Disco619
  • Disco619
Am I right about the answer you get when you add up to Infinity? Math Experts Only, Anyone else is Welcome too.
Mathematics
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Disco619
  • Disco619
\[1+2+4+8+16... = \infty \] Fact \[1 \times x = x\] Fact \[(1)(1+2+4+8+16...) = 1+2+4+8+16...\] Fact \[2-1 = 1\] Fact \[(2-1)(1+2+4+8+16...) = 1+2+4+8+16...\] ALSO \[(2-1)(1+2+4+8+16...) = 2+4+8+32...-1-2-4-8-16-32...\] Almost everything on the right side cancels out, guess what's left behind? \[= -1\] \[1+2+4+8+16... = -1\] \[\infty = -1\] How about that?
Disco619
  • Disco619
Correction: \[(2-1)(1+2+4+8+16...) = 2+4+8+16+32...-1-2-4-8-16-32...\] |dw:1433714372127:dw|
anonymous
  • anonymous
I may be wrong, but your sum would be equivalent to \[\sum_{n=0}^{\infty}2^{n}\] A rearrangement of a series would only converge to the same value if it were absolutely convergent, but this series is divergent. Thus I'm sure you could do a manipulation and get any result you want.

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Disco619
  • Disco619
Fair enough, but this side of Maths is still interesting though, it isn't so 'Perfect'? If that's how someone would put it.
anonymous
  • anonymous
Of course, a lot of this stuff is fascinating in my eyes. It's what makes it so fun to study :)
Disco619
  • Disco619
I'll just leave this Question open for a bit and Bump it for others to see. Thanks!
ybarrap
  • ybarrap
If the series is not absolutely convergent then the addition of infinite sums is not commutative. You might want to take a look at https://en.wikipedia.org/wiki/Alternating_series#Rearrangements Here we are talking about creating a convergent series from a divergent one. Since the original series is not absolutely convergent, then commutation is not allowed.

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