Determine whether the sequence \( \left\{(-2.1)^{\mathrm{n}}\right\} \) converges or diverges and describe whether it does so monotonically or by oscillation. Give the limit if the sequence converges. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The sequence \( \left\{(-2.1)^{n}\right\} \) converges monotonically to the limit \( L=\square \). B. The sequence \( \left\{(-2.1)^{n}\right\} \) diverges by oscillation. C. The sequence \( \left\{(-2.1)^{\mathrm{m}}\right\} \) converges by oscillation to the limit \( L=\square \). D. The sequence \( \left\{(-2.1)^{n}\right\} \) diverges monotonically.

The sequence \(\{(-2.1)^{n}\}\) diverges by oscillation. This is because the base is negative and greater than 1 in absolute value, meaning that as \(n\) increases, the terms alternate in sign and their absolute values grow larger. Specifically, for even \(n\), the terms are positive and increasing, and for odd \(n\), the terms are negative and increasing in magnitude but decreasing in absolute value. Thus, we can conclude that the sequence \(\{(-2.1)^{n}\}\) does not settle toward a single limit but rather oscillates between increasingly large positive and negative values. Therefore, the correct choice is: B. The sequence \(\{(-2.1)^{n}\}\) diverges by oscillation.