The given problem presents the following series:

$\sum \left( 1 + \frac{1}{n} \right)^{-n^2}$

We are tasked with determining the nature of this series: whether it is convergent, divergent, oscillatory, or none of the above.

Approach:

1. Examine the general term of the series: The general term of the series is $\left( 1 + \frac{1}{n} \right)^{-n^2}$.

2. Evaluate the behavior as $n$ approaches infinity: We know that for large $n$, the expression $\left( 1 + \frac{1}{n} \right)$ approaches 1. Specifically: $\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^{-n^2} = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^{-n^2} = 0$ because the term inside the parentheses approaches 1, but the exponent $-n^2$ grows rapidly, causing the overall term to shrink to 0.

3. Convergence Check: The fact that each term approaches 0 quickly as $n \to \infty$ suggests that the series converges.

Thus, the correct answer is likely Convergent (option a).

Would you like a more detailed explanation on this? Or if you have any further questions or other problems, feel free to ask!

Related Questions:

1. How do we apply the limit comparison test to determine convergence?
2. What is the role of the exponent in the convergence of a series?
3. Can the behavior of a series change if we modify the exponent in the general term?
4. What is the difference between a convergent series and a divergent series?
5. How do you check the convergence of a series involving terms like $\left( 1 + \frac{1}{n} \right)^{-n^2}$?

Tip: When analyzing series, always consider the limit of the general term as $n \to \infty$. This can often give you an immediate sense of whether the series converges or diverges.