Hi, My attempt at the right part of the inequality is as follows : $1 + x + x^2 + ... + x^{n-1} = x^n$ $\Rightarrow$ $\frac{x^n-1}{x-1} = x^n$ $\Rightarrow$ $x^{n+1} + 1 = 2 x^n$ $\Rightarrow$ $x + \frac{1}{x^n} = 2$ $\Rightarrow$ $x < 2$ $\Rightarrow$ $\frac{2}{x} > 1$. Now, $x + \frac{1}{x^n} = 2$ $\Rightarrow$ $2^n x + (\frac{2}{x})^n = 2^{n+1}$ $\Rightarrow$ $2^n x = 2^{n+1} - (\frac{2}{x})^n$ $\Rightarrow$ $2^n x < 2^{n+1} - 1$ (since, $\frac{2}{x} > 1$) $\Rightarrow$ $x < 2 - \frac{1}{2^n}$ I am trying to figure out the left part of the inequality. ************************************************************************************ Sambit

Let n be a natural number $ \geq 2$, and let u be an nonnegative real number satisfying $ 1 + u + u^2 + ... + u^{n - 1} = u^n$. Then prove that $ 2 - \frac {1}{2^{n - 1}} < u < 2 - \frac {1}{2^n}$. I will use calculus in my solution (this is rather justified, since the number u generally can't be written as a sum of radicals). Multiplying the equation $ 1 + u + u^2 + ... + u^{n - 1} = u^n$ with 1 - u, we get $ (1 + u + u^2 + ... + u^{n - 1})(1 - u) = u^n(1 - u)$, or, in other words, $ 1 - u^n = u^n - u^{n + 1}$, or, equivalently, $ u^{n + 1} - 2u^n + 1 = 0$. Hence, the number u is a zero of the function $ f(x) = x^{n + 1} - 2x^n + 1$. Now I will prove that this function $ f(x) = x^{n + 1} - 2x^n + 1$ has at most two nonnegative zeros. In fact, one zero is clearly x = 1, but we are not interested in this zero, since u cannot be 1 (because the equation $ 1 + 1 + 1^2 + ... + 1^{n - 1} = 1^n$ is not possible with $ n\geq 2$; we have actually let in this zero x = 1 when we multiplied our equation with 1 - u). This will yield that there is only one nonnegative number u satisfying the equation $ 1 + u + u^2 + ... + u^{n - 1} = u^n$. Actually, since the derivative of the function f(x) is $ f^{\prime}(x) = (n + 1)x^n - 2nx^{n - 1}$, it is easy to see that f(x) is monotonically decreasing for x lying in the interval $ [0;\;\frac {2n}{n + 1}]$ and monotonically increasing for x lying in the interval $ [\frac {2n}{n + 1};\;\infty[$. In each of these two intervals, the function f(x) can have only one zero; hence, it can have at most two nonnegative zeros. As we have said above, one zero is x = 1. We are interested in the other zero. Now, our problem requires to show that this second zero lies in the interval $ ]2 - \frac {1}{2^{n - 1}};\;2 - \frac {1}{2^n}[$. We will prove that $ f(2 - \frac {1}{2^{n - 1}}) < 0$ and that $ f(2 - \frac {1}{2^n}) > 0$; since the function f(x) is continuous, this will yield that there is a zero of the function f(x) in the interval $ ]2 - \frac {1}{2^{n - 1}};\;2 - \frac {1}{2^n}[$, and since 1 doesn't belong to this interval, it follows that the zero we are seeking lies in this interval, so the problem is solved. Hence, what remains is to prove $ f(2 - \frac {1}{2^{n - 1}}) < 0$ and that $ f(2 - \frac {1}{2^n}) > 0$. At first, we will establish $ f(2 - \frac {1}{2^{n - 1}}) < 0$. In fact, $ f(2 - \frac {1}{2^{n - 1}}) = (2 - \frac {1}{2^{n - 1}})^{n + 1} - 2(2 - \frac {1}{2^{n - 1}})^n + 1 $ $ = 1 - (2 - \frac {1}{2^{n - 1}})^n \frac {1}{2^{n - 1}}$, so proving $ f(2 - \frac {1}{2^{n - 1}}) < 0$ is equivalent to proving $ 1 - (2 - \frac {1}{2^{n - 1}})^n \frac {1}{2^{n - 1}} < 0$, i. e. $ 1 < (2 - \frac {1}{2^{n - 1}})^n \frac {1}{2^{n - 1}}$. This rewrites as $ 2^{n - 1} < (2 - \frac {1}{2^{n - 1}})^n$. Upon division by $ 2^n$, this becomes $ \frac12 < (1 - \frac {1}{2^n})^n$. But this is easily seen, since Bernoulli's inequality (with $ - \frac {1}{2^n} > - 1$) yields $ (1 - \frac {1}{2^n})^n = (1 + ( - \frac {1}{2^n}))^n > 1 + n( - \frac {1}{2^n}) = 1 - \frac {n}{2^n} = 1 - \frac12 \cdot \frac {n}{2^{n - 1}}$, while $ n\geq 2$ yields $ n\leq 2^{n - 1}$, so that $ \frac {n}{2^{n - 1}} \leq 1$ and $ (1 - \frac {1}{2^n})^n > 1 - \frac12 \cdot \underbrace{\frac {n}{2^{n - 1}}}_{\leq 1} \geq 1 - \frac12 = \frac12$, qed.. Thanks to Peter Scholze for this nice argument. Remains to prove $ f(2 - \frac {1}{2^n}) > 0$; but this is yet easier: Noting that ${{{{ f(2 - \frac {1}{2^n}}) = (2 - \frac {1}{2^n}})^{n + 1} - 2(2 - \frac {1}{2^n}})^n + 1 = 1 - (2 - \frac {1}{2^n}})^n \frac {1}{2^n}$, we see that all we have to show is ${ 1 - (2 - \frac {1}{2^n}})^n \frac {1}{2^n} > 0$, i. e. ${ 1 > (2 - \frac {1}{2^n}})^n \frac {1}{2^n}$, or, equivalently, ${ 2^n > (2 - \frac {1}{2^n}})^n$, what is absolutely trival. $ \blacksquare$ Darij

For $n=1$, the unique root is $x=1$, which satisfies $1\le x\le \frac{3}{2}$. Assume $n>1$. By Descartes' rule of signs, the polynomial $x^n-x^{n-1}-\cdots -x-1=0$ has one positive root. Hence we have \[\left(1-\frac{1}{2^n}\right)^n\ge\left(1-\frac{1}{2^n}\right)^{2n-2}=\left(1-\frac{1}{2^{n-1}}+\frac{1}{2^{2n}}\right)^{n-1}>\left(1-\frac{1}{2^{n-1}}\right)^{n-1}\]. So iterating we have $\left(1-\frac{1}{2^n}\right)^n>\frac{1}{2} (*)$. Now, let $f(x)=x^n-x^{n-1}-\cdots -x-1$, and let $g(x)=(x-1)f(x)=x^{n+1}-2x^n+1$. We have \[g(2-1/2^n) = -(1/2^n) (2 - 1/2^n)^n + 1 = - (1 - 1/2^{n+1})^n + 1 > 0\] since $0<1-1/2^{n+1}<1$. Hence $f(2-1/2^n)>0$. Similarly, \[g(2 - 1/2^{n-1}) = -(1/2^{n-1}) (2 - 1/2^{n-1})^n + 1 = -2(1 - 1/2^n)^n + 1 < 0\] by $(*)$. Hence $f(2-1/2^{n-1})<0$. $\Box$