radius of convergence of a given series How could I find the radius of convergence for the following power series:
$$
\sum\limits_{n = 1}^\infty  {\frac{{2^{n - 1} x^{2n - 1} }}{{(4n - 3)^2 }}} .
$$
From what I read I need to find: $$
\mathop {\lim \sup }\limits_{n \to  + \infty } \sqrt[n]{{a_n }}
$$ and then the field is (-1/Answer,1/Answer)
Note: $a_n$ is the scalar multiplied by $x$.
But that didn't work here. (I found that R=0.5 which is wrong)
 A: The main thing to note about these power series is that they're usually close to a geometric series. These converge iff the ratio is in $(-1,1)$.
Polynomials (in $n$) such as the $(4n-3)^{-2}$ in the denominator don't matter unless the geometric ratio is precisely $-1$ or $1$.
So, first off, take out everything that's not an exponential (you can rigourously show this with the limit you've written down):
$$a_n=\frac{1}{2x}\frac{1}{(4n-3)^2}\left(2x^2\right)^n$$
where the first part is just adjusting, second part is the polynomial (function of $n$ to a constant power), third part is the true exponential (constant to the power $n$). Note the $x^2$ inside the exponential as you have $x^{2n-1}$!
Clearly, it must converge if $2x^2\in(-1,1)$, so $|x|<\frac{1}{\sqrt{2}}$. The series clearly cannot converge on $|x|>\frac{1}{\sqrt{2}}$ as you have an exponential $>1$.
Right at $\pm\frac{1}{\sqrt{2}}$ it gets more interesting. At precisely there $x=\pm\frac{1}{\sqrt{2}}$ what you have is
$$\sum_{n=1}^{\infty}\frac{1}{\sqrt{2}}\frac{1}{(4n-3)^2}$$
and you should know by comparison with say
$$\sum_{n=1}^{\infty}\frac{1}{n^2}$$
that this converges.
The interval of convergence is $\left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right]$, and the radius is $\frac{1}{\sqrt{2}}$.
PS: finding $L=\lim_{n\rightarrow\infty}\sqrt[n]{a_n}$ is fine. What's not is blindly plugging in to say radius of convergence is $(-1/L,1/L)$. You're using the fact that a geometric series converges iff $r<1$, so you need to solve the resulting equation $L<1$.
A: By Cauchy-Hadamard, $r=1/\limsup_{n\to\infty}\sqrt[2n-1]{2^{n-1}/(4n-3)^2}=1/\limsup(2^{(n-1)/(2n-1)}/(4n-3)^{2/(2n-1)}=\limsup 1/\sqrt2/1=1/\sqrt2$.
A: Both the ratio test and the root test say
$$|2x^2|<1$$ and this is conclusive.

For $2x^2=1$, the denominator is $\sim\dfrac1{16n^2}$ and the series converges (by comparison with the Basel series).
