Why is the ratio test not working for this series? Show that the series $\frac{1}{1+x} + \frac{2}{1+x^2} + ... + \frac{2^n}{1+x^{2n}} + ...$ converges when $|x|>1$, and find it's sum.
I tried using the ratio test, but I first rewrote the above series as $\sum_{n=0}^{\infty}\frac{2^n(1-x^{2^n})}{1-x^{2^{n+1}}}$. The ratio test on this series gets us $\lim_{n\to\infty} \frac{2}{x^2}$ after disregarding the terms that disappear and computing it. However, in this case, $|x|>\sqrt2$, which is incorrect.
Where did I go wrong? I'm pretty sure I rewrote the series correctly, so this should give the correct answer. The solution is apparently to find a formula after testing the first few terms and proving by induction, but I am much more curious to why my method did not work, as that may show me some pitfalls that I am falling into when I am using the ratio test. 
Thank you!
 A: Let $a_n$ be the term of the series. For the series to converge,
$$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{2^n}{1+x^{2n}}=\lim_{n\to\infty}\frac{1}{\frac{1}{2^n}+\left(\frac{x^2}{2}\right)^n}=0 $$
has to be true. That's a necessary condition. This implies that $x^2/2>1\Leftrightarrow |x|>\sqrt{2}$ (think about $\lim_{n\to\infty}q^n$ for $|q|<1, |q|=1$, and $|q|>1$). The ratio test shows that $|x|>\sqrt{2}$ is also a sufficient condition:
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{2}{x^2}<1\,\,\text{for }|x|>\sqrt{2} $$
Hence, the series converges iff $|x|>\sqrt{2}$. And the ratio test works here.
A: The ratio test gives
$$\lim_{n\to\infty}\frac{2}{x^2}$$
the series converges if
$$|x|>\sqrt{2}.$$
If $|x|=\sqrt{2},$ the general term becomes
$$\frac{2^n}{1+2^n}$$
which goes to $1$, the series diverges.
If $|x|<\sqrt{2}$, it diverges.
A: Definition of the Ratio Test taken from Gilbert Strang's "Calculus" testbook:

Apply the "Ratio Test" to the problem: $$\frac{a_{n+1}}{a_{n}}=\frac{2^{n+1}}{1+x^{2n+1}}\cdot \frac{1+x^{2n}}{2^{n}}=\frac{2\left ( 1+x^{2n} \right )}{1+x^{2n+1}}$$
$$\lim_{n\rightarrow \infty }\frac{2\left ( 1+x^{2n} \right )}{1+x^{2n+1}}=2\lim_{n\rightarrow \infty }\frac{ 1+x^{2n}}{1+x^{2n+1}}=2\lim_{n\rightarrow \infty }\frac{1/x^{2n}+1}{1/x^{2n}+x}......\left(1\right)$$
$$\left | x \right |>1$$
Expression (1) = $2\lim_{n\rightarrow \infty }\frac{1}{x}=\frac{2}{x}$
The only thing that is certain about $\frac{2}{x}$ is that it is smaller than $2$, since $\left | x \right |>1$. There is no guarantee that it is smaller than $1$. In other words, the ratio test failed.
Try the "Root test":
$$\sqrt{\frac{x^{2n}}{1+x^{2n}}}< \frac{x^n}{1+x^{n}}$$
$$\lim_{n\rightarrow \infty }\frac{x^{n}}{1+x^{n}}=\lim_{n\rightarrow \infty }\frac{1}{1/x^{n}+1}......\left ( 2 \right )$$
$\left | x \right |> 1$ The limit in Expression (2) is $1$
That means $\lim_{n\rightarrow \infty }\sqrt{\frac{x^{2n}}{1+x^{2n}}}< 1$ if $\left | x \right |>1$ By the Root Test, the series converges if $\left | x \right |>1$.
A: A quick check:
$\dfrac{2^n}{1+x^{2n}} \lt (\dfrac{2}{x^2})^n=:a_n$.
Ratio test, or recalling criteria for geometric series:
1) $x^2 >2$, $\sum a_n$  converges 
By comparison test the original series converges for $x^2>2$.
2) Let $x^2\le 2$:
$\dfrac{2^n}{1+x^{2n}} \ge\dfrac{2^n}{1+2^n}:=b_n$.
$\lim_{n \rightarrow \infty} b_n=1 \not =0$, original series diverges.
