\begin{enumerate}
\item Assuming \( \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = a\) , proof \( \lim_{n \to \infty} \sqrt[n]{a_n} = a\).
\par \emph{Hint:}
\( \lim_{n \to \infty} a_n = a \Rightarrow \lim_{n \to \infty} (a_1 a_2 \cdots a_n)^{1/n} = a\)
\item Proof \( \lim_{n \to \infty} \frac{1}{n}(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}) = 0\).
\par \emph{Hint:}
\( \lim_{n \to \infty} a_n = a \Rightarrow \lim_{n \to \infty} (a_1 + a_2 +\cdots+ a_n) = a\)
\item Let \(p(n)\) be the number of prime factors of n, proof \( \lim_{n \to \infty} \frac{p(n)}{n} = 0\).
\par
\emph{Proof:} \[n = p_1^{s_1} p_2^{s_2} \cdots p_m^{s_m}, \quad m=p(n)\]
\[m<s_1+s_2+s_3+\cdots+s_m\]
Since 2 is the smallest prime number,
\[2^{s_1+s_2+s_3+\cdots+s_m}<n\]
thus,
\[ s_1+s_2+s_3+\cdots+s_m<\frac{\ln{n}}{\ln{2}} \]
\[ \frac{p(n)}{n} < \frac{\ln{n}}{n \ln{2}} \]
\item
\end{enumerate}
谢谢你解决我下面这个问题!