Consider the function 
defined by
![\[ f(n) = na^n \qquad \text{where} \quad |a| < 1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-fe216ddef6700d27517b19fecb42a6a7_l3.png "Rendered by QuickLaTeX.com")
Determine whether the sequence 
converges or diverges, and if it converges find the limit.
We use the definition of the exponential and some tricks to get into a form we know how to deal with,
![\begin{align*} \lim_{n \to \infty} f(n) &= \lim_{n \to \infty} na^n \\[9pt] &= \lim_{n \to \infty} \left( n^{\frac{1}{n}} a \right)^n \\[9pt] &= \lim_{n \to \infty} e^{n \log \left(a n^{\frac{1}{n}}\right)} \\[9pt] &= \exp \left( \lim_{n \to \infty} \left( n \log a + \log n \right) \right) \\[9pt] &= \exp \left( \lim_{n \to \infty} n \left( \log a + \frac{\log n}{n} \right) \right) \\[9pt] &= 0. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1e48237c2af837586d4108b63b669425_l3.png "Rendered by QuickLaTeX.com")
The final line follows since
![\[ \lim_{n \to \infty} \left(\log a + \frac{\log n}{n}\right) < 0 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9c4a8450d56b893eed76bcf3823aa343_l3.png "Rendered by QuickLaTeX.com")
since
![\[ \lim_{n \to \infty} \frac{\log n}{n} = 0 \qquad \text{and} \qquad |a| < 1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1d368ba7926ce95c5a87e36774c5ba73_l3.png "Rendered by QuickLaTeX.com")
Therefore,
![\[ n \left( \log a + \frac{\log n}{n} \right) \to -\infty \qquad \text{as} \qquad n \to \infty.\]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-389b95138998457ad22eb4b59bdeaa53_l3.png "Rendered by QuickLaTeX.com")
Thus, converges to the limit 0.