Attempt. Clearly, the root is $x=a^{1/n}$. Now, in the Newton method, the sequence given iteratively by $$ x_{n+1} = x_n - \frac{ f(x_n) }{f'(x_n)} $$ should converge to the root. We have $$ x_{n+1} = x_n - \frac{ x_n^n - a }{n x_n^{n-1} } $$ and so $$ x_{n+1} = x_n - \frac{x_n}{n} + \frac{a x_n^{1-n} }{n} $$ How can we show that $x_n$ indeed converges to $a^{1/n}$? Im trying to show monotonicty and boundedness but it seems that it will take some work with the given iteration. Is my approach correct?
Why don't you look at the proof of convergence in the general case? Since this is just a special case of the general case, it should not be too difficult to tailor the general proof to the more specific one. Check out the section labeled "Proof of quadratic convergence for Newton's iterative method" on https://en.wikipedia.org/wiki/Newton%27s_method. This proof uses Taylor's Theorem with additional assumptions to prove convergence. So, you can use Taylor's Theorem as in the general proof, and show that $f(x)$ satisfies the criteria listed in the proof.