a limit of recurrence relation

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.

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 $$\lim_{n\rightarrow +\infty} \frac {x_n^a}n = 1$$

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