1. ## Quickie #9

Evaluate the infinite product: . $3^{\frac{1}{3}}\cdot9^{\frac{1}{9}}\cdot27^{\frac{ 1}{27}}\,\cdots\,(3^n)^{\frac{1}{3^n}}\,\cdots$

2. $
\left( {3^n } \right)^{\frac{1}{{3^n }}} = 3^{\frac{n}{{3^n }}}
$

So:

$
\prod\limits_{n = 1}^\infty {\left( {3^n } \right)^{\frac{1}{{3^n }}} } = \prod\limits_{n = 1}^\infty {3^{\frac{n}{{3^n }}} } = 3^{\sum\limits_{n = 1}^\infty {\frac{n}{{3^n }}} }
$

Which gives:

$
\sum\limits_{n = 1}^\infty {\frac{n}{{3^n }}} = \frac{3}{4} \Rightarrow \prod\limits_{n = 1}^\infty {\left( {3^n } \right)^{\frac{1}{{3^n }}} } = 3^{\frac{3}{4}}
$

3. TD! was not clear in explaining how the infinite sum gives the value it gives, perhaps it is well known. Here is my explanation.

We want to find,
$\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...$
This is absolutely convergent we can write,
$\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...$
+
$\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...$
+
$\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+...$
+....

Each is a geometric series hence the sum respectively are,
$\frac{1}{2}$
$\frac{1}{6}$
$\frac{1}{18}$
+....

Thus the answer is, again geometric
$\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+... =\frac{3}{4}$

4. I have worked through this before, Soroban.

Rewrite:

$9^{\frac{1}{9}}=3^{\frac{2}{9}}$

$27^{\frac{1}{27}}=3^{\frac{3}{27}}$

and so forth.

Then we have:

$3^{\frac{1}{3}}\cdot{3^{\frac{2}{9}}}\cdot{3^{\fra c{3}{27}}}...$

$3^{\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+....}$

Now, we have $\sum_{n=1}^{\infty}\frac{n}{3^{n}}$ in the exponent. Which is easily found to be 3/4

So, the answer is $3^{\frac{3}{4}}\approx{2.279507061...}$

I had this problem back in 'Seminars of Mathematics'.

5. Originally Posted by ThePerfectHacker
TD! was not clear in explaining how the infinite sum gives the value it gives, perhaps it is well known. Here is my explanation.
It had to be a quickie, so I omitted the steps of evaluating that sum

Other possibility, from the standard geometric series with |x|<1, we use absolute convergence to differentiate term-wise:

$
\sum\limits_{i = 0}^\infty {x^i } = \frac{1}{{1 - x}} \Rightarrow \left( {\sum\limits_{i = 0}^\infty {x^i } } \right)^\prime = \left( {\frac{1}{{1 - x}}} \right)^\prime \Rightarrow \sum\limits_{i = 0}^\infty {ix^{i - 1} } \frac{1}{{\left( {1 - x} \right)^2 }}
$

So:

$
\boxed{\sum\limits_{i = 0}^\infty {ix^i } = \frac{x}{{\left( {1 - x} \right)^2 }}} \Rightarrow \sum\limits_{i = 0}^\infty {i\left( {\frac{1}{3}} \right)^i } = \frac{{\frac{1}{3}}}{{\left( {1 - \frac{1}{3}} \right)^2 }} = \frac{1}{3}\frac{9}{4} = \frac{3}{4}
$

Where the boxed expression is a nice general result as well

6. Nice work, everyone! .You are all correct.

There is a algebraic approach to that sum of exponents.

. . . . . . . . Let . $S \;=\;\frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + \hdots$

Multiply by $\frac{1}{3}\!:\;\frac{1}{3}S \;=\qquad\frac{1}{3^2} + \frac{2}{3^3} + \frac{3}{3^4} + \hdots$

. . . Subtract: . $\frac{2}{3}S \;= \;\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \hdots$

The right side is a geometric series with first term $a = \frac{1}{3}$ and common ratio $r = \frac{1}{3}$
. . Its sum is: . $\frac{\frac{1}{3}}{1 - \frac{1}{3}} \:=\:\frac{1}{2}$

Therefore: . $\frac{2}{3}S \:=\:\frac{1}{2}\quad\Rightarrow\quad S \:=\:\frac{3}{4}$