$ \{x_n\}_{n\in\mathbb{N}}: x_1 = 2, x_{n+1} =\sqrt{x_n+\frac{1}{n}}$, find $ \lim x_n^n\ $.

Question

Romanian Math Olympiad 2009

$$\text{ The sequence } (x_n) : x_1 = 2, \; x_{n+1} = \sqrt{x_n + \frac{1}{n}} \; \forall n \ge 1. \text{ Find } \lim x_n^n. $$

My Approach

$$ \text{Denote } \lim_{n \to \infty} n(x_n - 1) = a \text{ and } L = \lim_{n \to \infty} x_n^n $$

$$ \text{ } \lim x_n^{2n} = \lim \left(x_{n-1} + \frac{1}{n-1} \right)^n = \lim \left( 1 + x_{n-1} - 1 + \frac{1}{n-1} \right)^n = \lim e^{n\left(x_{n-1} - 1 + \frac{1}{n-1} \right)} = e^{1 + \lim_{n \to \infty} n(x_{n-1} - 1)} = e^{1 + a} $$

$$ \Rightarrow L = e^{\frac{1 + a}{2}} $$

$$ \text{ On the other hand } L = \lim_{n \to \infty} x_n^n = \lim_{n \to \infty} \left(1 + x_n - 1 \right)^n = e^{\lim_{n \to \infty} n(x_n - 1)} = e^a $$

$$ \text{ So we have } e^{\frac{1 + a}{2}} = e^a \Rightarrow a = 1 \text{ or } a = \infty \Rightarrow L = e \text{ or } L = \infty $$

I precisely need help with this-

$1$.How do I prove $a$ exists finitely because my whole solution is based on the assumption that $a$ exists. please help me with that.

thank You!

Answer 1

I added another solution. See Added below.

Let $y_n:=n(x_n-1)$.

In order to prove that $\displaystyle\lim_{n\to\infty}y_n$ exists and is finite, it is sufficient to prove that for $n\ge 3$, $$0\lt y_{n+1}\lt y_n\tag1$$

Let us prove $(1)$ by induction on $n$.

For $n=3$, $(1)$ holds.

Suppose that $(1)$ holds for $n$.

Then, we have

$$\begin{align}y_{n+2}&=(n+2)(x_{n+2}-1) \\\\&=(n+2)\bigg(\sqrt{x_{n+1}+\frac{1}{n+1}}-1\bigg) \\\\&=(n+2)\frac{x_{n+1}-1+\frac{1}{n+1}}{\sqrt{x_{n+1}+\frac{1}{n+1}}+1} \\\\&=(n+2)\frac{\frac{y_{n+1}}{n+1}+\frac{1}{n+1}}{\sqrt{x_{n+1}+\frac{1}{n+1}}+1} \\\\&\gt 0\end{align}$$

and

$$\begin{align}&y_{n+1}-y_{n+2} \\\\&=(n+1)(x_{n+1}-1)-(n+2)(x_{n+2}-1) \\\\&=(n+1)(x_{n+1}-1)-(n+2)\bigg(\sqrt{x_{n+1}+\frac{1}{n+1}}-1\bigg) \\\\&=(n+1)x_{n+1}+1-(n+2)\sqrt{x_{n+1}+\frac{1}{n+1}} \\\\&=\frac{((n+1)x_{n+1}+1)^2-(n+2)^2(x_{n+1}+\frac{1}{n+1})}{(n+1)x_{n+1}+1+(n+2)\sqrt{x_{n+1}+\frac{1}{n+1}}} \\\\&=\frac{(n+1)\bigg((n+1) x_{n+1} + 1\bigg) \bigg(x_{n+1}-\frac{n^2+3n+3}{(n+1)^2}\bigg)}{(n+1)x_{n+1}+1+(n+2)\sqrt{x_{n+1}+\frac{1}{n+1}}}\end{align}$$

This is positive since $$x_{n+1}-\frac{n^2+3n+3}{(n+1)^2}\gt 0\tag2$$

Here, $(2)$ does hold since $$\begin{align}y_{n+1}\lt y_n&\iff (n+1)(x_{n+1}-1)\lt n(x_n-1) \\&\iff (n+1)(x_{n+1}-1)\lt n(x_{n+1}^2-\frac 1n-1) \\&\iff x_{n+1}(nx_{n+1}-n-1)\gt 0 \\&\iff x_{n+1}\gt \frac{n+1}{n}\end{align}$$ and $$\frac{n+1}{n}-\frac{n^2+3n+3}{(n+1)^2}=\frac{1}{n (n + 1)^2}\gt 0$$

---

if possible , for completeness post your full solution

The following proof uses your idea with a few additional explanations.

