<s>[INST] Find the value of the infinite product $(3)^{\frac{1}{3}} (9)^{\frac{1}{9}} (27)^{\frac{1}{27}}$.... 
I'm not sure if this is meant to be a contradiction but if a product is an infinite product it does not mean that the value if infinity? Or is the word infinite product just misleading.
I let:
$ (3)^{\frac{1}{3}}(9)^{\frac{1}{9}}(27)^{\frac{1}{27}}....=x$
If i cube each side: I get:
$(3)(9)^{\frac{1}{3}}(27)^{\frac{1}{9}}...=x^3$
If you notice you can rewrite the inside as:
$(3)(3^2)^{\frac{1}{3}}(3^3)^{\frac{1}{9}}...=x^3$
and simplifying the last part, we get:
$(3)(3^2)^{\frac{1}{3}}(3)^{\frac{1}{3}}...=x^3$
However, by substitution we have:
$(3)(3^2)^{\frac{1}{3}}x...=x^3$
$(3)(3^2)^{\frac{1}{3}}...=x^2$
$(3)^{\frac{3}{3}} (3)^{\frac{2}{3}}=x^2$
$(3)^{\frac{5}{3}}=x^2$
I got up to here but for some reason it doesnt look right.
 [/INST] Do all your work in the exponent:
\begin{align}
\prod_{k=1}^\infty (3^k)^\frac{1}{3^k}
    & = \prod_{k=1}^\infty 3^\frac{k}{3^k} \\
    & = 3^{\sum_{k=1}^\infty \frac{k}{3^k}} \\
    & = 3^{\frac{3}{4}} \\
    & = \sqrt[4]{27}
\end{align}
The sum can be obtained as follows:
\begin{align}
\sum_{k=1}^\infty k \sigma^k
    & = \sigma \sum_{k=1}^\infty k \sigma^{k-1} \\
    & = \sigma \sum_{k=1}^\infty \frac{d}{d\sigma} \sigma^k \\
    & = \sigma \frac{d}{d\sigma} \sum_{k=1}^\infty \sigma^k \\
    & = \sigma \frac{d}{d\sigma} \frac{\sigma}{1-\sigma} \\
    & = \frac{\sigma}{(1-\sigma)^2}
\end{align}
When $\sigma = 1/3$, the sum is $3/4$.
ETA: As Arthur points out in the comments, this requires logarithms to hold in the infinite case.  In this instance, this should be fine because the logarithms in this case are all positive (so the series is absolutely convergent).
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