Is a positive sequences of real numbers that converge to zero eventually decreasing?

Question

I wonder if a sequence of real numbers that converges to zero, where every term is positive is eventually decreasing. Is this true ? I have an idea on how to prove it by contradiction. If this isn’t true, we can always find a term in the sequence bigger than any term before it. We choose a subsequence of these “peaks”, and show it converges to something greater than zero, which is a contradiction. Is this a valid proof ?

Answer 1

Consider the sequence $$x_n = \frac{2+(-1)^n}{2^n}$$ This alternates between larger and smaller values, so it never becomes decreasing.

Answer 2

You can take the sequence $a_1 = 1, a_2 = \dfrac{1}{2}, a_3 = \dfrac{3}{4}, a_4 = \dfrac{1}{4}, a_5 = \dfrac{3}{8}, ...$. You can see the pattern yourself and the sequence does converge to $0$ while having the property that: $a_1 >a_2 < a_3 > a_4 < a_5....$

Answer 3

Here are many counterexamples.

Start with any positive sequence $(a_n)$ that converges to zero.

Now define a new sequence: $$b_n = \begin{cases} a_k &\quad\text{if $n=2k-1$ is odd} \\ 2a_k &\quad\text{if $n=2k$ is even} \end{cases} $$ Every even term of the sequence $(b_n)$ will be greater than the preceding odd term so $(b_n)$ is certainly not decreasing, in fact it is not even "eventually decreasing". But $(b_n)$ is a positive sequence converging to zero.

Answer 4

Consider $$a_n={\sin^2(n) \over n}$$

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