gobathegreat wrote: Let $a_1=1$ and $a_{n+1}=a_{n}+\frac{1}{2a_n}$ for $n \geq 1$. Prove that $a)$ $n \leq a_n^2 < n + \sqrt[3]{n}$ $b)$ $\lim_{n\to\infty} (a_n-\sqrt{n})=0$ The second statement follows from the first one, because $\sqrt{n}\leq a_n < \sqrt{n+\sqrt[3]{n}}$ and $$\sqrt{n+\sqrt[3]{n}}-\sqrt{n}= \frac{\sqrt[3]{n}}{\sqrt{n+\sqrt[3]{n}}+\sqrt{n}}\to 0.$$For the first statement we use induction. $a_{n+1}^2=a_n^2+\frac{1}{4a_n^2}+1$, so 1. $a_{n+1}^2\geq n+\frac{1}{4a_n^2}+1 > n+1$. 2. $a_{n+1}^2\leq n+\sqrt[3]{n}+\frac{1}{4n}+1=(n+1)+\sqrt[3]{n}+\frac{1}{4n} < (n+1) + \sqrt[3]{n+1}$, because $$\sqrt[3]{n+1}-\sqrt[3]{n}= \frac{1}{\sqrt[3]{(n+1)^2}+\sqrt[3]{n(n+1)}+\sqrt[3]{n^2}} > \frac{1}{3n} >\frac{1}{4n},$$when $n > 2$.