(1)
\(\frac{{{a}_{n+1}}}{{{a}_{n}}}=\left( \frac{n+1}{n} \right)a\)
(2)
\(n=1\)時，\(\left| {{a}_{1}} \right|\le \left| a \right|\)成立
設\(n=k\)時，\(\left| {{a}_{k}} \right|\le {{2}^{k-1}}{{\left| a \right|}^{k}}\)成立
\(\begin{align}
& {{a}_{k+1}}=\left( \frac{k+1}{k} \right)a\times {{a}_{k}} \\
& \left| {{a}_{k+1}} \right|=\left| \frac{k+1}{k} \right|\times \left| a \right|\times \left| {{a}_{k}} \right| \\
& \le \left| 1+\frac{1}{k} \right|\times \left| a \right|\times {{2}^{k-1}}{{\left| a \right|}^{k}} \\
& \le 2\times {{2}^{k-1}}{{\left| a \right|}^{k+1}} \\
& ={{2}^{k}}{{\left| a \right|}^{k+1}} \\
\end{align}\)
(3)
\(\begin{align}
& 0<a<\frac{1}{2} \\
& \underset{n\to \infty }{\mathop{\lim }}\,{{a}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,n{{a}^{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{{{\left( \frac{1}{a} \right)}^{n}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{\left( -\ln a \right){{\left( \frac{1}{a} \right)}^{n}}}=0 \\
& -\frac{1}{2}<a<0 \\
\end{align}\)
改寫一下即可
