A sequence $c_{1}, c_{2}, \ldots, c_{n}, \ldots$ is given, where $c_{1}=1$ and

$$
c_{n+1}=\sqrt{2 \sqrt[3]{3 \sqrt[5]{5 c_{n}}}} \quad(n=1,2, \ldots)
$$

Is this sequence convergent?
The formation law of the sequence can also be written as:

$$
c_{n+1}=a \cdot c_{n}^{1 / 30}, \text { where } a=\sqrt[30]{2^{15} \cdot 3^{5} \cdot 5}(>1)
$$

According to this, the initial terms of the sequence are:

$$
c_{1}=1=a^{0}, \quad c_{2}=a, \quad c_{3}=a^{1+1 / 30}, \quad c_{4}=a^{1+1 / 30+1 / 900}
$$

These are all of the form $a^{\lambda}$, so it seems reasonable to reformulate the formation rule in terms of the $\lambda$ exponent. If $c_{n}$ is equal to the power of the base $a$ with exponent $\lambda_{n}$, then according to the formation rule, $c_{n+1}$ is also a power of the base $a$, specifically with the exponent

$$
\lambda_{n+1}=1+\frac{1}{30} \lambda_{n}
$$

Thus (as can be easily proven by induction), the exponent $\lambda_{n+1}$ is the sum of an $n$-term geometric series:

$$
\lambda_{n+1}=1+\frac{1}{30}+\ldots+\frac{1}{30^{n-1}}
$$

This can also be stated as follows: $\lambda_{n+1}$ is the sum of the first $n$ terms of an infinite geometric series with the first term 1 and common ratio $q=\frac{1}{30}$, from which it follows that the sequence $\lambda_{n}$ is convergent, and its limit is equal to the sum of this geometric series:

$$
\lim _{n \rightarrow \infty} \lambda_{n}=\frac{1}{1-q}=\frac{30}{29}
$$

We also know that the exponential function is continuous. If $\lambda_{n}$ approaches $\frac{30}{29}$, then the power of the number $a$ with exponent $\lambda_{n}$ approaches the number

$$
A=a^{30 / 29}=\sqrt[29]{2^{15} \cdot 3^{5} \cdot 5}=1.831
$$

Thus, the sequence $c_{n}$ is convergent, and this number $A$ is its limit. [^0]

[^0]:    ${ }^{1}$ See Czapáry E.-Horvay K.-Pálmay L.: Mathematics for secondary and specialized secondary schools, 1st edition, 1968, p. 239.