Bernoulli's inequality and an unexpected limit

Solution 1:

Major Edit: The last attempt had a fatal flaw.

Let $J$ be the set of odd integers greater than $1$, and let $p_n(x)=x^n-nx+(n-1)$ for $n\in J$. Calculating derivatives $$p_n^{\prime}(x)=n(x^{n-1}-1)\text{ and }p^{\prime\prime}_n(x)=n(n-1)x^{n-2}$$ tells us that $p_n(-1)=2(n-1)$ is a local max, $p_n(1)=0$ is a local min, there are no other extrema, $p_n$ is strictly increasing on $(-\infty,-1]$, and $p_n(x)\geq0$ for all $x\geq a_n$, where $a_n$ is the least real root of $p_n$.

We will need this: $$p_n\left(-\frac{n+1}{n}\right)=\left(-\frac{n+1}n\right)^n-n\left(-\frac{n+1}n\right)+(n-1)\\=-\left(1+\frac1n\right)^n+2n>2n-e>0.$$

Thus $a_n<-\left(1+\frac1n\right)$. (This is the problem with my first attempt. The roots are bounded away from $-1$ by a sequence that doesn't converge particularly fast.)

We factor, using "$x^a-1=(x^{a-1}+x^{a-2}+\ldots+x+1)(x-1)$" once between the first and second lines and $n-1$ times between the third and fourth lines. $$p_n(x)=x^n-1-n(x-1)\\=\left(\left(\sum_{k=1}^nx^{n-k}\right)-n\right)(x-1)\\=\left(\sum_{k=1}^{n-1}(x^{n-k}-1)\right)(x-1)\\=\left(\sum_{k=0}^{n-2}(n-1-k)x^k\right)(x-1)^2.$$

Now we evaluate $$p_{n+2}(a_n)=\left(\sum_{k=0}^{n}(n+1-k)a_n^k\right)(a_n-1)^2\\=\left(a_n^2\left(\sum_{k=2}^{n}(n+1-k)a_n^{k-2}\right)+na_n+(n+1)\right)(a_n-1)^2\\=\left(a_n^2\left(\sum_{k=0}^{n-2}(n-1-k)a_n^{k}\right)+n a_n+(n+1)\right)(a_n-1)^2\\=(a_n^2\cdot0+n a_n+(n+1))(a_n-1)^2\\<\left(n\left(-\frac{n+1}{n}\right)+(n+1)\right)(a_n-1)^2=0.$$

Second to third line was a reindex. The term the cancelation from the third to fourth line is a recognition that the term is $p_n(a_n)(a_n-1)^{-2}$. Fourth line to fifth is using $a_n<-\frac{n+1}n$, which was established above.

Since $p_{n+2}$ is increasing, $a_{n+2}>a_n$. Thus, the sequence $a_n$ is strictly increasing. Let $a=\lim_{n\in J}a_n$. Since $a_n<-1$, $a\leq-1$. Because $a_n<a$, we have $p_n(a)>0$ for all $n\in J$. Thus $$\sqrt[n]{n(a-1)+1}<a\leq-1$$ for all $n\in J$. Letting $n\to\infty$ (keeping $n\in J$) gives $-1\leq a\leq-1$.

Define $q_n(x):=p_n(x+1)$ for each $n\in J$, and $x_n$ be the least real root of $q_n$. Note that $$q_n(x)=(1+x)^n-(1+nx)=p(x+1,n)\geq0$$ iff $x\geq x_n$. Hence, these $x_n$ are the same numbers as in your question. Also, $x_n=a_n-1$. Since $a_n\to-1$, $x_n\to-2$ as $n\to\infty$ (keeping $n$ odd).

Solution 2:

Edit: This is an answer for arbitrary $n.$ The OP has edited the question to ask for $n$ odd.

Note that for $n\ge 2$ even it is

$$(1+(-100))^n=(-99)^n\ge 1+(-99)\cdot n.$$

So, $x_n\le -100$ for all even $n\in\mathbb{N}.$ Indeed, $x_n=-\infty,$ at least, for $n$ even.

This contradicts the conclusion.