value of $\delta$ for which the sequence converges A sequence $a_n$ is defined by $a_1=1$ and the induction formula $a_{n+1}=\sqrt{1+{a_n}^{\delta}}$ where $\delta \gt 0$ and is a real number. What is the condition on $\delta$ for which the sequence converges?
If the sequence converges then suppose $\lim_{n\to \infty}a_n=l$,then $l^2=1+l^{\delta}$ which is not helping. 
Thanks in advance!! 
 A: The equation $l^2=1+l^{\delta}$ is helpful. For it is clear that $a_n\ge 1$ for all $n$. Thus the limit, if it exists, is $\ge 1$. But if $\delta\ge 2$, then $x^2=1+x^{\delta}$ has no solution $\ge 1$. Thus we cannot have convergence if $\delta\ge 2$.
We show that if $0\lt \delta\lt 2$, we have convergence.  First we show the sequence $(a_n)$ is increasing. Certainly $a_2\gt a_1$. The rest is done by induction. Suppose that for a certain $k$ we have $a_{k+1}\lt a_{k}$. Then 
$$a_{k+2}=\sqrt{1+a_{k+1}^{\delta}}\gt \sqrt{1+a_{k}^{\delta}}=a_{k+1}.$$
It remains to show that if $0\lt \delta\lt 2$ then the sequence $(a_n)$ is bounded. Let $r$ be the positive root of $x^2=1+x^\delta$. We show  that $a_n\lt r$ for all $n$, or equivalently that $a_n^2\lt 1+a_n^\delta$. 
So suppose that $a_k\lt r$, meaning that $a_k^2\lt 1+a_k^\delta$. Then 
we need to show that $a_{k+1}^2 \lt 1+a_{k+1}^\delta$. So we need to show that $1+a_k^\delta\lt 1+a_{k+1}^\delta$. This has already been done, since we have seen that the sequence $(a_n)$ is increasing. 
Since for $0\lt \delta\lt 2$, the sequence $(a_n)$ is increasing and bounded above, the sequence  converges for these $\delta$. 
