# 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

• I know that we can use another Test to prove the convergence of the first series .. but i want it in this way .. with thanks – Yaser Tarek Dec 24 '17 at 16:09
• Hint: consider the even and odd subseries... – gniourf_gniourf Dec 24 '17 at 16:10
• yes i have dont that before .. – Yaser Tarek Dec 24 '17 at 16:12
• And what's your conclusion? – gniourf_gniourf Dec 24 '17 at 16:13
• it turns out that my opinion is True .. Am I Right ? – Yaser Tarek Dec 24 '17 at 16:15

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.

• Happy New Year!. And when you have enough reputation points, feel free to up vote as you see fit. ;-)) – Mark Viola Jan 11 '18 at 19:39