(We use $\displaystyle\lim_{n\to\infty}x_n=1$ since it seems you already know this.)

Let $a:=\displaystyle\lim_{n\to\infty}y_n$ where we already know that $a$ exists and is finite.

Let $z_n:=x_{n-1}-1+\frac{1}{n-1}$.

We have $$\begin{align}x_n^n&=((x_n)^2)^{n/2} \\\\&=\bigg(x_{n-1}+\frac{1}{n-1}\bigg)^{n/2} \\\\&=(1+z_n)^{n/2} \\\\&=\bigg((1+z_n)^{1/z_n}\bigg)^{nz_n/2} \\\\&=\bigg((1+z_n)^{1/z_n}\bigg)^{y_{n-1}/(2-2/n)+1/(2-2/n)}\end{align}$$

Since $\lim z_n=0$, we have $$\lim_{n\to\infty}x_n^n=e^{\frac a2+\frac 12}\tag3$$

Also, we have $$x_n^n=(1+x_n-1)^n=((1+x_n-1)^{1/(x_n-1)})^{y_n}$$ from which we get $$\lim_{n\to\infty}x_n^n=e^{a}\tag4$$

From $(3)(4)$, we have $\frac a2+\frac 12=a$ which implies $a=1$, and so we finally obtain $$\color{red}{\lim_{n\to\infty}x_n^n=e}$$

---

Added :

I'm going to write another solution.

Let us first prove by induction that, for $n\ge 4$, $$\frac{n}{n-1}\lt x_n\lt \frac{n}{n-1}+\frac{1}{(n-1)^2}$$

For $n=4$, it holds.

Suppose that it holds for $n$.

Then, we have

$$\begin{align}x_{n+1}-\frac{n+1}{n}&=\sqrt{x_n+\frac 1n}-\frac{n+1}{n} \\\\&=\frac{x_n+\frac 1n-(\frac{n+1}{n})^2}{\sqrt{x_n+\frac 1n}+\frac{n+1}{n}} \\\\&=\frac{\overbrace{(x_n-\frac{n}{n-1})}^{\text{positive}}+\overbrace{\frac{1}{(n - 1) n^2}}^{\text{positive}}}{\sqrt{x_n+\frac 1n}+\frac{n+1}{n}} \\\\&\gt 0\end{align}$$

and

$$\begin{align}&\frac{n+1}{n}+\frac{1}{n^2}-x_{n+1} \\\\&=\frac{n+1}{n}+\frac{1}{n^2}-\sqrt{x_n+\frac 1n} \\\\&=\frac{(\frac{n+1}{n}+\frac{1}{n^2})^2-(x_n+\frac 1n)}{\frac{n+1}{n}+\frac{1}{n^2}+\sqrt{x_n+\frac 1n}} \\\\&=\frac{\overbrace{(\frac{n}{n-1}+\frac{1}{(n-1)^2}-x_n)}^{\text{positive}}+\overbrace{\frac{(n - 3) n^3 + 1}{(n - 1)^2 n^4}}^{\text{positive}}}{\frac{n+1}{n}+\frac{1}{n^2}+\sqrt{x_n+\frac 1n}} \\\\&\gt 0.\ \square\end{align}$$

So, we have $$\bigg(\frac{n}{n-1}\bigg)^n\lt x_n^n\lt\bigg(\frac{n}{n-1}+\frac{1}{(n-1)^2}\bigg)^n$$ i.e. $$\bigg(1+\frac{1}{n-1}\bigg)\bigg(1+\frac{1}{n-1}\bigg)^{n-1}\lt x_n^n\lt\bigg(\bigg(1+\frac{1}{\frac{(n-1)^2}{n}}\bigg)^{\frac{(n-1)^2}{n}}\bigg)^{\frac{1}{1-2/n+1/(n^2)}}$$

Therefore, we finally get $\displaystyle\lim_{n\to\infty}x_n^n=e$.