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 answer is correct, and the two formulas are equivalent if $$\lim_ \limits{n \to \infty}|c_n|^{1/n}$$exists.
The ratio test is however the easier test in this case. Observe
$$\lim_ \limits{n \to \infty} \left| \frac{x^{n+1}}{(n+1)^{3/2}} \frac{n^{3/2}}{x^n}\right| = |x|,$$ and since ratio test implies absolute convergence when the ratio is less than $1$, we get a radius of convergence of $1$.
