If each term in this sequence is positive, the result follows. Claim: $a_{2k} > a_{2k+2} \implies a_{2k+1} < a_{2k+3}$ and $a_{2k+1} < a_{2k+3} \implies a_{2k+2} > a_{2k+4}$. Proof. Follows by conditions. $\blacksquare$ As such, if $a_{2k} > a_{2k+2}$ or $a_{2k+1} < a_{2k+3}$ ever holds, then we get monotonicity where the evens always decrease and the odds always increase eventually. Truncate terms so this always hold. Else, if this never holds, the evens always increase, and the odds always decrease. Truncate the first term if necessary such that $a_1 > a_3 > a_5 > \dots$ and $a_2 < a_4 < a_6 < \dots$. Claim: WLOG we can assume that $a_1$ and $a_2$ are the same sign. Proof. Note that $a_1a_2 < \max\{(a_1 - 10)(a_2 - N), (a_1 + N)(a_2 + N)\}$ holds for any $a_1, a_2$ and $N$ because \[ \frac{1}{2} ((a_1 - 10)(a_2 - 10) + (a_1 + 10)(a_2 + 10)) > a_1a_2. \]Then take $N > |a_1|, |a_2|$. It suffices to show the sequences $a_1 + N, a_2 + N, \dots$ and $a_1 - N, a_2 - N, \dots$ are bounded. Then WLOG flip the signs so they are positive. $\blacksquare$ Then if $a_{2k+1}$ is the first negative term, then $a_{2k+3}, a_{2k+5}, \dots$ are all negative. Since $a_{2n} > a_2$, this gives us the fact that the sign eventually alternates, so the bound of $0$ suffices eventually. Else, all $a_i$ are positive, so we finish by the bound $(a_1 + a_2)^2$.

Define $\{c_i\}_{i\geq 2}$ by $c_{2n}=a_{2n+1}-a_{2n-1}$ for all $n\geq 1$ and $c_{2n+1}=a_{2n}-a_{2n+2}$ for all $n\geq 1.$ The condition implies that $\{c_i\}$ is strictly increasing. $\textbf{Case 1: }$ $\{c_i\}$ is bounded Let $\ell$ be the upper bound. Then we have \begin{align*} a_{2n+1}a_{2n+2} &= (a_1+c_2+c_4+\dots+c_{2n})(a_2-c_3-c_5-\dots-c_{2n+1})\\ &\leq\max((a_1+c_2+c_4+\dots+c_{2n})(a_2-c_4-c_6-\dots-c_{2n+2}),(a_1+c_2+c_4+\dots+c_{2n})(a_2-c_2-c_4-\dots-c_{2n})\\ &\leq\max\left((\tfrac{a_1+a_2+c_2-c_{2n+2}}{2})^2,(\tfrac{a_1+a_2}{2})^2\right)\\ &\leq \left(\frac{|a_1+a_2|+|\ell-c_2|}{2}\right)^2, \end{align*}\begin{align*} a_{2n+1}a_{2n} &=(a_1+c_2+c_4+\dots+c_{2n})(a_2-c_3-c_5-\dots-c_{2n-1})\\ &\leq\max((a_1+c_2+c_4+\dots+c_{2n})(a_2-c_4-c_6-\dots-c_{2n}),(a_1+c_2+c_4+\dots+c_{2n})(a_2-c_2-c_4-\dots-c_{2n-2})\\ &\leq\max\left((\tfrac{a_1+a_2+c_2}{2})^2,(\tfrac{a_1+a_2+c_{2n}}{2})^2\right)\\ &\leq\left(\frac{|a_1+a_2|+|\ell|}{2}\right)^2, \end{align*}as needed. $\textbf{Case 2: }$ $\{c_i\}$ is unbounded For all $n$ sufficiently large, we have $a+c_2+c_4+\dots+c_{2n}>0$ and $a-c_3-c_5-\dots-c_{2n-1}<0.$ Therefore $a_{n}a_{n+1}<0$ for all sufficiently large $n,$ implying the conclusion.

