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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?

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would you mind to write out the product series in a formula? –  al-Hwarizmi Aug 4 '13 at 15:44
    
sorry i couldnt post the question on my account due to the stupid 24 hour limit –  Commander Shepard Aug 4 '13 at 15:46
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The last factor in the product, $3^{3^{-n}}$ doesn't go to 0, it approaches 1. –  Francis Adams Aug 4 '13 at 15:51
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How could it possibly be zero, given that each term is bigger than $1$? –  Thomas Andrews Aug 5 '13 at 0:20
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5 Answers

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

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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 $=\frac13<1$

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

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

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$ 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 } $

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First 3^0 = 1.

Second it will end with somethink like ...*3^(1/"a very large number")*3^(1/infinit)

After "a very large number" comes infinit.

1/infinit = 0

That makes ...*3^(1/"a very large number")*3^0

What makes ...*3^(1/"a very large number")*1

Since you can divide 1 infinitly by 3 the answer will be infinite as well.

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$1/\infty≠0$ Your intuition is wrong. –  Alizter Aug 5 '13 at 0:30
    
Also note that $\infty$ cannot be treated as a number. –  Alizter Aug 5 '13 at 0:31
    
The $\lim_{x\to\infty}1/x=0≠1/\infty$ –  Alizter Aug 5 '13 at 0:33
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