Let $a_{n+1}=\sqrt{a_n+n}$ with $a_1=1$ then find $\lim_{n \to \infty} (a_n -\sqrt{n})$

Question

Let $a_{n+1}=\sqrt{a_n+n}$ with $a_1=1$ then find $\displaystyle \lim_{n \to \infty} (a_n -\sqrt{n}).$

My Try: $$\lim_n( \sqrt{a_{n}+n}-\sqrt{n}) \times \dfrac{( \sqrt{a_{n}+n}+\sqrt{n}) }{( \sqrt{a_{n}+n}+\sqrt{n})}=\lim_n \dfrac{1}{( \sqrt{\dfrac{1}{a_{n}}+\dfrac{n}{a_{n}^2}}+\sqrt{\dfrac{n}{a_{n}^2}})}.$$ Now what do I do?

Answer 1

We first show by induction that for any $n\geq 1$, $$\sqrt{n}\leq a_n\leq \sqrt{n+2\sqrt{n}}\tag{1}.$$ The inequalities hold for $n=1$. As regards the inductive step we have that for $n\geq 1$, $$a_{n+1}=\sqrt{a_n+n}\geq \sqrt{\sqrt{n}+n}\geq \sqrt{1+n},$$ and $$a_{n+1}=\sqrt{a_n+n}\leq \sqrt{\sqrt{n+2\sqrt{n}}+n} \leq \sqrt{(\sqrt{n}+1)+n}\leq \sqrt{2\sqrt{n+1}+n+1}.$$ Hence, by (1) and by the Squeeze Theorem, $$1\leq \frac{a_n}{\sqrt{n}}\leq \sqrt{1+\frac{2}{\sqrt{n}}}\to 1\implies \lim_{n\to \infty}\frac{a_n}{\sqrt{n}}=1$$ Finally, as $n$ goes to infinity, $$ \begin{align} a_{n+1} -\sqrt{n+1}&=\sqrt{a_n+n} -\sqrt{n+1}=\frac{a_n-1}{\sqrt{a_n+n} +\sqrt{n+1}}\\&=\frac{\frac{a_n}{\sqrt{n}}-\frac{1}{\sqrt{n}}}{\sqrt{\frac{a_n}{n}+1} +\sqrt{1+\frac{1}{n}}}\to \frac{1}{2}. \end{align}$$

Answer 2

Let $b_n=a_n-\sqrt n$ therefore:$$b_{n+1}=\sqrt{b_n+\sqrt n+n}-\sqrt{n+1}=\dfrac{b_n+\sqrt n-1}{\sqrt{b_n+\sqrt n+n}+\sqrt{n+1}}=\dfrac{\dfrac{b_n-1}{\sqrt n}+1}{\sqrt{\dfrac{n+1}{n}}+\sqrt{\dfrac{b_n}{n}+\dfrac{1}{\sqrt n}+1}}\qquad\qquad\qquad(1)$$if we assume $\lim_{n\to\infty}b_n=l<\infty$ we have:$$l=\dfrac{1}{2}$$by substituting in $(1)$ and $n\to\infty$, but this is not a proof! We have now an intuition and a probable limit whether true or not. Instead here after we define $\epsilon_n=b_n-\frac{1}{2}$ and try to prove that $\lim_{n\to\infty}\epsilon_n=0$. By substituting $b_n=\epsilon_n+\frac{1}{2}$ in $(1)$ we obtain:$$\large\epsilon_{n+1}=\dfrac{\epsilon_n-\frac{1}{2}-\frac{1}{\sqrt n+\sqrt{n+1}}}{\sqrt{\epsilon_n+(\sqrt n+\frac{1}{2})^2+\frac{1}{4}}+\frac{1}{2}+\sqrt{n+1}}$$by taking absolute and applying some inequalities we have: $$\large|\epsilon_{n+1}|=\left |\dfrac{\epsilon_n-\frac{1}{2}-\frac{1}{\sqrt n+\sqrt{n+1}}}{\sqrt{\epsilon_n+(\sqrt n+\frac{1}{2})^2+\frac{1}{4}}+\frac{1}{2}+\sqrt{n+1}}\right |\le \dfrac{\left |\epsilon_n-\frac{1}{2}-\frac{1}{\sqrt n+\sqrt{n+1}}\right |}{\left |\frac{1}{2}+\sqrt{n+1}\right |}\le\dfrac{|\epsilon_n|+\frac{1}{2}+\frac{1}{\sqrt n+\sqrt{n+1}}}{\frac{1}{2}+\sqrt{n+1}}\qquad\qquad\qquad (2)$$So far we have bounded $|\epsilon_{n+1}|$. Now we need a trick: we know that $|\epsilon_2|=|b_2-\frac{1}{2}|=|a_2-\sqrt 2-\frac{1}{2}|=\frac{1}{2}\le 1$ so for any $n\ge 2$:$$|\epsilon_{n+1}|\le \dfrac{|\epsilon_n|+\frac{1}{2}+\frac{1}{\sqrt n+\sqrt{n+1}}}{\frac{1}{2}+\sqrt{n+1}}\le \dfrac{|\epsilon_n|+\frac{1}{2}+\frac{1}{\sqrt n+\sqrt{n+1}}}{2}\le\dfrac{|\epsilon_n|+1}{2}\le 1$$this show that for $n\ge 2$ we have $|\epsilon|\le 1$. Now substituting this in $(2)$ gives us:$$|\epsilon_{n+1}|\le \dfrac{|\epsilon_n|+\frac{1}{2}+\frac{1}{\sqrt n+\sqrt{n+1}}}{\frac{1}{2}+\sqrt{n+1}}\le \dfrac{\frac{3}{2}+\frac{1}{\sqrt n+\sqrt{n+1}}}{\frac{1}{2}+\sqrt{n+1}}\le\dfrac{4}{1+2\sqrt{n+1}}$$since $\lim_{n\to\infty}\dfrac{4}{1+2\sqrt{n+1}}=0$ so $\lim_{n\to\infty}|\epsilon_n|=0$ and this is what we wanted from beginning. Therefore:$$\Large\lim_{n\to\infty}a_n-\sqrt{n}=\frac{1}{2}$$