Cfrgtkky

up vote 1 down vote favorite Let $$a_1=3quad , quad a_{n+1}=frac{3a_n}{4}+frac{1}{a_n}$$ Show that $a_n$ converges. I know that I need to prove that $a_n$ is monotonic and bounded. I've assumed that the limit exists, made the calculation and get that $L=2$ , hence claiming two claims: $a_nge 2$ (I know that 0 is a simpler bound) $a_n$ is monotonically decreasing. Now, if I can prove the first claim. then I can write $$a_{n+1}=frac{3a_n}{4}+frac{1}{a_n}=frac{3a_n}{4}+frac{a_n}{a_n^2}le frac{3a_n}{4}+frac{a_n}{4}=a_n$$ which proves the second claim. My only problem is proving the first one. If it is not the right direction, please hint me. calculus share | cite | improve this question ...