The problem asks for the convergence behavior of the series:

$\sum_{n=1}^{\infty} \frac{1}{n} e^{\frac{n}{n+1}}$

To analyze this, let’s examine the behavior of the general term:

$a_n = \frac{1}{n} e^{\frac{n}{n+1}}$

As $n \to \infty$, observe that $\frac{n}{n+1} \approx 1$, meaning $e^{\frac{n}{n+1}} \approx e$, a constant. So, the term $a_n \approx \frac{e}{n}$, resembling a harmonic series, which diverges.

However, because the terms do not alternate, the series diverges due to the behavior of $\frac{1}{n}$. Thus, the answer is D: Divergent.

Would you like a detailed solution on divergence tests?

Related Questions:

1. What is the difference between conditional and absolute convergence?
2. How does the Ratio Test determine convergence?
3. Can divergence be determined if terms approach zero?
4. What other series are similar to the harmonic series?
5. How is the Comparison Test applied to series?

Tip:

For large $n$, approximating series terms can reveal convergence behavior efficiently.