# radius of convergence of $\sum \limits_{n = 1}^\infty \frac {(3n+1)^{10}x^n} {2n^n}$

How can I determine the radius of convergence of $\sum \limits_{n = 1}^\infty \frac {(3n+1)^{10}x^n} {2n^n}$?

I thought it should be easy by using the root test but I'm somehow failing to get a useful result.

• Can you show some of what you have tried? – Michael McGovern Jan 11 '17 at 1:42
• Yeah. Say for instance, have you tried the Ratio Test to establish the possible interval of convergence? Any thoughts? – Juniven Jan 11 '17 at 1:43
• @MichaelMcGovern after thinking about it again it could actually make sense. I get $0<1$. Would my R then be $\infty$ ? Sorry I'm new to this. – fatalError Jan 11 '17 at 2:30
• @fatalError You get $0$ by using the Ratio Test? – Juniven Jan 11 '17 at 2:38
• @juniven No I'm talking about the root test. That was my question and I still want to know how to solve it this way before I look at other variants sorry. – fatalError Jan 11 '17 at 2:39

$$R = \frac{1}{\lim_{n \to \infty} \sqrt[n]{\frac {(3n+1)^{10}} {2n^n}}} = \frac{1}{\lim_{n \to \infty} \frac{(3n+1)^{10/n}}{\sqrt[n]{2}n}} = \infty$$ as $\lim_{n \to \infty} (3n+1)^{10/n}$ is finite and $\lim_{n \to \infty} \sqrt[n]{2}n = \infty$.

I think the ratio test is the one that in most cases give the easier answer :

Let be $\sum a_nx^n$ with $a_n=\frac{(3n+1)^{10}}{2n^n}$ and $\frac1R=lim|\frac{a_{n+1}}{a_n}|$.

$\frac{a_{n+1}}{a_n}=\frac{(3n+4)^{10}2n^n}{(3n+1)^{10}2(n+1)^{n+1}}=\frac{(3n+4)^{10}}{(3n+1)^{10}}\times\frac{n^n}{n^{n+1}(1+\frac1n)^{n+1}}\sim1\times\frac{1}{ne}\to 0$ so the radius of convergence is infinite.

For the n-root test look at positrono answer, I should add a little bit of detail :

This $(3n+1)^\frac{10}{n}=\exp(\frac{10}{n}\,\ln(3n+1))=\exp(30\,\frac{\ln(3n+1)}{3n})\to 1$ since $\frac{\ln(u)}{u}\to_{+\infty} 0$.

And $\sqrt[n]{2n^n}=n\times\sqrt[n]2=n\times\exp(\frac{\ln(2)}{n})\sim n\times 1\sim n\to +\infty$ since $\frac{\ln(2)}{n}\to_{+\infty}0$

This converges everywhere since $n^n \ge n!$, the $(3n+1)^{10}$ does not affect convergence, and this converges at least as well as $e^x$ which converges everywhere.