To evaluate the series

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

Analysis:

1. Series Expression: Each term of the series is given by $\frac{e^{1/n}}{n}$, where $e^{1/n} = \exp\left(\frac{1}{n}\right)$.

2. Behavior of $e^{1/n}$ as $n \to \infty$:

   • As $n$ increases, $1/n \to 0$, so $e^{1/n} \to e^0 = 1$.
   • For large $n$, the term behaves approximately as $\frac{1}{n}$, which resembles the harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ (divergent).

3. Comparison with a Divergent Series:

   • The harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges.
   • Since $e^{1/n} > 1$ for $n > 1$, it implies that $\frac{e^{1/n}}{n} > \frac{1}{n}$, indicating this series diverges more strongly than the harmonic series.

Conclusion:

The series $\sum_{n=1}^\infty \frac{e^{1/n}}{n}$ diverges because it grows faster than the harmonic series, which is known to diverge.

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Would you like to explore the divergence further or need help with a related topic?

Expanding Questions:

1. What are the criteria for determining divergence of series?
2. How does the comparison test apply to series like this?
3. Can you compute the first few terms of this series numerically?
4. What happens if we modify the series to $\sum \frac{e^{-n}}{n}$?
5. How does the $e^{1/n}$ term affect the divergence compared to the harmonic series?

Tip: When working with series, use convergence tests (e.g., Comparison Test, Ratio Test, Integral Test) to determine whether the series converges or diverges.