datargumme
Rank #3726 on Comments
Level 270 Comments: Ninja Pirate Offline
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Date Signed Up:  11/21/2011 
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I could might as well have constructed it like this:
1
+1/2 +1/3+1/4
+1/5+1/6+1/7+1/8 +1/9+1/10+1/11+1/12+1/12+1/13+1/14+1/15+1/16
...
+1/(2^{n1}+1)+1/(2^{n1}+2)+1/(2^{n1}+3)+...+1/{2^n)
Now there is an infinite ammount of parts, where they all have the value 1. An infinite ammount of ones, diverges.
I would suggest you do some research on the subject instead of simply calling it b.s. logic.
The first proof of it was done by Nicole Oresme, who was born in the 1300's, you can look up his proof, it is solid. A huge number of mathematicians have done diffrent proofs for the same thing, and they all conclude that the series diverges. You can try for yourself to contruct a proof that the series converges, i can tell you allready that you will fail.
But there are sums which converges, an example is 1+1/2+1/4+1/8+1/16+...+1/n^2, that sum converges towards 2.
It might be counter intuitive, but you can find a lot of people on the internet who can give a more rigorous explanation on why 1/n diverges. I kinda skipped the last part where you have to prove that every segment of my series is greater than 1/2, which is doable but i cant bother right now, you an look it up on the internet if you want the whole proof.
I could might as well have constructed it like this:
1
+1/2 +1/3+1/4
+1/5+1/6+1/7+1/8 +1/9+1/10+1/11+1/12+1/12+1/13+1/14+1/15+1/16
...
+1/(2^{n1}+1)+1/(2^{n1}+2)+1/(2^{n1}+3)+...+1/{2^n)
Now there is an infinite ammount of parts, where they all have the value 1. An infinite ammount of ones, diverges.
I would suggest you do some research on the subject instead of simply calling it b.s. logic.
The first proof of it was done by Nicole Oresme, who was born in the 1300's, you can look up his proof, it is solid. A huge number of mathematicians have done diffrent proofs for the same thing, and they all conclude that the series diverges. You can try for yourself to contruct a proof that the series converges, i can tell you allready that you will fail.