Find all complex numbers 
such that the series
![\[ \sum_{n=1}^{\infty} \left( 1 + \frac{1}{5n+1} \right)^{n^2} |z|^{17n} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-48a47ce9c425c9e1fee1dbd06b0028de_l3.png "Rendered by QuickLaTeX.com")
converges.
We look to apply the root test, letting
![\[ a_n = \left( 1 + \frac{1}{5n+1} \right)^{n^2} |z|^{17n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3610350c2cb782a6badad48a2fd0a421_l3.png "Rendered by QuickLaTeX.com")
Then we have
![\begin{align*} \lim_{n \to \infty} a_n^{\frac{1}{n}} &= \lim_{n \to \infty} \left( 1 + \frac{1}{5n+1} \right)^n |z|^{17} \\[9pt] &= \lim_{n \to \infty} e^{\frac{1}{5}} \cdot |z|^{17}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-89ccba24d87ce52e279a6e18d019eef0_l3.png "Rendered by QuickLaTeX.com")
Since this series converges if this limit is less than 1 and diverges if it is greater than 1, we have the series is absolutely convergent for
![\[ e^{\frac{1}{5}} \cdot |z|^{17} < 1 \quad \implies \quad |z|<e^{-\frac{1}{85}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-20f0299ebc4b9b4caff0aa90a31d8d91_l3.png "Rendered by QuickLaTeX.com")
And about |z|=e^(-1/85)? In this case, I know that the root test is inconclusive, but is there a way to guarantee the convergence or divergence?
With |z|=e^(-1/85), the limit of the terms of the serie goes to e^(-3/50) as n-> infinity. Therefore, the series diverge in this case.
Nice point, Rafael. But I get e^(17n/85-17n/85)=e^0=1, so the sequence goes to 1.
Hello, good day to you fellow mathematicians!
The limit is in fact
. Tom you must take the limit of the first expression, before the 
-th root. To evaluate that limit, use the Taylor expansion of log with 
changed to 
.
Cheers, and happy problem solving!