## 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
japiga
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### Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

And moreover it works for any expression in the form k/(n+k+2), e.g. 32/(1/2+34), 33/(1/2+35), 34/(1/2+36,... and I assume for all others expressed in this form. Do you think so?

Martingale
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### Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

And moreover it works for any expression in the form k/(n+k+2), e.g. 32/(1/2+34), 33/(1/2+35), 34/(1/2+36,... and I assume for all others expressed in this form. Do you think so?
It does..though it looks like for your problem you are only going up to k=31...therefore there is an n that works

japiga
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### Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Yes, I accept it as a Game of God with us, ordinary people! Since, we are affected by restriction where restriction doesn’t exist!

japiga
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### Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

This task we need to bring back. I recently discovered that your previous formula k/(n+k+2) gives a proper answer only if n= k - 3. May you confirm it?

Martingale
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### Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Yes, I accept it as a Game of God with us, ordinary people! Since, we are affected by restriction where restriction doesn’t exist!
This task we need to bring back. I recently discovered that your previous formula k/(n+k+2) gives a proper answer only if n= k - 3. May you confirm it?
I have to admit that I'm only understanding a tiny fraction of what you are saying. If you have specific questions that you can articulate in a meaningful way I'd be happy to help.

japiga
Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm
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### Re: 7/(n+9), 8/(n+10),9/(n+11),…. 31/(n+33)

Maybe I have solved it, but I need only confirmation. Above expression is k/(n+k+2) . We need to find n (the smallest one) such as those expressions not to be further divided. And if you put n=k-7, this doesn’t work since for 9/(4+11)=9/15 what is divided by 3. Similar situation we have for n=k-6, n=k-5, n=k-4, n=k-2. Only it works for n=k-3 and we are looking for smallest "n" and we end up on this n=k-3, it doesn't matter what is going with n=k-1. Moreover this n=k-3 valid for all k belong to N, not only for restricted k €(7,8,9,...31) as it is given by this assignment above.
May you help me to prove it. I think it is induction method which should be useful to be applied, but I am not so familiar how to proceed it.