(a) $\textbf{Claim}:$ $x_n \geq n^2$. $\textbf{Proof:}$ Clear for $n=1$. Assume for $n=k$. For $n=k+1$, we have: $x_{k+1} \geq x_k+3\sqrt{x_k} \geq k^2+3k \geq (k+1)^2$. Thus, $0<\frac{n}{x_n} \leq \frac{n}{n^2}=\frac{1}{n}$ and hence as $n \to \infty$, $\frac{n}{x_n} \to 0$ (b) Partial progress: Using claim from part (a), $x_{n+1} \leq x_n+3\sqrt{x_n}+1 \leq \left(\sqrt{x}_n+\frac{3}{2}\right)^2$. We know $x_1 =1$ and thus by induction one can easily observe that $x_n \leq \left(\frac{3n-1}{2}\right)^2 $and thus we have $1 \geq \lim_{n \to \infty} \frac{n^2}{x_n} \geq \frac{4}{9}$

Shreehari_Bodas wrote: (a) $\textbf{Claim}:$ $x_n \geq n^2$. $\textbf{Proof:}$ Clear for $n=1$. Assume for $n=k$. For $n=k+1$, we have: $x_{k+1} \geq x_k+3\sqrt{x_k} \geq k^2+3k \geq (k+1)^2$. Thus, $0<\frac{n}{x_n} \leq \frac{n}{n^2}=\frac{1}{n}$ and hence as $n \to \infty$, $\frac{n}{x_n} \to 0$ Very nice solution ,and for b) Stolz-Cesaro.

teomihai wrote: Shreehari_Bodas wrote: (a) $\textbf{Claim}:$ $x_n \geq n^2$. $\textbf{Proof:}$ Clear for $n=1$. Assume for $n=k$. For $n=k+1$, we have: $x_{k+1} \geq x_k+3\sqrt{x_k} \geq k^2+3k \geq (k+1)^2$. Thus, $0<\frac{n}{x_n} \leq \frac{n}{n^2}=\frac{1}{n}$ and hence as $n \to \infty$, $\frac{n}{x_n} \to 0$ Very nice solution ,and for b) Stolz-Cesaro. Hi! Can you please explain what Stolz-Cesaro is? Thanks!

Shreehari_Bodas wrote: teomihai wrote: Shreehari_Bodas wrote: (a) $\textbf{Claim}:$ $x_n \geq n^2$. $\textbf{Proof:}$ Clear for $n=1$. Assume for $n=k$. For $n=k+1$, we have: $x_{k+1} \geq x_k+3\sqrt{x_k} \geq k^2+3k \geq (k+1)^2$. Thus, $0<\frac{n}{x_n} \leq \frac{n}{n^2}=\frac{1}{n}$ and hence as $n \to \infty$, $\frac{n}{x_n} \to 0$ Very nice solution ,and for b) Stolz-Cesaro. Hi! Can you please explain what Stolz-Cesaro is? Thanks!

teomihai wrote: Exist $ \lim_{n} \frac{(n+1)^2-n^2}{x_{n+1}-x_n}\leq{\lim_{n}\frac{2n+1}{(n+1)^2}}=0$ This is incorrect

BUMP! Can anyone please complete part (b)? Thanks!

$\lim{\frac{(n+1)-n}{\sqrt{x_{n+1}}-\sqrt{x_n}}}=\lim{\frac{1}{\sqrt{x_n+3\sqrt{x_n} + \frac{n}{\sqrt{x_n}}}-\sqrt{x_n}}}$ $=\lim{\frac{\sqrt{x_n+3\sqrt{x_n} + \frac{n}{\sqrt{x_n}}}+\sqrt{x_n}}{3\sqrt{x_n} + \frac{n}{\sqrt{x_n}}}}$ $=\lim{\frac{\sqrt{1+\frac{3}{\sqrt{x_n}} + \frac{n}{x_n\sqrt{x_n}}}+1}{3 + \frac{n}{x_n}}}=\frac{1+1}{3}=\frac{2}{3}$ Stolz's theorem $\implies \lim\frac{n}{\sqrt{x_n}}=\frac{2}{3} \implies \lim\frac{n^2}{x_n}=\frac{4}{9}$

tait1k27 wrote: $\lim{\frac{(n+1)-n}{\sqrt{x_{n+1}}-\sqrt{x_n}}}=\lim{\frac{1}{\sqrt{x_n+3\sqrt{x_n} + \frac{n}{\sqrt{x_n}}}-\sqrt{x_n}}}$ $=\lim{\frac{\sqrt{x_n+3\sqrt{x_n} + \frac{n}{\sqrt{x_n}}}+\sqrt{x_n}}{3\sqrt{x_n} + \frac{n}{\sqrt{x_n}}}}$ $=\lim{\frac{\sqrt{1+\frac{3}{\sqrt{x_n}} + \frac{n}{x_n\sqrt{x_n}}}+1}{3 + \frac{n}{x_n}}}=\frac{1+1}{3}=\frac{2}{3}$ Stolz's theorem $\implies \lim\frac{n}{\sqrt{x_n}}=\frac{2}{3} \implies \lim\frac{n^2}{x_n}=\frac{4}{9}$ beautiful solution! All ,Happy new year!

