$f_{n}(x)= x^n+n\sqrt x-1 $. Prove that $a_{n+1} < a_{n}$

Question

$f_{n}(x)= x^n+n\sqrt x-1$

We already proved the following:

• $f_{n+1}(a_n)= a_n^{n+1} + \sqrt{a_n} \cdot(1-a_n^{n-1}\sqrt{a_n}) $
• $a_{n}$ is a solution to $f_{n}(x)=0$
• $0 < a_{n} < 1$

1- Prove that $a_{n+1} < a_{n}$

Then, we also prove that $0 < a_{n} < 1/n^2$

2- Prove that: $\frac{1}{n}$ - $\frac{1}{n^2+1}$ < $\sqrt{a_n}$ < $\frac{1}{n}$

here is an image of the exercise to help: image

Thank you so much in advance

Answer 1

We can prove $a_{n+1}<a_n$ by directly substituting $x=a_n$ into $f_n(x)$ and with what was already proven by OP,

\begin{align}(a_n)^{n+1}+(n+1)\sqrt{a_n}-1&=(a_n)^{n+1}+\sqrt{a_n}(1-(a_n)^{n-1}\sqrt{a_n})\\(a_n)^n&=n\sqrt{a_n} -1\\(a_n)^n-n\sqrt{a_n}-1&=0\tag{*}\end{align} To prove $a_{n+1}<a_n$, we can show that for increasing values of $n$, that $a_n$ will need to decrease to satisfy equation $(*)$. Here is a graphical interpretation (as you move the slider for $n$; try it out for yourself!):

Answer 2

As the function $f_{n+1}$ is strictly increasing, to show that $a_{n+1} < a_n$, we only need to show that $f_{n+1}(a_n) >0$. This follows from : $$f_{n+1}(a_n) = a_n^{n+1} + \sqrt{a_n} (1-\sqrt{a_n} a_n^{n-1})$$ as $\sqrt{a_n} a_n^{n-1} <1$.

Likewise, we can show that $a_n < \frac{1}{n^2}$ by remarking that $f_n(\frac{1}{n^2}) = \frac{1}{n^{2n}}>0$.