A positive sequence must monotonically converges for large enough index. This question is rather simple,
Let $a_n$ be a sequence that converges to zero, exists a $N$ such that for all $n>N$ the following $a_{n+1}\le a_{n}$
Is the theorem above correct? I am confused since I used it in an exam and the professor said that this does not necessarily happen, looking for a counter-example or something?
 A: 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.
A: 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 $a_{2p+1}<a_{2p+2}$.
A: This claim is not true.
Hint: consider the sequence defined by $x_{2n}=2^{-n}$ and $x_{2n+1}=3^{-n}$.
A: A more generic method to construct such a sequence: Take an arbitrary convergent sequence that strictly monotonic decreases, that is, $a_{n+1}<a_n$ for all $n$. Then you can construct a new sequence $b_n$ by $b_{2n}=a_{2n+1}$, $b_{2n+1}=a_{2n}$. It is not hard to check that $b_n$ converges to the same limit as $a_n$, but never gets monotonic as by construction $b_{2n+1}>b_{2n}$.
Also note that from each monotonic decreasing convergent sequence that doesn't eventually get constant, you can get a strictly monotonic decreasing one with the same limit by simply omitting the repetitions.
Also note that when considering the discrete topology (e.g. by considering integer sequences only), your assumption is true, as there only the sequences that eventually get constant are convergent, and the constant sequence is monotonic decreasing.
A: 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.