Here is an elementary solution for b) (i.e. without using Stolz-Cèsaro): Claim 1: $x_n \ge n^2$. Proof by induction. $n=1$ is clear. In the induction step, we have \[x_{n+1} \ge x_n+3\sqrt{x_n} \ge n^2+3n \ge (n+1)^2.\]Claim 2: $x_n \le \frac{9n^2}{4}$. Proof by induction. $n=1$ is clear. In the induction step, we have \[x_{n+1} \le x_n+3\sqrt{x_n}+1 \le \frac{9n^2}{4}+\frac{9n}{2}+1 \le \frac{9}{4}\left(n+1\right)^2.\]This already shows that the limit (if it exists) is at most $\frac{9}{4}$. To finish, fix any $0<\beta<\frac{3}{2}$. We claim that there is some integer $k>0$ such that $x_n \ge \beta^2(n-k)^2$ for all $n \ge 2k$. Indeed, any $k>\frac{\frac{2}{3}-\beta^2}{3\beta-2\beta^2}$ does the job. Proof by induction. For $n=2k$, we have $x_{2k} \ge 4k^2 \ge \beta^2 k^2$ since $\beta<2$. Now in the induction step we have \[x_{n+1} \ge x_n+3\sqrt{x_n}+\frac{2}{3} \ge \beta^2(n-k)^2+3\beta(n-k)+\frac{2}{3}=\beta^2(n-k+1)^2+(3\beta-2\beta^2)(n-k)+\frac{2}{3}-\beta^2 \ge \beta^2(n-k+1)^2+(3\beta-2\beta^2)k+\frac{2}{3}-\beta^2 \ge \beta^2(n-k+1)^2\]by our choice of $k$. Hence $\liminf_{n \to \infty} \frac{x_n}{n^2} \ge \frac{9}{4}$ and since we already proved $x_n \le \frac{9n^2}{4}$, the limit exists and must equal $\frac{9}{4}$.

Here is my solution for this problem Solution a) It's easy to see that: $x_n > 0, \forall n \in \mathbb{N}$ Since: $x_{n + 1} - x_n = 3\sqrt{x_n} + \dfrac{n}{\sqrt{x_n}} > 0$ then: $x_{n + 1} > x_n > 0$ or $(x_n)$ is increasing Suppose $(x_n)$ has finite limit, let $\lim x_n = L > 0$ We have: $x_{n + 1} = x_n + 3\sqrt{x_n} + \dfrac{n}{\sqrt{x_n}} > x_n + 3 \sqrt{x_n}, \forall n \in \mathbb{N}$ So: $\lim x_{n + 1} \ge \lim (x_n + 3 \sqrt{x_n})$ or $L > L + 3 \sqrt{L}$ Hence: $\sqrt{L} < 0$, which is a contradiction Then: $\lim x_n = + \infty$ By Stolz theorem, we have: $\lim \dfrac{n}{x_n} = \lim \dfrac{n + 1 - n}{x_{n + 1} - x_n} = \lim \dfrac{1}{3 \sqrt{x_n} + \dfrac{n}{\sqrt{x_n}}} = 0$ b) By Stolz theorem, we have: $\lim \dfrac{n}{\sqrt{x_n}} = \lim \dfrac{n + 1 - n}{\sqrt{x_{n + 1}} - \sqrt{x_n}} = \lim \dfrac{\sqrt{x_{n + 1}} + \sqrt{x_n}}{x_{n + 1} - x_n} = \lim \dfrac{\sqrt{x_n + 3 \sqrt{x_n} + \dfrac{n}{\sqrt{x_n}}} + \sqrt{x_n}}{3 \sqrt{x_n} + \dfrac{n}{\sqrt{x_n}}} = \lim \dfrac{\sqrt{1 + \dfrac{3}{\sqrt{x_n}} + \dfrac{n}{x_n}} + 1}{3 + \dfrac{n}{x_n}} = \dfrac{2}{3}$ So: $\lim \dfrac{n^2}{x_n} = \dfrac{4}{9}$

Thank you guys.