My Math Forum playing around with limits and infinite series.

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 October 18th, 2016, 06:35 PM #1 Newbie   Joined: Oct 2016 From: Mexico Posts: 2 Thanks: 0 playing around with limits and infinite series. Ok, so I've had this idea going around for a while now, and I would like to show it to you so we can get some answers (spoiler: primes ahead!). So one day I was just goofing around, and I came up with this infinite sum of fractions red1.png nothing very special, and after stirring it up a little I came up with this red2.png so everything is cool in the hood, right? But I just wanted to take it another step up, so I started to play with primes. It's important for me now to specify that I call the value of the infinite sum "delta x", where x is the number of my choosing. So, for example, delta 5 would be .3125, because red3.png Ok? Let's move on! So if we get the delta values of prime numbers and start subtracting them, we'll get something like this: red4.jpg and if we start adding the results of the subtractions, we finally start getting somewhere. Somewhere close to the number 2. Reaaaaally close. Actually, I would even dare to say that the Limit of the sum of the subtractions of immediate delta prime values is two. But, why two? we're dealing with prime numbers now, "random" as they can be. Is there some kind of order in them? It get's even weirder. If you try the same thing, but with a cubic root instead of square, you'll get the same answer, but it'll escalate much more quickly (here I mean the first equation), so we get that red4.jpg Why stop with the power of three? Go nuts! Power of four is the same thing, and five... what is this? if you followed me up to this point, thank you. Anything that I didn't explain very well, please leave a comment, and I'll be sure to answer. Last edited by skipjack; October 18th, 2016 at 09:37 PM.
 October 18th, 2016, 09:24 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,968 Thanks: 1850 After cancelling identical terms, your sum (for $n$ pairs of terms) is $\displaystyle \Delta2 - \Delta p_{n+1}$, where the $p_i$ are the primes. As $\Delta2 = 2$ and $\Delta p_{n+1}\to0$ as $n\to\infty$, your limit is exactly 2. Thanks from red
 October 18th, 2016, 11:03 PM #3 Newbie   Joined: Oct 2016 From: Mexico Posts: 2 Thanks: 0 wow, sounds pretty obvious when you explain it! thanks, I didn't see it that way... so I could just think of it as -3 and +3 cancelling out? same thing with the 5s and 7s and so on with every single prime, infinetly? Last edited by red; October 19th, 2016 at 12:01 AM.

 Tags infinite, limits, playing, primes, series

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