12
$\begingroup$

This question is inspired by What would happen to Bernoulli's inequality if $x<-1$?.

Let $x_n=\min\{x\in{\bf R}:(1+x)^n\geq 1+nx\}$, where $n$ is natural and odd (my mistake in the first version, observed by @mfl). Is it true, that $$ \lim_{n\to\infty}x_n=-2? $$ Numerical computations suggest the positive answer.

$\endgroup$
  • $\begingroup$ reWard (editing seems to be impossible) $\endgroup$ – Przemysław Scherwentke Dec 19 '14 at 3:23
4
+50
$\begingroup$

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).

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.