Question

Let $a_n=16,4,1,\dots .$ be a geometric sequence. Define $P_n$ as the product of the first $n$ terms. Then the value of $\frac{1}{4}\sum _{n=1}^{\infty }\sqrt[n]{P_n}$ is

Answer 1

Answer: 8

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