Infinite Product - Seems to telescope

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Evaluate

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

It looks like this product telescopes: the denominators cancel out (except the last one) and the numerators all become 3.

What would my answer be?

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we have the following identity (which affirms that the product telescopes): $$\left (1+\frac{2}{3^n+1}\right)=3\cdot\frac{3^{n-1}+1}{3^n+1}=\frac{1+3^{-(n-1)}}{1+3^{-n}}$$

(as denoted in the comment by Thomas Andrews)and as a result: $$\prod_{k=1}^n \left (1+\frac{2}{3^k+1}\right)=\frac{2\cdot 3^n}{3^n+1}=\frac{2}{1+3^{-n}}$$

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Hint: The general form of a factor is $$ 1+\frac{2}{3^n+1}=\frac{3^n+3}{3^n+1}=\frac{3\cdot(3^{n-1}+1)}{3^n+1} $$ Now, can you find the general form of a partial product?