问题：已知\(\displaystyle \left\{ a_n \right\}\)为递增数列，且\(\displaystyle a_1=1\)，\(\displaystyle a_2=2\)，\(\displaystyle a_3=5\)，且\(\displaystyle a_{n+1}=3a_n-a_{n-1}\left( n=2,3,\cdots \right)\)，判别级数\(\displaystyle \sum_{n=1}^{\infty}{\frac{1}{a_n}}\)的敛散性。

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过程如下：\(\displaystyle \left\{ a_n \right\}\)为递增数列，所以有\(\displaystyle a_{n+1}\geqslant a_n\)

所以有

\[a_{n+1}=3a_n-a_{n-1}\geqslant 3a_{n-1}-a_{n-1}=2a_{n-1}\left( n\geqslant 2 \right) \]

一方面

\[\frac{1}{a_{2n+1}}\leqslant \frac{1}{2a_{2n-1}}\leqslant \frac{1}{2^2a_{2n-3}}\leqslant \cdots \leqslant \frac{1}{2^na_{2n-\left( 2n-1 \right)}}=\frac{1}{2^n} \]

另一方面

\[\frac{1}{a_{2n}}\leqslant \frac{1}{2a_{2n-2}}\leqslant \frac{1}{2^2a_{2n-4}}\leqslant \cdots \leqslant \frac{1}{2^{n-1}a_{2n-2\left( n-1 \right)}}=\frac{1}{2^n} \]

综合可得

\[\frac{1}{a_n}\leqslant \frac{1}{\left( \sqrt{2} \right) ^{n-1}} \]

利用比较判别法易知级数\(\displaystyle \sum_{n=1}^{\infty}{\frac{1}{a_n}}\)收敛。