Question

lim_(n->\infty )(1+3+5+...+(2n-1))/(1+2+3+...+n)

Answer 1

$\displaystyle 1+3+5+...+(2n-1) = n\cdot(2n-1+1)/2 = n^2\\1+2+3+4...+n = n(n+1)/2 = (n^2+n)/2\\\\\\\lim\limits_{n\rightarrow\infty}\frac{1+3+...+(2n-1)}{1+2+3+...+n} = \lim\limits_{n\rightarrow\infty}\frac{2n^2}{n^2+n} = \\\\ =\lim\limits_{n\rightarrow\infty}\frac{2}{1+1/n}= 2$