tapir1729 wrote: An infinite sequence $a_1$, $a_2$, $a_3$, $\ldots$ of real numbers satisfies \[ a_{2n-1} + a_{2n} > a_{2n+1} + a_{2n+2} \qquad \mbox{and} \qquad a_{2n} + a_{2n+1} < a_{2n+2} + a_{2n+3} \]for every positive integer $n$. Prove that there exists a real number $C$ such that $a_{n} a_{n+1} < C$ for every positive integer $n$. Merlijn Staps $a_{2n-1} + a_{2n} > a_{2n+1} + a_{2n+2}$ (1) $a_{2n+2}+a_{2n+3}>a_{2n}+a_{2n+1}$ (2) Ading (1),(2) we get that : $a_{2n-1}+a_{2n+3}>2a_{2n+1}$ (3) In (1) for $n\rightarrow n+1$ and adding (2) we get that $2a_{2n+2}>a_{2n+4}+a_{2n}$ Soppuse that there exist $i$ such that $a_{2i+1}<a_{2i+3}$ then for (3) we have that $a_{2j+1}$ is increasing from this point and then while from (1) we get that $a_{2j}$ is decresing. Now for (4) we get that $a_{2n+2}-a_{2n}>a_{2n+4}-a_{2n+2}$ so at some point we get that $a_{2k}<0$ and so $a_ia_{i+1}<0$ and is decresing. Soppuse now that there is no such $i$ exist then $a_{2i+1}$ is decresing and from (2) we get that $a_2i$ is incresing. If there exist $e$ such that $a_{2n+1}-a_{2n+3}>e$ for every $n$ then there exist $k$ such that $a_{2k+1}<0$ and we are done. If no such $e$ exist then we have that $\lim_{n\rightarrow +\infty }a_{2n+1}-a_{2n+3}=0$ and from (1) we get that: $\lim_{n\rightarrow +\infty }(a_{2n+1}-a_{2n+3})=0\geqslant \lim_{n\rightarrow +\infty }(a_{2n+2}-a_{2n})\geqslant 0\Rightarrow \lim_{n\rightarrow +\infty }(a_{2n+2}-a_{2n})=0$ So for $n>N$ we get that $a_{2n+1}$ belongs $[x,x+a]$ and $a_{2n}$ belongs to $[y,y+b]$ so we are done.

Let $d_n=a_n-a_{n-2}$. The conditions become $$0>d_{2n+1}+d_{2n+2} \;\;\; \text{and} \;\;\; 0<d_{2n+2}+2_{2n+3}$$Adding shifted versions of these yields that the odd $d_i$ are strictly increasing and even $d_i$ are strictly decreasing. Thus there must exist a point where the even $d_i$ and the odd $d_i$ stay the same sign. Notice that $a_{2k}=a_2+d_4+d_6+\dots+d_{2k}$ and $a_{2k+1}=a_1+d_3+d_5+\dots+d_{2k+1}$. Thus there must exist a point where the even $a_i$ and the odd $a_i$ stay the same sign (it doesn't hurt to assume this happens right away). If they are different in sign, then were clearly done. If they are both positive then the first inequality implies that $a_i<a_1+a_2$ so we can easily chose a $C$. If they are both negative then the second inequality implies that $a_i>a_2+a_3$ so we can easily chose a $C$.

Claim: Eventually the odd indices are increasing while the even indices are decreasing, or the other way around. Proof. By the given conditions we have $a_{2n+1} - a_{2n-1} < a_{2n} - a_{2n+2} < a_{2n+3} - a_{2n+1}$, so if $a_{2n+1} > a_{2n-1}$ once, then the odds will increase will evens will decrease from then on. If this never happens, then the odds will decrease while evens will increase. Now, we do a casework bash to finish the problem. Case 1. $a_1, a_2 > 0$ or $a_1, a_2 < 0$ Note that in the first subcase, the first condition implies $a_i < a_1+a_2$ for all i, meaning we can take $C = (a_1+a_2)^2$, because the sequences cannot both become negative. Similarly, for the second subcase, the second condition implies $a_i > a_2 + a_3$ for all $i$, implying that $C = (a_2+a_3)^2$ works, because the sequences cannot both become positive. Case 2. $a_1, a_2$ are opposite signs If the two sequences keep their initial signs or they both end up swapping their signs, then selecting $C$ is trivial because $a_ia_{i+1} < 0$ for large $i$. If two sequences end up being the same sign after some time, then we can finish by the first case.

