A positive sequence must monotonically converges for large enough index.

This is false. From the question title, it's better to give a simple example using a sequence with positive entries. Counterexample:

Consider the sequence $(a_n)$

$$2,\frac12,\dots,\frac{2}{2n-1},\frac{1}{2n},\dots$$

Since each term $a_n$ has a denominator which tends to infinity as $n\to+\infty$, $a_n\to0$ as $n\to+\infty$.

$$a_{2n-1}-a_{2n} = \frac{2}{2n-1}-\frac1{2n}=\frac{2n+1}{2n(2n-1)}>0 \text{, but}$$ $$a_{2n}-a_{2n+1} = \frac1{2n}-\frac{2}{2n+1}=\frac{-2n+1}{2n(2n+1)}<0$$

Therefore, the sequence $(a_n)$ converges to zero while fluctuating.

The claim is incorrect.

Consider the sequence $a_k = \frac{2 + (-1)^k}{k}$. This sequence is positive as requested in the title and converges to zero but contradicts the conclusion of the claim.

It's false, as can be seen by this counter-example $$a_n=\frac1n+\dfrac{(-1)^n}{n^2}.$$ This sequence has positive terms but is not monotonically convergent to $0$. One checks $a_{2p}>a_{2p+1}$ and $a_{2p+1}<a_{2p+2}$.