Limit of sequence $x_n^n$ 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 $\lim\limits_{n\to\infty}x_n^n$.
 A: 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.

A: For $1$, first observe that if $x_n>1$ then $x_{n+1}=\sqrt{x_n+\frac{1}{n}}>\sqrt{1+\frac{1}{n}}>1$, so the sequence $(x_n)$ is bounded below by $1$. Note that
$$\begin{eqnarray}x_n>x_{n+1}&\iff&x_n>\sqrt{x_n+\frac{1}{n}}\\
&\iff&x_n^2>x_n+\frac{1}{n}\\
\end{eqnarray}$$
and that this is clearly true when $n=1$. If it holds for $n$ then
$$\begin{eqnarray}x_{n+1}^2&=&x_n+\frac{1}{n}\\
&=&\sqrt{x_n^2}+\frac{1}{n}\\
&>&\sqrt{x_n+\frac{1}{n}}+\frac{1}{n}\\
&>&x_{n+1}+\frac{1}{n}\\
&>&x_{n+1}+\frac{1}{n+1}\\
\end{eqnarray}$$
so it holds for $n+1$, thus by induction it holds for all $n$. Thus we have a decreasing sequence which is bounded below, so it must come to some limit. Since $x_{n+1}=\sqrt{x_n+\frac{1}{n}}$ is a continuous function of $x_n + \frac{1}{n}$, the limit must satisfy $L=\sqrt{L}$, so it must be either $1$ or $0$, and since the sequence is bounded below by $1$ the limit must be $1$.
For $2$, I suggest using the fact that
$$x_{n+1}^{n+1}=x_{n+1}\sqrt{\left(x_n+\frac{1}{n}\right)^{n}}$$
and applying the binomial theorem and looking at the resulting terms.
A: First question. $0\leq x_{n+1}=\sqrt{x_{n}+\frac{1}{n}}\leq x_{n} $ (decreasing sequence for $n\geq 2$). Then the sequence admits limit, this limit is precisely its infimum. $x_{n}\rightarrow l\in [0,\infty)$. Also: $\sqrt{x_{n}+\frac{1}{n}}\geq 1$, then:
$l=\lim_{n\rightarrow \infty} x_{n+1}=\lim_{n\rightarrow \infty} \sqrt{x_{n}+\frac{1}{n}}=\sqrt l$. It follows that $l^{2}-l=0$ if $l=0$ or $l=1$ and the first solution is not acceptable. 
A: The key thing here (I would say) is to prove that the sequence is bounded below by 1. This you can prove by induction on $n$. 
Clearly $x_1\geq 1$. Assume that $x_k\geq1$ some $k\in N$, then $x_{k+1}=\sqrt{x_k+1/k}>\sqrt{1+1/k}>1$. Thus, by the axiom of induction the above statement is true for the set of natural numbers. 
Knowing this you can prove that the sequence is decreasing. Try, and ask me if it does not work out. 
I think you are able to figure out the rest?    
