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Let be the sequence $(x_{n})$, $n \geq 0$ with $x_{0}>0$, and for any $n$ natural number, $x_{n+1}=x_{n}+\frac{1}{\sqrt{x_{n}}}$. I'm required to calculate: $$ \lim_{n\rightarrow\infty} \frac{{x_{n}}^{3}}{{n}^{2}}$$ I still try to figure out what part I should start with. Any solution is very welcomed. |
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The associated differential equation is $\xi'(t)=\frac1{\sqrt{\xi(t)}}$, whose solutions are $\xi(t)^{3/2}=\frac32t+C$. This suggests to look at the sequence $(z_n)$ defined by $z_n=x_n^{3/2}$. Thus, $$ z_{n+1}=z_n(1+z_n^{-1})^{3/2}. $$ First, $(1+u)^{3/2}\geqslant1+\frac32u$ for every $u\geqslant0$ hence $z_{n+1}\geqslant z_n+\frac32$, thus $z_n\geqslant\frac32n+z_0$ and in particular $z_n\to+\infty$. On the other hand, $(1+u)^{3/2}\leqslant1+\frac32u+\frac38u^2$ for every $u\geqslant0$ hence $z_{n+1}\leqslant z_n+\frac32+\frac38z_n^{-1}$. This shows that $z_n\leqslant\frac32n+z_0+\frac38t_n$ with $t_n=\sum\limits_{k=0}^{n-1}z_k^{-1}$. Since $z_n\to+\infty$, $t_n=o(n)$ and $z_n\leqslant\frac32n+o(n)$. Finally, $\frac{z_n}n\to\frac32$ hence $$\lim\limits_{n\to\infty}\frac{x_n^3}{n^2}=\frac94. $$ |
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You had a typo in the question, so let's explore a slightly more general solution. Suppose there is a positive finite limit of $\lim_{n\to\infty}\dfrac{x_n}{n^k} = c$ for some $c$ and $k$. Then for large $n$, $x_n \approx c n^k$ and $x_{n+1}-x_n \approx \dfrac{1}{\sqrt{c} n^{k/2}}$ but taking $n$ as real $\frac{dx}{dn} \approx kcn^{k-1}$. To make these two correspond you need $k-1=-k/2$ and $kc=1/\sqrt{c}$ which have the solutions $k=\dfrac{2}{3}$ and $c=\sqrt[3]{\dfrac94}$. So the limit of $\dfrac{x}{n^{2/3}}$ is $\sqrt[3]{\dfrac94}$ and of $\dfrac{x^3}{n^{2}}$ is ${\dfrac94}$. For a formal proof see did's answer. |
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