Leaps and bounds

If we write $u_n = n(x_n - 1)$ , then $\big(1 + \tfrac{1}{n}u_n\big)^n \; = \; \tfrac{1}{n^2}u_n^2 + \tfrac{3}{n}u_n + 3 \hspace{2cm} n \ge 1$ If $n \ge 4$ then $\begin{aligned} \tfrac{1}{n^2}u_n^2 + \tfrac{3}{n}u_n + 3 & = \; \big(1 + \tfrac{1}{n}u_n\big)^n \; > \; 1 + n \times \tfrac{1}{n}u_n + \tfrac{n(n-1)}{2} \times \tfrac{1}{n^2}u_n^2 \\ 2 & \ge \; \tfrac{n-3}{n}u_n + \tfrac{n(n-1) - 2}{2n^2}u_n^2 \; \ge \; \tfrac14u_n \end{aligned}$ and hence $0 < u_n \le 8$ for all $n \ge 4$ . Thus it is now clear that $\lim_{n\to\infty}\big(3 + \tfrac{3}{n}u_n + \tfrac{1}{n^2}u_n^2\big) \, = \, 3$ , and hence $\lim_{n \to \infty} \big(1 + \tfrac{1}{n}u_n\big)^n \; = \; 3$ so that $\lim_{n \to \infty} n\ln\big(1 + \tfrac{1}{n}u_n\big) \; = \; \ln3$ Note that $x - \tfrac12x^2 < \ln(1+x) < x$ for $x > 0$ , and hence $\begin{aligned} u_n - \tfrac{1}{2n}u_n^2 & < \; n\ln\big(1 + \tfrac{1}{n}u_n\big) \; < \; u_n \\ \left|u_n - n\ln\big(1 + \tfrac{1}{n}u_n\big)\right| & \; < \; \tfrac{1}{2n}u_n^2 \; \le \; \tfrac{32}{n} \end{aligned}$ for $n \ge 4$ . Thus we deduce that $\begin{aligned} \lim_{n \to \infty} \left(u_n - n\ln\big(1 + \tfrac{1}{n}u_n\big)\right) & = \; 0 \\ \lim_{n \to \infty} u_n \; = \; \lim_{n \to \infty}n\ln\big(1 + \tfrac{1}{n}u_n\big) & = \; \ln3 \end{aligned}$ making the answer $\boxed{3}$ .

Proposition: Let $x_n\in\mathbb{R}$ be a positive root of $x^n=x^2+x+1$ . Then, $e^{\lim_{n\rightarrow\infty}n(x_n-1)}=3$ .

Proof:

First, note that for $x=1$ ,

$1^n<1^2+1+1, \forall n\in\mathbb{R}$

Now let $\varepsilon>0$ . Because $(1+\varepsilon)^n$ is an increasing exponential, this means that given an $\varepsilon>0$ , $\exists N$ such that for all $n\geq N$

$(1+\varepsilon) ^n> (1+\varepsilon) ^2+ (1+\varepsilon) +1$

Moreover, since the left hand side changes from less than to greater than as we go from $1$ to $1+\varepsilon$ , by the continuity of polynomials we know that there must be a positive real root $x_n\in(1, 1+\varepsilon)$ .

Thus, we can see that for $n\geq N$ ,

$1^2+1+1<x_n^n< (1+\varepsilon)^2+ (1+\varepsilon)+1\\ \Rightarrow 3^{\frac{1}{n}}<x_n<(3+3\varepsilon +\varepsilon^2)^{\frac{1}{n}}\\ \Rightarrow n(3^{\frac{1}{n}}-1)< n(x_n-1)< n((3+3\varepsilon +\varepsilon^2)^{\frac{1}{n}}-1)$

Therefore, $\text{}^\alpha$

$\lim_{n\rightarrow\infty}n(3^{\frac{1}{n}}-1)\leq \liminf_{n\rightarrow\infty} n(x_n-1)\leq\limsup_{n\rightarrow\infty} n(x_n-1)\leq \lim_{n\rightarrow\infty}n((3+3\varepsilon +\varepsilon^2)^{\frac{1}{n}}-1)$

Now, by using L'Hospital's rule, we can see that

$\lim_{n\rightarrow\infty}n(3^{\frac{1}{n}}-1)=\lim_{n\rightarrow\infty}\dfrac{(3^{\frac{1}{n}}-1)}{\frac{1}{n}}\\ = \lim_{n\rightarrow\infty}3^{\frac{1}{n}}\ln{3}\\ =\ln{3}$

Similarly,

$\lim_{n\rightarrow\infty}n((3+3\varepsilon +\varepsilon^2)^{\frac{1}{n}}-1)=\ln{(3+3\varepsilon +\varepsilon^2)}$

This means that

$\ln{3}\leq \liminf_{n\rightarrow\infty} n(x_n-1)\leq\limsup_{n\rightarrow\infty} n(x_n-1)\leq \ln{(3+3\varepsilon +\varepsilon^2)}$

If we now take the limit as $\varepsilon\rightarrow 0$ , we get

$\ln{3}\leq \liminf_{n\rightarrow\infty} n(x_n-1)\leq\limsup_{n\rightarrow\infty} n(x_n-1)\leq \ln{3}$

Thus, by the Squeeze Theorem,

$\liminf_{n\rightarrow\infty}n(x_n-1)=\limsup_{n\rightarrow\infty}n(x_n-1)=\ln{3}\\ \Rightarrow \lim_{n\rightarrow\infty}n(x_n-1)=\ln{3}\\ \Rightarrow e^{\lim_{n\rightarrow\infty}n(x_n-1)}=3$

$Q.E.D.$

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$\alpha$ Contribution by Mark Hennings

lets define the series $u_n=n(x_n-1)\to x_n=1+\dfrac{u_n}{n}$ when we plug it in the equation we have $\left(1+\dfrac{u_n}{n}\right)^n=\dfrac{u_n^2}{n^2}+3\dfrac{u_n}{n}+3$ we can see that if $u_n$ converges then using the defination of $e$ $e^u=\boxed{3}$ . now all we need to do is show $u_n$ converges. notice that if the series $x_n$ is constant, the equation cannot be satisfied. same goes for polynomial series for $x_n$ as the LHS diverges too fast.(exponential vs polynomial). so ofcourse that means the laurent series for $x_n$ must not have any positive powers.and the independent term must be 1 otherwise we would have exponential vs constants or zero vs non zero constants. $x_n=1+\sum_{k=1}^\infty a_k n^{-k}\to u_n=a_1+\sum_{k=2}^\infty a_k n^{1-k}$ ofcourse at infinity $u_n=a_1$ showing a convergence.

Numerical solution: (matlab code) for i = 3:1000 a = zeros(1,i-3); p = [1, a, -1, -1, -1]; exp(i * (max(roots(p)) - 1)) end