,b,a). Oct 20 revised Prove the inequality$\int_0^{+\infty} {\sin x \over x}dx<\int_0^\pi {\sin x \over x}dx$ added 391 characters in body Oct 20 revised Prove the inequality$\int_0^{+\infty} {\sin x \over x}dx<\int_0^\pi {\sin x \over x}dx$ added 1 character in body; edited title Oct 20 answered Prove the inequality$\int_0^{+\infty} {\sin x \over x}dx<\int_0^\pi {\sin x \over x}dx$ Oct 20 comment The sequence $\sin \left({n\pi}\over 6\right)$ has the superior limit $L=1\dots$ @bd1251252 Re the period, note that $\sin((n+12)\pi/6)=\sin((n\pi/6)+2\pi)=\sin(n\pi/6)$. Re the 5, see my answer. Oct 20 comment How to derive this inequality Added the argument. Oct 20 revised How to derive this inequality added 262 characters in body Oct 20 comment The sequence $\sin \left({n\pi}\over 6\right)$ has the superior limit $L=1\dots$ Alternatively, compute explicitely the first values of the sequence and observe what you get... Oct 20 comment The sequence $\sin \left({n\pi}\over 6\right)$ has the superior limit $L=1\dots$ @bd1251252 My previous comment deals only with the number of distinct values and does not touch on limsup and liminf. If you want to understand where 12 and 5 are coming from, posting cryptic comments is not the best way to proceed. Re 12: what might be the period of the sequence of general term sin(n.pi/6), according to you? Oct 20 comment The sequence $\sin \left({n\pi}\over 6\right)$ has the superior limit $L=1\dots$ The period is 12, a priori this makes for 12 numbers. But amongst these, 5 values are achieved twice and 2 values are achieved only once. Thus 5+2 distinct values. Oct 20 answered How to derive this inequality Oct 20 comment The sequence $\sin \left({n\pi}\over 6\right)$ has the superior limit $L=1\dots$ Example of what? Oct 20 revised The sequence $\sin \left({n\pi}\over 6\right)$ has the superior limit $L=1\dots$ added 107 characters in body Oct 20 comment The sequence $\sin \left({n\pi}\over 6\right)$ has the superior limit $L=1\dots$ @bd1251252 ?? Know what a periodic sequence is? Oct 20 comment The sequence $\sin \left({n\pi}\over 6\right)$ has the superior limit $L=1\dots$ @bd1251252 Period 12 and 5 values achieved twice in each period make for 12-5=7 distinct values. Oct 20 answered The sequence $\sin \left({n\pi}\over 6\right)$ has the superior limit $L=1\dots$