Consider the sequence $\{b_i\}_{i=1}^\infty$ defined by $b_i=(-1)^ia_i$. Let $d_i=b_{i+2}-b_i$. The conditions simplify to $d_i>d_{i+1}$ for all positive integers $i$, and we want to show that $b_nb_{n+1}$ is bounded below since we flipped the sign of exactly one of $a_n$ and $a_{n+1}$. It suffices to prove that the product is bounded for all sufficiently large $n>N$ as we can take the minimum of that bound and $b_nb_{n+1}$ for $n\leq N$. If $d$ is eventually nonpositive, say $d_k$ is the first $d$ that is $\leq0$, then all $d_i$ are at most $d_{k+1}<0$. This means that the $b$ sequence is bounded above by a strictly decreasing linear function, and thus is eventually always negative. So $b_nb_{n+1}>0$ for all sufficiently large $n$. Otherwise, $d$ is always positive, which means the $b$ sequence is increasing. If $b$ is bounded, the final condition is clearly true. Otherwise, at least one of the partial sums $d_1+d_3+\dots+d_{2k-1}$ and $d_2+d_4+\dots+d_{2k}$ is unbounded in the positive direction, but the $d_i$ decreasing condition implies that both must be unbounded (the partial sums can be bounded by each other up to some constant). This means that $b$ is eventually positive, so $b_nb_{n+1}$ is eventually positive for sufficiently large $n$.

Write the condition as $a_1 + a_2 > a_3+a_4 > \cdots$ and $a_2 + a_3 < a_4 + a_5 < \cdots$. If $a_{2k-1} < a_{2k+ 1}$ or $a_{2k-2} > a_{2k}$, then we have $a_{2k-1} < a_{2k+1} < \cdots$ and $a_{2k-2} > a_{2k} > \cdots$. If this happens, we just remove terms so that $a_1 < a_3 < \cdots$ and $a_2 > a_4 > \cdots$. If $a_1, a_2$ are the same signs, $|a_i - a_{i+2}|$ is increasing and nonzero, so we get that eventually for some $N$, we have $a_{2k+1} > 0$, $a_{2k} < 0$ for $k > N$. This means that there are only finite many $a_i a_{i+1} > 0$, so it is bounded. So we assume that $a_1 > a_3 > \cdots$ and $a_2 < a_4 < \cdots$. Note that if they have different signs, then we have that $a_ia_{i+1} < 0$. $a_i, a_{i+1}$ must share a sign at some $N$ otherwise we take $C = 0$. However, if $a_N, a_{N+1}$ are the first consecutive terms to share a sign, we just take bound $(a_N + a_{N+1})^2$ to finish.

By adding the two conditions we can see that \[ a_1<a_3<a_5 \dots \qquad \mbox{and} \qquad a_2>a_4>a_6 >\dots\] Let's focus on $a_1$ and $a_2$. Case 1: Both are positive or both are negative. If we have that both are positive, We have that $a_1+a_2 > a_3+a_4 > a_5+a_6 \dots $ by spamming the first condition of the problem. So we get that $a_i<a_2+a_1$. Hence setting $C=(a_1+a_2)^2$ is sufficient. Similarly for both being negative we can set $C=(a_2+a_3)^2$. Case 2: They have opposite signs. Then we have that $a_na_{n+1}<0$, now if for sufficiently large $n$ they get same sign, then that’s dealt with in case one, hence we are done!

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