suppose n is a natural number , prove equation $x^n+nx-1=0$ exist an unique real positive root $x_n$

Question

suppose n is a natural number

prove :

equation $x^n+nx-1=0$ exist an unique real positive root $x_n$ ; and when $a>1$,$\sum_{n=1}^{\inf}x^a_n$ converges.

Answer 1

I think you problem may add this $x$ domain,following I add $x>0$ $$f(x)=x^n+nx-1\Longrightarrow f'(x)=n(x^{n-1}+1)>0$$ and $$f(0)=-1<0,f(\dfrac{1}{n})=\dfrac{1}{n^n}>0$$ so this equation has unique real postive..

and $$f(\dfrac{1}{n+1})=\left(\dfrac{1}{n+1}\right)^n-\dfrac{1}{n+1}<0$$ so $$\dfrac{1}{n+1}<x_{n}<\dfrac{1}{n}$$ so $$\sum_{n=1}^{\infty}x^a_{n},a>1$$converges