Special Case in Ratio Test for Series We all know the Ratio Test  in series which is one of many tests that are  used to determine whether a Series is Converges or Not (Most Of The Time Work).
Here is a link  that describe Ratio test "In Short".
 Ratio Test
I have example : Suppose we have $U_n$=$2^{-n+({-1})^{n+1}}$ and we want to know whether the series $$\sum_{n=1}^\infty U_n$$  Is converges Or Not.
$\\$
If we Use the Ratio Test then $\frac{U_{n+1}}{U_n}$=$2^{2({-1})^{n}-1}$, and we can Rewrite it as$$ \frac{U_{n+1}}{U_n} =
\begin{cases}
2,  & \text{if $n$ is even} \\
\frac{1}{8}, & \text{if $n$ is odd}
\end{cases}$$
 And I think if the Proudct of the two number $2$ and $\frac{1}{8}$ is exactly less than $1$ then the series is Converges . in general Form 
if the fraction $\frac{U_{n+1}}{U_n}$ Can be written of the Form of 
 $$ \frac{U_{n+1}}{U_n} =
\begin{cases}
A_n,  & \text{if $n$ is even} \\
B_n, & \text{if $n$ is odd}
\end{cases}$$
and we have  $$\lim_{n\to \infty}{A_n}= a  \text{ and} \lim_{n\to \infty}{B_n}= b  $$
then if   $ 0 < a.b < 1  $then the series $\sum U_n$ Is Converges.
What do you think in this idea ?
for Now it is just true when $U_n >0$, I did not study it when $U_n <0$
So Give me your Opinion whether my idea is True Or False 
 A: Note that 
$$\sum_{n=1}^{2N} U_n=\sum_{n=1}^N U_{2n}+\sum_{n=1}^N U_{2n-1}\tag1$$
We assume that $U_n\ge 0$,
If the limits $\lim_{n\to \infty}\frac{U_{2n+2}}{U_{2n+1}}$ and $\lim_{n\to \infty}\frac{U_{2n+1}}{U_{2n}}$ exist and are finite, then the ratio test guarantees that the first sequence of partial sums on the right-hand side of $(1)$ converges when 
$$\begin{align}
\lim_{n\to\infty}\frac{U_{2(n+1)}}{U_{2n}}&=\lim_{n\to \infty}\left(\frac{U_{2n+2}}{U_{2n+1}}\frac{U_{2n+1}}{U_{2n}}\right)\\\\
&=\left(\lim_{n\to \infty}\frac{U_{2n+2}}{U_{2n+1}}\right)\,\left(\lim_{n\to \infty}\frac{U_{2n+1}}{U_{2n}}\right)
\\\\&<1\tag 2
\end{align}$$
Similary, if the limits $\lim_{n\to \infty}\frac{U_{2n}}{U_{2n-1}}$ and $\lim_{n\to \infty}\frac{U_{2n+1}}{U_{2n}}$ exist and are finite, then the ratio test guarantees that the second sequence of partial sums on the right-hand side of $(1)$ converges when 
$$\begin{align}
\lim_{n\to\infty}\frac{U_{2(n+1)-1}}{U_{2n-1}}&=\lim_{n\to \infty}\left(\frac{U_{2n+1}}{U_{2n}}\frac{U_{2n}}{U_{2n-1}}\right)\\\\
&=\left(\lim_{n\to \infty}\frac{U_{2n+1}}{U_{2n}}\right)\,\left(\lim_{n\to \infty}\frac{U_{2n}}{U_{2n-1}}\right)
\\\\&<1\tag3
\end{align}$$
Hence, if $(2)$ and $(3)$ hold, then the sequence of partial sums on the left-hand side of $(1)$ converges.

We conclude that if the product of the limits $\lim_{n\to\infty}\frac{U_{n+1}}{U_n}$ for $n$ even and $\lim_{n\to\infty}\frac{U_{n+1}}{U_n}$ for  $n$ odd is less than $1$, then the series $\sum_{n=1}^\infty U_n$ converges.

