Is the number \[
\left( 1 + \frac12 \right) \left( 1 + \frac14 \right) \left( 1 + \frac16 \right)\cdots\left( 1 + \frac{1}{2018} \right)
\] greater than, less than, or equal to $50$?
1. Consider the product 
   \[
   P = \left( 1 + \frac12 \right) \left( 1 + \frac14 \right) \left( 1 + \frac16 \right) \cdots \left( 1 + \frac{1}{2018} \right).
   \]
2. To simplify the analysis, multiply \(P\) by another product:
   \[
   Q = \left( 1 + \frac{1}{1} \right) \left( 1 + \frac{1}{3} \right) \left( 1 + \frac{1}{5} \right) \cdots \left( 1 + \frac{1}{2017} \right).
   \]
3. The product of \(P\) and \(Q\) is:
   \[
   P \cdot Q = \left( 1 + \frac12 \right) \left( 1 + \frac14 \right) \left( 1 + \frac16 \right) \cdots \left( 1 + \frac{1}{2018} \right) \left( 1 + \frac{1}{1} \right) \left( 1 + \frac{1}{3} \right) \left( 1 + \frac{1}{5} \right) \cdots \left( 1 + \frac{1}{2017} \right).
   \]
4. Notice that each term in \(P\) is of the form \(1 + \frac{1}{2k}\) and each term in \(Q\) is of the form \(1 + \frac{1}{2k-1}\). Therefore, the product \(P \cdot Q\) includes all terms from \(1 + \frac{1}{1}\) to \(1 + \frac{1}{2018}\):
   \[
   P \cdot Q = \prod_{k=1}^{2018} \left( 1 + \frac{1}{k} \right).
   \]
5. The product \(\prod_{k=1}^{2018} \left( 1 + \frac{1}{k} \right)\) can be simplified using the property of telescoping products:
   \[
   \prod_{k=1}^{n} \left( 1 + \frac{1}{k} \right) = \frac{n+1}{1}.
   \]
   For \(n = 2018\), we have:
   \[
   \prod_{k=1}^{2018} \left( 1 + \frac{1}{k} \right) = \frac{2019}{1} = 2019.
   \]
6. Therefore, we have:
   \[
   P \cdot Q = 2019.
   \]
7. To compare \(P\) and \(Q\), note that each term in \(P\) is less than the corresponding term in \(Q\) because:
   \[
   1 + \frac{1}{2k} < 1 + \frac{1}{2k-1}.
   \]
8. Since each term in \(P\) is less than the corresponding term in \(Q\), it follows that:
   \[
   P < Q.
   \]
9. Given that \(P \cdot Q = 2019\) and \(P < Q\), we can conclude that:
   \[
   P^2 < 2019.
   \]
10. Taking the square root of both sides, we get:
    \[
    P < \sqrt{2019}.
    \]
11. Since \(\sqrt{2019} < 50\), we have:
    \[
    P < 50.
    \]

The final answer is \(\boxed{< 50}\).