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?
 A: 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 $1$.
A: 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, $R = 1$.
A: First you have to calculate
$$
 \limsup_{n \to \infty} \left( \frac{1}{n \sqrt{n}} \right)^{1/n}
 = \limsup_{n \to \infty} \left( \frac{1}{n^{3/2}} \right)^{1/n}
 = \limsup_{n \to \infty} \left( \frac{1}{n^{1/n}} \right)^{3/2}.
$$
Because $\lim_{n \to \infty} n^{1/n} = 1$ it follows that
$$
 \limsup_{n \to \infty} \left( \frac{1}{n^{1/n}} \right)^{3/2}
 = \lim_{n \to \infty} \left( \frac{1}{n^{1/n}} \right)^{3/2}
 = \left( \lim_{n \to \infty} \frac{1}{n^{1/n}} \right)^{3/2}
 = 1
$$
because the limit exists. The radius of convergence is now given by
$$
 \frac{1}{\limsup_{n \to \infty} \left( \frac{1}{n \sqrt{n}} \right)^{1/n}}
 = \frac{1}{1}
 = 1.
$$
A: What you wrote is correct, from the root test the radius of convergence is $R=1$.
You may observe that both series $\displaystyle \sum_{n} \frac{(-1)^n}{n\sqrt{n}}$ and $\displaystyle \sum_{n} \frac{1}{n\sqrt{n}}$ are also absolutely convergent:
$$
\left|\sum_{n\geq1} \frac{(-1)^n}{n\sqrt{n}}\right|\leq \sum_{n\geq1}\left| \frac{1}{n\sqrt{n}}\right|=\sum_{n\geq1}\frac1{n^{3/2}}<\infty
$$ as a $p$-series with $p>1$.
A: $$\lim_{n\to\infty} \left|x^nn^{-\dfrac32}\right|^{\dfrac1n} = |x|\lim_{n\to\infty} \exp\left(-\dfrac 32\dfrac {\log n}n\right) = |x|\exp(0) = |x|.$$
