Power Series Trouble I was going over some power series problems when I noticed this one :- $ \sum \frac {3^n}{\sqrt{n}}. x^{2n+1}$ find the radius of convergence of the power Series.
Immediately i noticed that I needed to be careful while applying root test as this is a kind of Sub Sequential Power Series ! So instead of finding  $n$th root we have to find the $2n+1$th root of the corresponding coefficient and then take the LimSup of it !
But i can't finish the computation of the limit ! Can you be kind enough to help me through ? 
 A: Use the Ratio Test instead:
$$\lim_{n \longrightarrow+\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n \longrightarrow+\infty}|\frac{\frac{3^{n+1}}{\sqrt{n+1}}x^{2(n+1)+1}}{\frac{3^n}{\sqrt{n}}x^{2n+1}}|=\lim_{n \longrightarrow+\infty}3x^2\sqrt\frac{n}{n+1}=3x^2<1 \Longrightarrow |x| < \frac{1}{\sqrt{3}}$$
So it's radius is $\frac{1}{\sqrt{3}}$.
A: Note that with $a_n = 3^n/\sqrt{n}$ we have $\displaystyle\sum_{n=1}^\infty a_n x^{2n+1} = \sum_{k=1}^\infty b_k x^k$ where $\displaystyle b_k = \begin{cases} a_n, & k = 2n+1\\ 0, & k \neq 2n+1 \end{cases}$
Applying the root test to find the radius of convergence we consider
$$\limsup_{k \to \infty}\, |b_kx^k|^{1/k} = \limsup_{n \to \infty} |a_n|^{1/(2n+1)} |x| < 1,$$
which implies that the radius of convergence is 
$$R=\left(\limsup_{n \to \infty} |a_n|^{1/(2n+1)}\right)^{-1} = \left(\limsup_{n \to \infty} \frac{3^{n/(2n+1)}}{n^{1/(2(2n+1))}}\right)^{-1} = \frac{1}{\sqrt{3}}$$ 
More simply, the ratio test gives
$$\lim_{n \to \infty} \frac{3^{n+1}|x|^{2n+3}}{\sqrt{n+1}}\frac{\sqrt{n}}{3^n|x|^{2n+1}}< 1 \implies 3|x|^2 < 1 \implies R = \frac{1}{\sqrt{3}}$$
A: No problem.
Write it as
$\sum \frac {3^n}{\sqrt{n}}. x^{2n+1}
=x\sum \frac {3^n}{\sqrt{n}}. x^{2n}
$
and use the fact that
$n^{1/n} \to 1$.
Then the $n$-th root is
$3x^2$.
The only thing 
powers of $n$ and
$(-1)^n$ do
is possibly affect convergence
at the boundaries.
The radius of convergence is not affected.
A: $$\limsup_n(a_n)^{1/n}=\limsup_k (\frac{3^{2k-1}}{\sqrt{2k-1}})^{\frac{1}{2k-1}}=3$$
Note that $$\sqrt{2k-1}^{\frac{1}{2k-1}}=(2k-1)^{\frac{1}{4k-2}}=e^{\frac{\log{(2k-1)}}{4k-2}} \to 1$$
