Determine if the following series converges or diverges and justify your decision.
![\[ \sum_{n=1}^{\infty} e^{-n^2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-291f6e3ca6d8b1787c9397c99454f188_l3.png "Rendered by QuickLaTeX.com")
The series converges by the root test since
![\begin{align*} \lim_{n \to \infty} a_n^{\frac{1}{n}} &= \lim_{n \to \infty} \left( e^{-n^2} \right)^{\frac{1}{n}} \\[9pt] &= \lim_{n \to \infty} e^{-n} \\[9pt] &= 0 < 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-354ffa4670e2c420647fee85250bd23e_l3.png "Rendered by QuickLaTeX.com")
Hence, the series converges.
Tag: Root Test
Test the given series for convergence or divergence
Determine if the following series converges or diverges and justify your decision.
![\[ \sum_{n=1}^{\infty} \left( n^{\frac{1}{n}} - 1 \right)^n. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-51b5e3efa7faade456842e9e7c4b5b2c_l3.png "Rendered by QuickLaTeX.com")
The series converges by the root test since
![\begin{align*} \lim_{n \to \infty} a_n^{\frac{1}{n}} &= \lim_{n \to \infty} \left( \left( n^{\frac{1}{n}} - 1\right)^n \right)^{\frac{1}{n}} \\[9pt] &= \lim_{n \to \infty} \left(n^{\frac{1}{n}} - 1\right) \\[9pt] &= 0 < 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-39b101f16a8d29c383db72443e402f54_l3.png "Rendered by QuickLaTeX.com")
Hence, the series converges.