General term of a sequence and its limit. Having a sequence $\;\sqrt[3]5\;,\;\sqrt[3]{5\sqrt[3]5}\;,\;\sqrt[3]{5\sqrt[3]{5\sqrt[3]5}}\;,\;\ldots\;,\;$ what could be the general term and its limit ?
Note that the second term is cube root of $5$ and inside that there is cube root of $5$ again and the third term is cube root of $5$ and two cube root of $5$ inside the outer cube root of $5$.
I tried to define it recursively as
$c_{n+1}=\sqrt[3]{5c_n}\;$ where $\;c_1=\sqrt[3]5\;.$
 A: Observe the pattern that, if the $n$th term of the sequence is denoted $c_n$, then
$$c_n = 5^{1/3 + (1/3)^2 + (1/3)^3 + \cdots + (1/3)^n}$$
To see this, notice that, for instance:
\begin{align*}
c_2 &= \sqrt[3]{ 5 \sqrt[3] 5} = \left( 5 \cdot 5^{1/3} \right)^{1/3} = 5^{1/3} \cdot 5^{1/3^2} = 5^{1/3 + 1/3^2} \\
c_3 &= \sqrt[3]{ 5 \sqrt[3]{5\sqrt[3]{5}}} = \left( 5 \cdot \left(5 \cdot 5^{1/3}\right)^{1/3} \right)^{1/3} = 5^{1/3} \cdot 5^{1/3^2} \cdot 5^{1/3^3} = 5^{1/3 + 1/3^2 + 1/3^3} \\
\end{align*}
and so on and so forth. The exponent of $c_n$ is a geometric series with ratio $1/3$ (just starting at $1/3$ and not $1$), and thus
\begin{align*}
\sum_{k=1}^n \left( \frac 1 3 \right)^k 
&= \frac{(1/3)^{n+1} - 1}{1/3 - 1} - 1\\
&= \left( -\frac 3 2 \right)\cdot \frac{1}{3^{n+1}} - \left( -\frac 3 2 \right)\cdot1 - 1\\
&= \frac 1 2 \left( 1 - \frac{1}{3^n} \right)
\end{align*}
and so
$$c_n = 5^{(1 - 3^{-n})/2}$$
Taking the limit $n \to \infty$ is trivial enough, and gives the limit $c_n \to 5^{1/2} = \sqrt 5$.
