Determine the radius of convergence 
of the power series:
![\[ \sum_{n=1}^{\infty} \left( 1 + \frac{1}{n} \right)^{n^2} z^n. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-efc084a78cd919b0d8f2355c3a97144d_l3.png "Rendered by QuickLaTeX.com")
Test for convergence at the boundary points if is finite.
We have
![\[ a_n = \left( 1 + \frac{1}{n} \right)^{n^2} z^n \quad \implies \quad a_n^{\frac{1}{n}} = \left( 1 + \frac{1}{n} \right)^n z. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b7d3acbe25504713a025cdbf0d900c9f_l3.png "Rendered by QuickLaTeX.com")
Therefore,
![\begin{align*} \lim_{n \to \infty} a_n^{\frac{1}{n}} &= \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n z \\[9pt] &= e\cdot z. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9c8f68f57eca24b4ca2988ef32360673_l3.png "Rendered by QuickLaTeX.com")
Hence, 
and the series converges for 
.
Determine the radius of convergence of the power series:
Test for convergence at the boundary points if is finite.
We have
Therefore,
Hence, and the series converges for .