The given series is:

$\sum_{n=1}^\infty \left(1 + \frac{1}{n}\right)^{-n^2}.$

We are tasked with determining the convergence of this series.

Step 1: Examine the general term

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

Let us simplify and examine the behavior of $a_n$ as $n \to \infty$. Using the approximation: $\ln(1 + x) \approx x \quad \text{for small } x,$ we have: $\ln\left(1 + \frac{1}{n}\right) \approx \frac{1}{n}.$

Thus: $\ln\left[\left(1 + \frac{1}{n}\right)^{-n^2}\right] = -n^2 \cdot \ln\left(1 + \frac{1}{n}\right) \approx -n^2 \cdot \frac{1}{n} = -n.$

Therefore: $\left(1 + \frac{1}{n}\right)^{-n^2} \approx e^{-n}.$

This implies: $a_n \approx e^{-n}.$

Step 2: Behavior of the series

The general term $a_n \approx e^{-n}$ decreases extremely rapidly as $n \to \infty$. In fact, $e^{-n}$ is so small that the series converges. To confirm this rigorously:

1. Compare $a_n$ with a known convergent series. Since $e^{-n}$ (or equivalently, $\frac{1}{e^n}$) is smaller than $\frac{1}{n^p}$ for any $p > 0$ (which converges for $p > 1$), the comparison test confirms convergence.

Conclusion

The series: $\sum_{n=1}^\infty \left(1 + \frac{1}{n}\right)^{-n^2}$ converges.

Would you like further clarification or detailed steps on specific parts of the solution?

Related Questions:

1. What is the formal definition of a convergent series?
2. How does the comparison test work for proving convergence of series?
3. Can we estimate the sum of this series numerically?
4. What is the relationship between the convergence of exponential terms and p-series?
5. How does the approximation $\ln(1 + x) \approx x$ work for small $x$?

Tip:

Always simplify the general term of a series and analyze its behavior as $n \to \infty$ before applying formal convergence tests!