Question

Let $a_{n}=16,4,1, \ldots .$ be a geometric sequence. Define $P_{n}$ as the product of the first $n$ terms. Then the value of $\frac{1}{4} \displaystyle\sum_{n=1}^{\infty} \sqrt[n]{ P _{ n }}$ is

Answer 1

For the G.P. a, ar, ar $^{2}, \ldots$
$P _{ n }= a (\text{ar})\left(\text{ar}^{2}\right) \ldots\left(\text{ar}^{ n -1}\right)= a ^{ n } \cdot r ^{ n ( n -1) / 2}$
$\therefore S =\displaystyle\sum_{ n =1}^{\infty} \sqrt[n]{ P _{ n }}=\displaystyle\sum_{ n =1}^{\infty} ar ^{( n -1) / 2}$
Now, $\displaystyle\sum_{ n =1}^{\infty} ar ^{( n -1) / 2}= a [1+\sqrt{ r }+ r + r \sqrt{ r }+\ldots .+\infty]$
$=\frac{a}{1-\sqrt{r}}$
Given $a =16$ and $r =1 / 4$
$\therefore S =\frac{16}{1-(1 / 2)}=32$