TITLE: Limit of sequence $x_n^n$
QUESTION [6 upvotes]: Let $x_1=2$, $x_{n+1}=\sqrt{x_n+\frac{1}{n}}$ for all $n\geq 1$. Prove that $\lim\limits_{n\to\infty}x_n=1$ and evaluate $\lim\limits_{n\to\infty}x_n^n$.

REPLY [2 votes]: For completeness, we will provide full proofs for both parts, different from the other answers and both using the same lemma.
The Lemma we are talking about:

Lemma: Suppose $\displaystyle z_n \gt 0$ is a sequence and $\displaystyle \alpha \gt 1$ is a real number such such that  $$\lim_{n \to \infty} \frac{z_{n+1}^{\alpha}}{z_n} = q$$ then $$\lim_{n \to
 \infty} z_n = q^{1/(\alpha -1)}$$

Proof: 
See the end of this answer.
$\square$

Now we apply this to our problem. Note: we use the case $\displaystyle \alpha = 2$ for our purposes.
Part i) 
Now it is clear that $\displaystyle x_n \gt 1$
We have that
$$\frac{x_{n+1}^2}{x_n} = 1 + \frac{1}{nx_n} \tag{1}$$
For any $\displaystyle a_n$ such that $\displaystyle |a_n| \gt M \gt 0$ we can easily show that $\displaystyle \frac{1}{n a_n} \to 0$ and thus $\displaystyle \frac{x_{n+1}^2}{x_n} \to 1$ and thus by the above lemma, $\displaystyle x_n \to 1$.
For part ii)
Let $\displaystyle y_n = x_n^n$.
and so, raising $(1)$ to the $\displaystyle n^{th}$ powers gives us
$$ \frac{y_{n+1}^2}{y_nx_{n+1}^2} = \left(1 + \frac{1}{nx_n}\right)^n$$
Now if $\displaystyle c_n \to c$, then $\displaystyle \left(1 + \frac{c_n}{n}\right)^n \to e^c$.
Thus 
$$\frac{y_{n+1}^2}{y_n} \to e$$
Thus by the above Lemma, $$x_n^n  = y_n \to e$$

Proof of Lemma
For the moment, assume that $\displaystyle q \gt 0$.
We have that, given an arbitrary $q \gt \varepsilon \gt 0$, there is some $n_0$ such that $\forall n \ge n_0$
$$ q - \varepsilon \lt \frac{z_{n+1}^\alpha}{z_n}  \lt q+\varepsilon$$
$$ \sqrt[\alpha]{q- \varepsilon}\lt \frac{z_{n}}{\sqrt[\alpha]{z_{n-1}}} \lt \sqrt[\alpha]{q+\varepsilon}$$
$$ \dots $$
$$ \left(q - \varepsilon\right)^{1/\alpha^{n-n_0}} \lt \frac{z_{n_0+1}^{1/\alpha^{n-n_0 + 1}}}{z_{n_0}^{1/\alpha^{n-n_0}}}  \lt\left(q + \varepsilon\right)^{1/\alpha^{n-n_0}} $$
Multiplying all and taking $\displaystyle \alpha^{th}$ root once gives us
$$C^{1/\alpha^{n-n_0+1}}\left(q - \varepsilon\right)^{1/(\alpha-1) - 1/\alpha^{n-n_0+1}}\lt z_{n+1} \lt C^{1/\alpha^{n-n_0+1}} \left(q + \varepsilon\right)^{1/(\alpha-1) - 1/\alpha^{n-n_0+1}}$$
Taking limits as $\displaystyle n \to \infty$ gives us
$$(q - \varepsilon)^{1/(\alpha-1)} \le \liminf z_{n} \le \limsup z_{n} \le (q+\varepsilon)^{1/(\alpha-1)}$$
Since $\displaystyle \varepsilon$ was arbitrary, we have that $\displaystyle \lim z_n = q^{1/(\alpha-1)}$.
If $\displaystyle q = 0$, all we need to do is replace the left hand side by $\displaystyle 0$ and the proof carries through.

Remark: The proof is similar to the textbook proof of $\lim
 \frac{a_{n+1}}{a_n} = \lim a_n^{1/n}$ which can be found in my answer
  here: https://math.stackexchange.com/a/116198/1102.