Consider the series
![\[ \sum_{n=1}^{\infty} (-1)^{\frac{n(n-1)}{2}} \left( \frac{n^{100}}{2^n} \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-cb74650c5fbdf50417582bcc639aa009_l3.png "Rendered by QuickLaTeX.com")
Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.
The series converges absolutely.
Proof. First, we have
![\[ \sum_{n=1}^{\infty} \left| (-1)^{\frac{n(n-1)}{2}} \left( \frac{n^{100}}{2^n} \right) \right| = \sum_{n=1}^{\infty} \left( \frac{n^{100}}{2^n} \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b92d8b16d696f398fe920ac106642c5c_l3.png "Rendered by QuickLaTeX.com")
Then using the ratio test we have
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(n+1)^{100}}{2^{n+1}} \right) \left( \frac{2^n}{n^{100}} \right) \\[9pt] &= \lim_{n \to \infty} \left( \frac{(n+1)^{100}}{2n^{100}} \\[9pt] &= \frac{1}{2} < 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-54f51dc493ad43fd7ace70b478b53f68_l3.png "Rendered by QuickLaTeX.com")
Hence, the series converges absolutely