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I have and the following recurrence relaation:

(x_{n})_{n\geq 0}, x_{n+1}=x_{n}+\frac{1}{a}\cdot x_{n}^{1-a},a\geq 1,n\geq 0,x_{0}=1

I need to solve \lim_{n\rightarrow \infty }\frac{x_{n}^{a}}{n}

I found that this string is increasing and I tried to find z and w from general form x_{n+1}=z\cdot x_{n}+w

I tried to factorize xn but I didn't get too far.

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You can prove by induction that $x_n\geq 1$. This implies that for all $n\geq 0$, $$x_{n+1}-x_n\geq \frac 1 a$$ Summing this up leads to $$x_n\geq 1 + \frac n a$$ So, clearly, $x_n\rightarrow +\infty$ as $n\rightarrow +\infty$.

The trick is to look for an estimate of $u_n = x_n^a$. Note that just like $(x_n)_{n\geq 0}$, $(u_n)_{n \geq 0}$ tends to $+\infty$, so we can use the appropriate Taylor expansion: $$\begin{split} u_{n+1} &= \left (x_n + \frac 1 a x_n^{1-a}\right)^a\\ &= u_n \left (1 + \frac 1 a \frac 1 {u_n}\right)^{a}\\ &= u_n \left (1 + \frac 1 {u_n} + o\left (\frac 1 {u_n}\right)\right)\\ &= u_n +1 +o(1) \end{split}$$ So summing this up leads to $u_n=n+o(n)$, which implies that $$\lim_{n\rightarrow +\infty} \frac {x_n^a}n = 1$$

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  • $\begingroup$ Thanks a lot! :) $\endgroup$ – Vali RO Jan 7 '19 at 12:34
  • $\begingroup$ You're welcome! $\endgroup$ – Stefan Lafon Jan 7 '19 at 16:57

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