Cross Cross Cross

This infinite product can be written as

$\displaystyle\prod_{k=1}^{\infty} \left(\dfrac{3^{k} + 3}{3^{k} + 1}\right) = \prod_{k=1}^{\infty} \left(\dfrac{3\times(3^{k-1} + 1)}{3^{k} + 1}\right).$

We then have that each denominator cancels the $(3^{k-1} + 1)$ term in the following fraction. For the $N$ th partial product we are then left with the numerator $3 \times 2$ from the first term, $N - 1$ factors of $3$ left over from the following $N - 1$ numerators, and the denominator of the last term, i.e.,

$\dfrac{2\times3\times3^{N-1}}{3^{N} + 1} = 2\times\dfrac{3^{N}}{3^{N} + 1},$ which as $N \rightarrow \infty$ goes to $2*1 = \boxed{2}.$

Let,

$A =\left(1+\frac { 2 }{ 3+1 }\right ) \left(1+\frac { 2 }{ { 3 }^{ 2 }+1 }\right )\left (1+\frac { 2 }{ { 3 }^{ 3 }+1 } \right) \cdots \left (1+\frac { 2 }{ { 3 }^{ x }+1 } \right)$

For $x=1$ we have

$A =\left(1+\frac { 2 }{ 3+1 }\right )$

$A = \frac{3}{2} = 1.5$

For $x=2$ we have

$A =\left(1+\frac { 2 }{ 3+1 }\right ) \left(1+\frac { 2 }{ { 3 }^{ 2 }+1 }\right )$

$A = \frac {3}{2} \cdot \frac{6}{5}$

$A = \frac{9}{5} = 1.8$

For $x=3$ we have

$A =\left(1+\frac { 2 }{ 3+1 }\right ) \left(1+\frac { 2 }{ { 3 }^{ 2 }+1 }\right )\left (1+\frac { 2 }{ { 3 }^{ 3 }+1 } \right)$

$A = \frac {3}{2} \cdot \frac{6}{5} \cdot \frac{15}{14}$

$A = \frac{27}{14} = 1.929$

For $x=4$ we have

$A =\left(1+\frac { 2 }{ 3+1 }\right ) \left(1+\frac { 2 }{ { 3 }^{ 2 }+1 }\right )\left (1+\frac { 2 }{ { 3 }^{ 3 }+1 } \right) \left (1+\frac { 2 }{ { 3 }^{ 4 }+1 } \right)$

$A = \frac {3}{2} \cdot \frac{6}{5} \cdot \frac{15}{14} \cdot \frac{42}{41}$

$A = \frac{81}{41} = 1.976$

For $x=5$ we have

$A =\left(1+\frac { 2 }{ 3+1 }\right ) \left(1+\frac { 2 }{ { 3 }^{ 2 }+1 }\right )\left (1+\frac { 2 }{ { 3 }^{ 3 }+1 } \right) \left (1+\frac { 2 }{ { 3 }^{ 4 }+1 } \right) \left (1+\frac { 2 }{ { 3 }^{ 5 }+1 } \right)$

$A = \frac {3}{2} \cdot \frac{6}{5} \cdot \frac{15}{14} \cdot \frac{42}{41} \cdot \frac{123}{122}$

$A = \frac{243}{122} = 1.9918$

For $x=6$ we have $A = 1.9973$

For $x=7$ we have $A = 1.9991$

For $x=8$ we have $A = 1.9997$

For $x=9$ we have $A = 1.999898$

For $x=10$ we have $A = 1.999966$

So for $x = \infty$ we have $\boxed{A=2}$