1. \(a_n = \sum\limits_{i=1}^{n}\frac{1}{i} = 1 + \frac{1}{2} + ... + \frac{1}{n}\)，数列\(\{a_n\}\)发散
\[\begin{align} a_n &= \sum\limits_{i=1}^{n}\frac{1}{i} \\ &= 1 + \frac{1}{2} + ... + \frac{1}{n} \\ &= 1 + \frac{1}{2} + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}) + ... \\ &> 1 + \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}) + ... + \frac{1}{n} \\ &= 1 + \frac{1}{2} + \frac{1}{2} + ... \\ &= 1 + \frac{m}{2} \end{align}\]
其中\(m\)是\(\frac{1}{2}\)的个数，随着n的增大而增大
所以，
\[\lim\limits_{n\rightarrow\infty}a_n = \infty\]
2. \(|a_{n+p} - a_{n}| \le \frac{p}{n}, \forall n, p \in N^{*}\)，数列\(\{a_n\}\)发散
举反例，如数列\(\{1 + \frac{1}{2} + ... + \frac{1}{n}\}\)满足\(|a_{n+p} - a_{n}| \le \frac{p}{n}, \forall n, p \in N^{*}\)，但是其发散
3. \(|a_{n+p} - a_{n}| \le \frac{p}{n^2}, \forall n, p \in N^{*}\)，数列\(\{a_n\}\)收敛
\[\forall n, p \in N^{*}\]
\[\begin{align} |a_{n+p} - a_{n}| &\le |a_{n+p} - a_{n+p+1}| + |a_{n+p-1} - a_{n+p-2}| + ... + |a_{n+1} - a_{n}| \\ &\le \frac{1}{(n+p-1)^2} + ... + \frac{1}{n^2} \\ &\le \frac{1}{(n + p - 1)(n + p - 2)} + ... + \frac{1}{n(n - 1)} \\ &= \frac{1}{n-1} - \frac{1}{n+p-1} \\ &< \frac{1}{n-1} \end{align}\]
于是
\[\forall p, n \in N^{*}: |a_{n+p} - a_{n}| < \frac{1}{n - 1} (n>1)\] 因此 \[\forall \epsilon > 0, \exists N = \lfloor\frac{1}{\epsilon}\rfloor + 2, \forall n > N, \forall p \in N^{*}: |a_{n+p} - a_{n}| < \epsilon\]
所以数列收敛