# Is this sequence monotonically decreasing?

Let $a_n = \frac{p_n - p_{n-1}}{p_n \log p_n}$ where $p_n$ denotes the $n$-th prime. Is this sequence decreasing (or decreasing after some $N$)?

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Very unlikely. If $p_n- p_{n-1}$ is a small prime gap, there will often come a much larger gap after that. If there are infinitely many twin primes, $p_n - p_{n-1} = 2$ happens infinitely often, and then $a_{n+1} > a_n$. – Daniel Fischer May 17 '14 at 12:17
I used ListPlot on mathematica to look at the graph of a_n from n = 2 to 700000 and it's really striking. It looks like there are a dozen (or more) bands of decreasing curves with an asymptote at the horizontal axis, concave up. The curves become more and more scattered along the top. – Braindead May 17 '14 at 12:23

The sequence is not decreasing.

You have for example:

$$a_4=\frac{2}{7 \cdot \log 7}=0,14 \dots$$ $$a_5=\frac{4}{11 \log 11}=0,15 \dots$$

Doing more examples,you will see that the sequence is getting in general smaller,but it is not monotone..

Here you can see a plot:

EDIT: Consider two twin primes,for example these ones: $3,5$ and $5,7$.

Then it will be like that: $$a_n=\frac{5-3}{5 \cdot \log{5}}=\frac{2}{5 \cdot \log{5}}$$

$$a_{n+1}=\frac{7-5}{7 \cdot \log{7}}=\frac{2}{7 \cdot \log{7}}$$

At this case, $a_{n}>a_{n+1}$.

So,you can't conclude that the sequence is decreasing,because then the relation $\frac{a_{n+1}}{a_n}>1$ would stand $\forall n \in \mathbb{N}$.

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Thank you for the data evinda. However, we still don't know if the sequence is eventually monotonic. – Mustafa Said May 17 '14 at 13:17
@MustafaSaid I edited my answer.. – evinda May 17 '14 at 14:08
I am asking if there exists an $N$ such that $a_{n+1} \leq a_n$ for all $n \geq N$. NOT "for all $n \in \mathbb{N}$". This is what I mean by "eventually monotonic." I should have been more specific about this point. Thanks evinda. – Mustafa Said May 17 '14 at 14:12