Consider the series
![\[ \sum_{n=1}^{\infty} (-1)^n \int_n^{n+1} \frac{e^{-x}}{x} \, dx. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-91524191994e4c6cfcd4940bb9b587dd_l3.png "Rendered by QuickLaTeX.com")
Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.
The given series converges absolutely.
Proof. The series is absolutely convergent since
![\begin{align*} \sum_{n=1}^{\infty} \left| (-1)^n \int_n^{n+1} \frac{e^{-x}}{x} \, dx \right| \\[9pt] &= \sum_{n=1}^{\infty} \int_n^{n+1} \frac{e^{-x}}{x} \, dx \\[9pt] &\leq \sum_{n=1}^{\infty} \int_n^{n+1} e^{-x} \, dx \\[9pt] &= \sum_{n=1}^{\infty} \left( e^{-n} - e^{-n+1}\right) \\[9pt] &= \frac{1}{e} - \lim_{n \to \infty} e^{-n} &(\text{telescoping series})\\[9pt] &= \frac{1}{e}. \qquad \blacksquare \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f79ed469d57d933910635a1d04669681_l3.png "Rendered by QuickLaTeX.com")
Consider the series
Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.
The given series converges absolutely.
Proof. The series is absolutely convergent since