Consider a sequence of positive numbers $a _ { 1 } , a _ { 2 } , a _ { 3 } , \cdots$ which satisfies

$$\begin{aligned}
a _ { 1 } & = 1 , \quad a _ { 2 } = 10 , \\
\left( a _ { n } \right) ^ { 2 } a _ { n - 2 } & = \left( a _ { n - 1 } \right) ^ { 3 } \quad ( n = 3,4 , \cdots ) .
\end{aligned}$$

We are to find $\lim _ { n \rightarrow \infty } a _ { n }$.

By finding the common logarithm of both sides of (1), we obtain

$$\mathbf { A } \log _ { 10 } a _ { n } + \log _ { 10 } a _ { n - 2 } = \mathbf { B } \log _ { 10 } a _ { n - 1 } .$$

When we set $b _ { n } = \log _ { 10 } a _ { n }$ $(n = 1,2 , \cdots)$, this equality is expressed as

$$\mathbf { A } b _ { n } + b _ { n - 2 } = \mathbf { B } b _ { n - 1 } .$$

By transforming (2), we have

$$b _ { n } - b _ { n - 1 } = \frac { 1 } { \mathbf { C } } \left( b _ { n - 1 } - b _ { n - 2 } \right) \quad ( n = 3,4 , \cdots ) ,$$

which gives

$$b _ { n } - b _ { n - 1 } = \left( \frac { 1 } { \mathbf { C } } \right) ^ { n - \mathbf { D } } \left( b _ { 2 } - b _ { 1 } \right) \quad ( n = 2,3 , \cdots ) .$$

Since $b _ { 1 } = \mathbf { E }$ and $b _ { 2 } = \mathbf { F }$, from (3) we get

$$b _ { n } = \sum _ { k = 2 } ^ { n } \left( \frac { 1 } { \mathbf { C } } \right) ^ { k - \mathbf { G } }$$

and hence

$$b _ { n } = \mathbf { H } - \left( \frac { 1 } { \mathbf { C } } \right) ^ { n - \mathbf{I} }$$

Finally, we obtain

$$\lim _ { n \rightarrow \infty } a _ { n } = \mathbf { J K L } .$$