Radius of convergence of $ \sum_{n} \frac{x^n}{n\sqrt{n}}$ Trying to find the radius of convergence for $ \displaystyle \sum_{n} \frac{x^n}{n\sqrt{n}}$ 
I apply the root test: $\displaystyle \lim_{n \to \infty} \left(\bigg|\frac{x^n}{n\sqrt{n}}\bigg|\right)^{1/n} =\lim_{n \to \infty} \frac{|x|}{n^{1/n}n^{1/2n}} = |x|. $
The series converges absolutely if $|x| < 1$ therefore $R = 1$. 
Is the above correct? On Wikipedia I find the formula:
$R = 1/\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}}$
How does one use this to find $R$ in the above case?

The formula $R = 1/\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}}$ is for the series $\displaystyle\sum_{n = 0}^{\infty}c_nx^n$. 
Note that $c_n$ is the coefficient of the $x^n$ term, not the $n$-th term itself. 
So, here $c_n = \dfrac{1}{n\sqrt{n}} = n^{-3/2}$, and we need to compute $\displaystyle\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}} = \displaystyle\limsup_{n \rightarrow \infty}n^{-\tfrac{3}{2n}}$. 
By taking the logarithm and using L'Hoptital's Rule (or whatever other method you prefer) you get $\displaystyle\lim_{n \rightarrow \infty}n^{-\tfrac{3}{2n}} = 1$. 
Since $\displaystyle\lim_{n \rightarrow \infty}n^{-\tfrac{3}{2n}}$ exists, we have $\displaystyle\limsup_{n \rightarrow \infty}n^{-\tfrac{3}{2n}} = \displaystyle\lim_{n \rightarrow \infty}n^{-\tfrac{3}{2n}} = 1$, and thus, $R = 1$.
