3. Consider the sequence $\left(a_{n}\right)_{n \geq 1}$ defined by $a_{n}=\lim _{\mathrm{x} \rightarrow 0}\left(\frac{1^{x}+2^{x}+\cdots+n^{x}}{n}\right)^{\frac{1}{x}}$.

Calculate:

$$
\lim _{n \rightarrow \infty} \frac{\ln a_{n}}{\ln n}
$$
3.For $n=1$ we get $a_{1}=1$. For $n \geq 2$ it follows that

$\lim _{x \rightarrow 0}\left(\frac{1^{x}+2^{x}+\cdots+n^{x}}{n}\right)^{\frac{1}{x}}=\lim _{x \rightarrow 0}\left(1+\frac{\left(2^{x}-1\right)+\cdots+\left(n^{x}-1\right)}{n}\right)^{\frac{1}{x}}=e^{\lim _{x \rightarrow 0} \frac{\left(2^{x}-1\right)+\cdots+\left(n^{x}-1\right)}{n x}}=e^{\frac{\ln 2+\cdots+\ln n}{n}}=$ $=e^{\ln \sqrt[n]{n!}}=\sqrt[n]{n!}$, so $a_{n}=\sqrt[n]{n!}$

$.4 p$

$\lim _{n \rightarrow \infty} \frac{\ln a_{n}}{\ln n}=\lim _{n \rightarrow \infty} \frac{\ln \sqrt[n]{n!}}{\ln n}=\lim _{n \rightarrow \infty} \frac{\ln 1+\ln 2+\cdots+\ln n}{n \ln n}$. Applying the Stolz-Cesaro lemma will

result in $\lim _{n \rightarrow \infty} \frac{\ln a_{n}}{\ln n}=\lim _{n \rightarrow \infty} \frac{\ln 1+\ln 2+\cdots+\ln n}{n \ln n}=\lim _{n \rightarrow \infty} \frac{\ln (n+1)}{(n+1) \ln (n+1)-n \ln n}=$

$\lim _{n \rightarrow \infty} \frac{\ln (n+1)}{n[\ln (n+1)-\ln n]+\ln (n+1)}=\lim _{n \rightarrow \infty} \frac{\ln (n+1)}{\ln \left(1+\frac{1}{n}\right)^{n}+\ln (n+1)}=1$