<s>[INST] 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 $\lim\limits_{n\to\infty}x_n^n$.
 [/INST] 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.
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