Show that $\lim\limits_{n\to\infty} a_n=a^{1/k}$ let $a\in \mathbb{R}:a>0$ and $k\in \mathbb{N}_{\geq 2}$. Furthermore $(a_n)_{n\geq 1} \subset \mathbb{R}$: $$a_1=1+a \text{ and } a_{n+1}=a_n\left(1+\frac{a-a_n^k} {ka_n^k}\right) \text{ for } n\in \mathbb{N}$$
show that $\lim\limits_{n\to\infty} a_n=a^{1/k}$. My problem is that I can't figure out what $a_n$ is equal to... Any ideas?
 A: This is Newton's iteration for
$f(x)
=x^k - a = 0
$.
$f'(x)
=kx^{k-1}
$.
Since the iteration is
$x_{n+1}
=x_n-\dfrac{f(x_n)}{f'(x_n)}
$,
$\begin{array}\\
x_{n+1}
&=x_n-\dfrac{f(x_n)}{f'(x_n)}\\
&=x_n-\dfrac{x_n^k - a }{kx_n^{k-1}}\\
&=x_n-x_n\dfrac{x_n^k - a }{kx_n^{k}}\\
&=x_n\left(1-\dfrac{x_n^k - a }{kx_n^{k}}\right)\\
&=x_n\left(1+\dfrac{a-x_n^k }{kx_n^{k}}\right)\\
\end{array}
$
I will now show that,
once $x_n^k$ is close to $a$,
then the iteration converges.
First,
$\begin{array}\\
x_{n+1}-a^{1/k}
&=x_n-a^{1/k}-\dfrac{f(x_n)}{f'(x_n)}\\
&=x_n-a^{1/k}-\dfrac{x_n^k - a }{kx_n^{k-1}}\\
&=x_n-a^{1/k}-\dfrac{(x_n - a^{1/k})\sum_{j=0}^{k-1}x_n^ja^{(k-1-j)/k} }{kx_n^{k-1}}\\
&=(x_n-a^{1/k})\left(1-\dfrac{\sum_{j=0}^{k-1}x_n^ja^{(k-1-j)/k} }{kx_n^{k-1}}\right)\\
&=(x_n-a^{1/k})\left(1-\frac1{k}\sum_{j=0}^{k-1}x_n^{j-k+1}a^{(k-1-j)/k} \right)\\
&=(x_n-a^{1/k})\left(1-\frac1{k}\sum_{j=0}^{k-1}(a^{1/k}/x_n)^{k-1-j} \right)\\
&=(x_n-a^{1/k})\left(1-\frac1{k}\sum_{j=0}^{k-1}(a^{1/k}/x_n)^j \right)\\
\end{array}
$
Then, if
$1-c
\lt a/x_n^k
\lt 1+c
$,
then
$1-c/k
\lt a^{1/k}/x_n
\lt 1+c/k
$,
so that,
by Bernoulli's inequality,
$1-cj/k < (a^{1/k}/x_n)^j < 1+cj/k
$,
so
$\sum_{j=0}^{k-1}(1-cj/k)
\lt \sum_{j=0}^{k-1}(a^{1/k}/x_n)^j 
\lt \sum_{j=0}^{k-1}(1+cj/k)
$,
or
$k-c(k-1)/2
\lt \sum_{j=0}^{k-1}(a^{1/k}/x_n)^j 
\lt k+c(k-1)/2
$,
or
$-c(k-1)/(2k)
\lt 1-\frac1{k}\sum_{j=0}^{k-1}(a^{1/k}/x_n)^j
\lt c(k-1)/(2k)
$,
so
$|1-\frac1{k}\sum_{j=0}^{k-1}(a^{1/k}/x_n)^j|
\lt c/2
$.
Therefore
$|x_{n+1}-a^{1/k}|
\lt (c/2)|x_n-a^{1/k}|
$
so that
the iteration will converge.
A: We have $a_1 = 1 + a$ and $a_{n+1} = h(a_n)$ where the function $h$ is defined on $(0, \infty)$ as
$$
h(x) = x \left( 1 + \frac{a-x^k}{kx^k}\right)\, .
$$
The idea is to show (in this order):


*

*$a_n \ge a^{1/k}$  for all $ n \in \Bbb N$.

*$(a_n)$ is monotonically decreasing.

*The limit $L = \lim_{n \to \infty} a_n$ exists.

*$L = a^{1/k}$. 


Proof of (1): For $n=1$ this follows from Bernoulli's inequality:
$$
 a_1^k = (1+a)^k \ge 1+ka \ge a \implies a_1 \ge a^{1/k} \, .
$$
For $n \ge 2$ we note that $h$ can be written in the form
$$
h(x) = \frac 1k \left( (k-1)x + \frac{a}{x^{k-1}}\right) \, .
$$
The expression on the right is the arithmetic mean of the $k$ numbers $(x, \ldots, x, \frac{a}{x^{k-1}})$, and that is greater or equal to the geometric mean:
$$
h(x) \ge \left( x^{k-1}\frac{a}{x^{k-1}}\right)^{1/k} = a^{1/k} 
$$
so that $a_n =h(a_{n-1})\ge a^{1/k} $ for all $n \ge 2$.
Proof of (2): $a_n^k \ge a$ implies
$$
a_{n+1} = a_n \left( 1 + \frac{a-a_n^k}{ka_n^k}\right) \le a_n \, .
$$
Proof of (3): This follows from (1) and (2) and the
monotone convergence theorem.
Proof of (4): The limit of the sequence must satisfy $L = h(L)$.
