# Alternating Sums of Two Sequences, One Termwise Dominating

Edited in light of Gerry Myerson's quick counterexample.

I have two finite sequences $(a_n)$ and $(b_n)$ satisfying the following:

• all terms are positive
• both sequences are strictly decreasing
• the $(b_n)$ strictly dominates $(a_n)$ (i.e. $b_n > a_n$ for all $n$)
• the sequence $\left(\frac{a_n}{b_n}\right)$ is strictly decreasing

I want to conclude $$\sum_{n=1}^N (-1)^{n-1}a_n \leq \sum_{n=1}^N (-1)^{n-1}b_n.$$

What additional properties might I seek to establish for the sequences to get the desired result?

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If this was true for decreasing sequences this would be true for nonincreasing ones. But $a_n=x^n$ with $x$ in $(0,1)$ and $b_n=1$ yields approximately $x/(1+x)$ on the LHS for large values of $N$, and alternatively $0$ and $1$ on the RHS. For every even $N$ this is a counterexample.
If one insists on the sequence $(b_n)$ being decreasing, one can consider $b_n=y^n$ with $y$ in $(0,1)$, in the regime $x\approx 1$, $y\approx1$, $N\gg1$ and $x^N\ll y^N$.
$$10-1\gt11-9$$ would seem to be a problem.