A Numerical Sequences proof. Let $ 0<a<1 $ be a real number, and let $ a_n \in \left ( -1,0 \right )$ defined by the relation $ \sqrt[n]{a} = 1+a_n$ ,   $ n \in \mathbb{N}$.
Show the following inequality:
$ |a_n| \le \frac{1}{n} \bigl( \frac{1-a}{a}\bigr)$,    $ n \ge 1$.
This problem was proposed by my professor of Mathematical Analysis $2$. The following is my personal solution, different from the one expected.
 A: By hypothesis it's $ \sqrt[n]{a} = 1 + a_n$ , in particular $ a_n=\sqrt[n]{a}-1<0$ $\to$ $|a_n|=1-\sqrt[n]{a}$.
Second member can be viewed as a division of polynomials:
$$ 1 - \sqrt[n]{a} = \frac{1 - a}{1 + \sqrt[n]{a} + \sqrt[n]{a^2} + \sqrt[n]{a^3} + ... +\sqrt[n]{a^{n-1}} }$$
Note that terms in denominator,
$$1 + \sqrt[n]{a} + \sqrt[n]{a^2} + \sqrt[n]{a^3} + ... +\sqrt[n]{a^{n-1}}$$
are all $\ge$ $a$: being $ 0<a<1$, indeed, elevating $a$ to an exponent $ \in \left [ 0,1 \right]$ (note that exponents of $ a$ in denominator are $\frac{1}{n}$, $\frac{2}{n}$, ... , $\frac{n-1}{n} \le 1$) will get a value  $\ge a$, i.e. closer to $1$.
There are $n$ terms in denominator (note that all powers of $ a $ from $0$ to $ n-1 $ are present, and $ n-1+1=n$).                                              Then it's
$$ |a_n| = 1 - \sqrt[n]{a} = \frac{1 - a}{1 + \sqrt[n]{a} + \sqrt[n]{a^2} + \sqrt[n]{a^3} + ... +\sqrt[n]{a^{n-1}} } \le \frac{1-a}{na}$$
from which follows: $ |a_n| \le \frac{1}{n} \bigl( \frac{1-a}{a} \bigr)$
$\square$
