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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?

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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.

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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.

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    $\begingroup$ The body doesn't repeat it, but the title seems to ask for a positive sequence. $\endgroup$ – Clement C. Mar 19 '17 at 16:03
  • $\begingroup$ @ClementC. I guess those who upvoted it can complete the argument from this sequence. Btw, I +1'ed b4 knowing how. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 19 '17 at 16:07
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    $\begingroup$ @ClementC Thanks for pointing that out. I missed that in the title. I've changed it to a slightly different sequence that is positive. $\endgroup$ – Qudit Mar 19 '17 at 19:39
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This claim is not true.

Hint: consider the sequence defined by $x_{2n}=2^{-n}$ and $x_{2n+1}=3^{-n}$.

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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 $a_{2p+1}<a_{2p+2}$.

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  • $\begingroup$ It's easily fixable while keeping the spirit, but the OP asks for a sequence going to 0. $\endgroup$ – Clement C. Mar 19 '17 at 16:08
  • $\begingroup$ @Clement C. : I thought the O.P. wanted a series with positive terms. I probably misread. I 'll add an example converging to $0$. $\endgroup$ – Bernard Mar 19 '17 at 17:20
  • $\begingroup$ OP also included positive in the title description. $\endgroup$ – Improve Mar 19 '17 at 17:22
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    $\begingroup$ OK, I eventually got it – and got a counter-example satisfying all constraints, if my computations are correct. $\endgroup$ – Bernard Mar 19 '17 at 17:48
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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.

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  • $\begingroup$ Please fix $b_{2n+1} = a_{2n}$. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 19 '17 at 16:04
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    $\begingroup$ @GNUSupporter: Fixed, thank you. I just spotted that I also had an actual math error, too, which I now also fixed. $\endgroup$ – celtschk Mar 19 '17 at 16:58

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