Find the value of $3^9\cdot 3^3\cdot 3\cdot 3^{1/3}\cdot\cdots$

Find the value of $3^9\cdot 3^3\cdot 3\cdot 3^{1/3}\cdot\cdots$

Doesn't this thing approaches 0 at the end? why does it approaches 1?

Solution 1:

Hint: $3^9\cdot3^3\cdot3^1\cdot\dots=3^{9+3+1+\cdots}$

Solution 2:

HINT:

Using Exponent Combination Laws, $$a^m\cdot a^n\cdot a^p\cdots=a^{m+n+p+\cdot},$$

$$\displaystyle 3^9\cdot 3^3\cdot3\cdot 3^\frac13\cdots=3^{\left(3^2+3+1+\frac13+\cdots\right)}$$

Observe that the power of $3$ is an infinite Geometric Series with the first Term $=9$ and common ratio $=\frac13<1$

Solution 3:

$$3^9 * 3^3 * 3 * 3^{\frac{1}{3}} * ...=$$

$$3^{9\sum_{n=0}^{\infty}3^{-n}}=$$ $$3^{9*1.5}=$$ $$3^{13.5}$$

Solution 4:

$ 3^{12} \times 3^{sum\ of\ geometric\ series }$

geometric series is

$ 1 + 1/3 + 1/9 + .... $

$ = 1 / (1-1/3) $

$ = 3/2 $

so

$= 3^ {12 + \frac{3}{2} } $

$= 3 ^{ 27/2 } $