Another form of the ratio test Let $a_k≠0, \forall k \in \mathbb{N}$. If $$\limsup_{k\to\infty}\left|\frac{a_{k+2}}{a_k}\right|<1$$ then the series $\sum_{k=1}^\infty a_k$ is absolutely convergent.
I am really interested into seeing a proof of this but I cannot find one. For the original ratio test, there are many proofs but this one is different. Has anyone an idea? Thanks.
 A: To follow leoli1's comment, you can do it as follows.
By the assumption and directly applying the ratio test you can show that the following two series are convergent:
$$
\sum_{k=1}^\infty |a_{2k}|,\quad \sum_{k=1}^\infty |a_{2k+1}|
$$
Now the partial sum of the original (absolute) series has the following estimate by the triangle inequality
$$
\sum_{k=1}^n|a_k|
\le 
\sum_{k=1\\k\textrm{ odd}}^n|a_k|
+\sum_{k=1\\k\textrm{ even}}^n|a_k|\tag{1}
$$
Taking $n\to\infty$, we are done.

Notes. One actually has equality in (1) and thus the triangle inequality is not needed. (Thanks to hardmath's comment below.)
A: The condition
$$\limsup_{k\to\infty}\left\vert\frac{a_{k+2}}{a_k}\right\vert<1$$ implies that it exists $N \ge 1$ and $0 \le K \lt 1$ such that
$$\left\vert\frac{a_{k+2}}{a_k}\right\vert \le K$$ for all $k \ge N$. Therefore
$$\left\vert\frac{a_{N+2p}}{a_N}\right\vert \le K^p, \left\vert\frac{a_{N+2p+1}}{a_{N+1}}\right\vert \le K^p$$ for all $p \ge 0$ and
$$\begin{aligned}\sum_{k \ge N} \vert a_k \vert &= \sum_{k \ge N\\k\textrm{ odd}} \vert a_k \vert + \sum_{k \ge N\\k\textrm{ even}} \vert a_k \vert\\
&\le \vert a_{N+1} \vert\sum_{k \ge 0} K^{p} + \le \vert a_{N} \vert\sum_{k \ge 0} K^{p}\\
&\le \frac{2}{1-K} \max(\vert a_N \vert, \vert a_{N+1} \vert)
\end{aligned}$$
Proving that $\sum a_k$ converges absolutely.
