absolute convergence of power series I'd like to determine the radius of convergence R of the power series
$$i) \sum_{k=1}^{\infty} \frac{x^{k}}{1+x^{k}}, \ x\in \mathbb{R} \ \ \ \ ii)\sum_{k=1}^{\infty} 2^{k}\cdot x^{k^{2}}, \ x\in \mathbb{R}$$ .
My ideas: 
i) Root test: $$ \sqrt[k]{\vert \frac{x^{k}}{1+x^{k}} \vert} = \frac{\vert x\vert}{\sqrt[k]{\vert 1+x^{k} \vert}} \Rightarrow \vert x \vert < \sqrt[k]{\vert 1+x^{k} \vert} $$
But I don't know how to determine $$ \vert x \vert $$
for convergence.
ii) Root test: $$ \sqrt[k]{\vert 2^{k}x^{k^{2}}\vert}=2\cdot \vert x \vert ^{k} \Rightarrow \limsup\limits_{k \rightarrow \infty}({2 \cdot \vert x \vert^{k} })<1 \Leftrightarrow \vert x \vert <1 \Rightarrow absolute \ convergence \ for \ \vert x \vert < 1$$ ?
 A: (i). Note that if $|x|<1$: 
$$\lim_{k\to\infty}\frac{\left|\frac{x^k}{1+x^k}\right|}{|x|^k}=1$$
Thus by limit comparison test the series is absolutely convergent
If $|x|>1$, then $\lim_{k\to\infty}\left|\frac{x^k}{1+x^k}\right|=1$, thus the series is divergent.
If $x=1$ then the series obviously diverges.
Lastly, $\frac{x^k}{1+x^k}$ is undefined if $x=-1$ and $k$ is odd.
Thus the series converges precisely when $|x|<1$.
(ii). Note that the coeffcients of the series are
$$a_n=\begin{cases}2^k, &\text{if }n=k^2\\
0, &\text{elsewhere}.\end{cases}$$
Thus
$$\limsup_{n\to\infty}|a_n|^{\frac{1}{n}}=\lim_{k\to\infty}2^{\frac{1}{k}}=1.$$
Thus if $|x|<1$ the series is absolutely convergent, and if $|x|>1$ the series is divergent.
Lastly the series obviously diverges if $x=\pm 1$.
A: $\lim_\limits{k\to\infty} \frac {|x|}{\sqrt[k]{1+x^k}} = |x|$ when $x<1$, passing the root test.
If $|x|\ge 1$ then $\lim_\limits{k\to\infty} \frac {|x|}{\sqrt[k]{1+x^k}} = 1$
and the root test is inconclusive.
but, if x>1 then $\lim_\limits{k\to\infty} \frac {x}{1+x^k} = 1$
It is necessary for the sequence to approach 0 in order for the series to converge.
|x|<1
ii) is a little easier... by the root test
$(2^kx^{k^2})^{\frac 1k} = (2 x^{k})$
$\lim_\limits{k\infty} 2x^k= 0$ when $x<1, 2$ when $x=1$ and $\infty$ when $x>1$
again $|x|<1$ 
A: Let's use the ratio test here for ii: We have $\sum_{n=1}^\infty2^k\cdot x^{k^2}$  Using the ratio test we get $$lim_{k \to\infty}\lvert \frac{2^{k+1} \cdot x^{(k+1)^2}}{2^k \cdot x^{k^2}}\rvert$$
This gives $$lim_{k \to\infty}\lvert 2 \cdot x^{2k+1}\rvert$$  
In order for this to converge $$\rvert x \lvert \lt1$$
This is only true when x is between -1 and 1.  Test endpoints and we see that the endpoints are not included.
