Let's go through each problem one by one:

Problem 1

The given series is: $\sum_{n=1}^{+\infty} e^{-n}$ This is a geometric series with the general term $a_n = e^{-n} = \left( \frac{1}{e} \right)^n$.

Since $\frac{1}{e} < 1$, this geometric series converges because, for geometric series, if the absolute value of the common ratio $r$ is less than 1, the series converges. The integral test is not necessary here as we can conclude convergence directly from the nature of the geometric series.

Answer: Συγκλίνει.

Problem 2

The given series is: $\sum_{n=1}^{\infty} \frac{\ln(n) \cdot (-1)^{n+1}}{n}$ This is an alternating series due to the $(-1)^{n+1}$ factor. We can consider the Leibniz criterion (Alternating Series Test) to check for convergence. The series converges if:

1. $\frac{\ln(n)}{n}$ is positive, decreasing, and

2. $\lim_{n \to \infty} \frac{\ln(n)}{n} = 0$.

3. For $n \geq 2$, $\frac{\ln(n)}{n}$ is positive and decreasing.

4. Taking the limit: $\lim_{n \to \infty} \frac{\ln(n)}{n} = 0$ by applying L'Hôpital's rule (since the limit yields an indeterminate form $\frac{\infty}{\infty}$).

Both conditions are satisfied, so the series converges by the Leibniz criterion.

Answer: Συγκλίνει σύμφωνα με το κριτήριο του Leibnitz.

Would you like further details on these solutions or explanations of the steps?

Related Questions

1. What are other convergence tests for series, and when are they used?
2. How does the Leibniz criterion apply to series with different types of terms?
3. What is the geometric series formula, and how is it derived?
4. How can we verify if a function is decreasing for convergence tests?
5. What other series would converge or diverge similarly?

Tip

When handling series with alternating signs, the Leibniz criterion is often useful, but always check if other convergence tests might simplify the process.